Answer:
they'll be black, because green absorbs red.
Explanation:
any color which absorbs red will appear "black", i.e. green, blue, black
The figure shows a system of five objects. Determine the magnitude of the gravitational force acting on the
object placed in the middle of the square.
Select one:
O a. 4Gm2/d2
O b. Gm2/d2
O c. 8Gm2/d2
Answer:
c. [tex]|F_T|=8\frac{Gm^2}{d^2}[/tex]
Explanation:
In order to calculate the gravitational force on the mass of the center, you take into account the following formula:
[tex]F=G\frac{m_1m_2}{r}[/tex] (1)
Furthermore, you take into account the components of the resultant vector.
By the illustration, you have that the force is given by:
[tex]F_T=F_1+F_2+F_3+F_4\\\\F_1=\frac{Gm_1m}{r^2}[-cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_2=\frac{Gm_2m}{r^2}[cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_3=\frac{Gm_3m}{r^2}[cos45\°\hat{i}-sin45\°\hat{j}]\\\\F_4=\frac{Gm_4m}{r^2}[-cos45\°\hat{i}-sin45\°\hat{j}][/tex]
where:
m1 = m
m2 = 2m
m3 = m
m4 = 4m
m: mass at the center of the system
The distance r is:
[tex]r=\sqrt{(\frac{d}{2})^2+(\frac{d}{2})^2}=\frac{d}{\sqrt{2}}[/tex]
You replace the values for all masses and sum the contributions of all forces:
[tex]F_1=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[-\hat{i}+\hat{j}]=\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}+\hat{j}]\\\\F_2=\frac{\sqrt{2}}{2}\frac{2Gm^2}{(\frac{d^2}{2})}[\hat{i}+\hat{j}]=2\sqrt{2}\frac{Gm^2}{d^2}[\hat{i}+\hat{j}]\\\\F_3=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[\hat{i}-\hat{j}]=\sqrt{2}\frac{Gm^2}{s^2}[\hat{i}-\hat{j}]\\\\F_4=\frac{\sqrt{2}}{2}\frac{4Gm^2}{(\frac{d^2}{2})}[-\hat{i}-\hat{j}]=4\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}-\hat{j}]\\\\F_T=-2\sqrt{2}\frac{Gm^2}{d^2}}[\hat{i}+\hat{j}][/tex]
and the magnitude is:
c. [tex]|F_T|=8\frac{Gm^2}{d^2}[/tex]
Two bullets are fired simultaneously parallel to a horizontal plane. The bullets have different masses and different initial velocities. Which one will strike the plane first?
a) The fastest one.b) The lightest one.c) The heaviest one.d) The slowest one.e) They strike the plane at the same time.
Answer:
Therefore, the answer is E. They strike the plane at the same time.
Explanation:
Here, it is seen that the time depends only on acceleration due to gravity (which is a constant) and vertical displacement, and not on velocity of the bullets or mass of the bullets.
Hence, the bullets that are fired simultaneously parallel to the horizontal plane will strike the plane at the same time.
using equation of motion for displacement
s= ut + ¹/₂gt²
here, g is the acceleration due to gravity along y- direction
U along y is 0
s = (0)t + ¹/₂gt²
s=¹/₂gt²
make t the subject of formula = [tex]\sqrt{\frac{2s}{g} }[/tex]
During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an angular displacement of 1.5758 rev. What is the angular acceleration of the CD
Answer:
126.01 rad/s^2
Explanation:
since it starts from rest, initial angular speed ω' = 0 rad/s
angular speed N = 477 rev/min
angular speed in rad/s ω = [tex]\frac{2\pi N}{60}[/tex] = [tex]\frac{2*3.142* 477}{60}[/tex] = 49.95 rad/s
angular displacement ∅ = 1.5758 rev
angular displacement in rad/s = [tex]2\pi N[/tex] = 2 x 3.142 x 1.5758 = 9.9 rad
angular acceleration [tex]\alpha[/tex] = ?
using the equation of angular motion
ω^2 = ω'^2 + 2[tex]\alpha[/tex]∅
imputing values, we have
[tex]49.95^{2} = 0^{2} + (2 *\alpha*9.9 )[/tex]
2495 = 19.8[tex]\alpha[/tex]
[tex]\alpha[/tex] = 2495/19.8 = 126.01 rad/s^2
A spherical balloon is being inflated. Find the rate of increase of the surface area (S = 4Ï€r2) with respect to the radius r when r is each of the following. (Answers in unit ft2/ft)
(a) 1 ft
(b) 3 ft
(c) 6 ft
Answer:
A) 8π ft²/ft
B) 24π ft²/ft
C) 48π ft²/ft
Explanation:
Surface area of the spherical balloon is not clear here but it is supposed to be;
S = 4πr²
where:
r is the radius of the spherical balloon
So thus, the rate of change of the surface area of the spherical balloon by its radius will be:
dS/dr = 8πr
A) at r = 1ft;
dS/dr = 8 × π × 1
dS/dr = 8π ft²/ft
B) at r = 3 ft;
dS/dr = 8 × π × 3
dS/dr = 24π ft²/ft
C) at r = 6ft;
dS/dr = 8 × π × 6
dS/dr = 48π ft²/ft
A 1.30-kg particle moves in the xy plane with a velocity of v with arrow = (4.50 î − 3.30 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.
Answer:
The angular momentum of the particle about the origin is [tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex].
Explanation:
Vectorially speaking, the angular momentum is given by the following cross product:
[tex]\vec l = \vec r \times m\vec v[/tex]
This cross product can be solved with the help of determinants and its properties, that is:
[tex]\vec l = \left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\m\cdot v_{x}&m\cdot v_{y}&0\end{array}\right|[/tex]
[tex]\vec l = m\left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\v_{x}& v_{y}&0\end{array}\right|[/tex]
The 3 x 3 determinant is solved by the Sarrus Law:
[tex]\vec l = m \cdot (r_{x}\cdot v_{y} - r_{y}\cdot v_{x})k[/tex]
If [tex]m = 1.30\,kg[/tex], [tex]\vec r = 1.50\,i + 2.20\,j\,[m][/tex] and [tex]\vec v = 4.50\,i-3.30\,j\,\left[\frac{m}{s} \right][/tex], the angular momentum of the particle about the origin is:
[tex]\vec l = (1.30\,kg)\cdot \left[\left(1.50\,m\right)\cdot\left(-3.30\,\frac{m}{s} \right)-\left(2.20\,m\right)\cdot\left(4.50\,\frac{m}{s} \right)\right]k[/tex]
[tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex]
The angular momentum of the particle about the origin is [tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex].
A rubber ball is attached to a string and whirled around in a circle. If the string is 1.0 m long (measured from the center of the baseball to the far end of the string) and the ball’s speed is 10 m/s, what is the ball’s centripetal acceleration?
Centripetal acceleration = (speed squared) / (radius)
Centripetal acceleration = (10 m/s)² / (1.0 m)
Centripetal acceleration = (100 m²/s²) / (1.0 m)
Centripetal acceleration = 100 m/s²
A spinning top initially spins at 16rad/s but slows down to 12rad/s in 18s, due to friction. If the rotational inertia of the top is 0.0004kg.m2, by how much the angular momentum of the top changes?
Answer:
The change in angular momentum is [tex]\Delta L = 0.0016 \ kgm^2/s[/tex]
Explanation:
From the question we are told that
The initial angular velocity of the spinning top is [tex]w_1 = 16 \ rad/s[/tex]
The angular velocity after it slow down is [tex]w_2 = 12 \ rad/s[/tex]
The time for it to slow down is [tex]t = 18 \ s[/tex]
The rotational inertia due to friction is [tex]I = 0.0004 \ kg \cdot m^2[/tex]
Generally the change in the angular momentum is mathematically represented as
[tex]\Delta L = I *(w_1 -w_2)[/tex]
substituting values
[tex]\Delta L = 0.0004 *(16 -12)[/tex]
[tex]\Delta L = 0.0016 \ kgm^2/s[/tex]
Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The objectstarts from rest and in 1.20 s experiences a displacement of (4.20 i - 3.30 j) m, where the direction of jis the upward vertical direction. Determine the other force.
Answer:
the other force= (16.3i + 14.6j)N
EXPLANATION:
Given:
Mass=2.80-kg
t= 1.2s
Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.
We will need to calculate the magnitude of the displacement using the below formula;
d = (1/2)at2 + v0t + d0
But note that
d0 = 0,( initial position)
v0 = 0( initial position)
a is the net acceleration
d = √[4.202 + (-3.30)2] m = 5.34 m
Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)
d = (1/2)at2
a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2
the net acceleration is 7.42 m/s2
Acceleration in terms of the vector can be calculated as
a=2(ri - r0)/t^2
Where t =1.2s which is the time
a= 2(4.2i - 3.30j)/ 1.2^2
a=( 5.83i - 4.58j)m/s
now the net force can now be calculated since we have known the value of acceleration, using the formula below;
F(x) = ma - mg
Where a = 5.83i - 4.58j)m/s and g= 9.8m/s
2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2
Therefore, the other force= (16.3i + 14.6j)N
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spaceship's diameter is 35 mm , and give your answer as the time needed for one revolut
Answer:
Time needed for one revolution is 0.38 s
Explanation:
The formula for the frequency of rotation of a spaceship, to create the desired artificial gravity, is as follows:
f = (1/2π)√(a/r)
where,
f = frequency of rotation = ?
a = artificial gravity required = 0.5 g
g = acceleration due to gravity on surface of Earth = 9.8 m/s²
r = radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m
Therefore,
f = (1/2π)√[(0.5)(9.8 m/s₂)/(17.5 x 10⁻³ m)]
f = 2.66 Hz
Now, for the time required for one revolution, is given as:
Time Period = T = 1/f
T = 1/2.66 Hz
T = 0.38 s
The time required for one revolution to simulate the desired gravity is 0.38 s.
The frequency can be calculate by the formula
[tex]\bold {f = (\dfrac {1}{2\pi})\sqrt{ar}}[/tex]
where,
f - frequency of rotation = ?
a- artificial gravity required = 0.5 g
g - gravitational acceleration on surface of Earth = 9.8 m/s²
r - radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m
Put the value in the equation,
[tex]\bold {f = \dfrac {1}{2\pi}\squrt {(0.5)(9.8\ m/s^2)}{(17.5 x 10^{-3} m)}}\\\\\bold {f = 2.66\ Hz}[/tex]
the time required for one revolution can be calculated as
[tex]\bold {T =\dfrac 1f}\\\\\bold {T = \dfrac 1{2.66}\ Hz}\\\\\bold {T = 0.38\ s}[/tex]
Therefore, the time required for one revolution to simulate the desired gravity is 0.38 s.
To know more about the Gravity,
https://brainly.com/question/12008506
Two parallel plates, each of area 7.37 cm2, are separated by 6.00 mm. The space between the plates is filled with air. A voltage of 7.55 V is applied between the plates. Calculate the magnitude of the electric field between the plates. Tries 0/20 Calculate the amount of the electric charge stored on each plate. Tries 0/20 Now distilled water is placed between the plates and the capacitor is charged up again to the same voltage as before. Calculate the magnitude of charge stored on each plate in this case. (Use κ = 80.0 for the dielectric constant of water.)
Answer:
- E = 1.25*10^3 N/C
- Q = 1.08*10^-12 C
- Q = 8.69*10^-11 C
Explanation:
- In order to calculate the magnitude of the electric field between the plates, you use the following formula:
[tex]E=\frac{V}{d}[/tex] (1)
V: potential difference between the plates = 7.55V
d: distance between the plates = 6.00mm = 6.00*10^-3m
You replace the values of the parameters n the equation (1):
[tex]E=\frac{7.55V}{6.00*10^{-3}m}=1.25*10^3\frac{N}{C}[/tex]
The magnitude of the electric field between the plates is 1.25*10^3N/C
- The charge on each plate is given by the following formula:
[tex]Q=CV[/tex] (2)
C: capacitance of the capacitor
The capacitance of a parallel plate capacitor is:
[tex]C=\epsilon_o \frac{A}{d}[/tex] (3)
You replace the previous formula into the equation (2) and solve for Q:
[tex]Q=(\epsilon_o k\frac{A}{d})(V)=(8.85*10^{-12}C^2/Nm^2)\frac{7.37*10^{-4}m^2}{6.00*10^{-3}m}\\\\Q=1.08*10^{-12}C[/tex]
The charge on each plate is 1.08*10^-12C = 1.08pC
- If water is placed in between the plates, the dielectric permittivity is changes by a factor of k = 80.0.
The capacitance of a parallel plate capacitor with a substance with a constant dielectric k, is given by:
[tex]C=\epsilon_o k\frac{A}{d}[/tex] (4)
You replace the previous formula in the equation (2) and replace the values of all parameters:
[tex]Q=(\epsilon_o k\frac{A}{d})(V)=(8.85*10^{-12}C^2/Nm^2)(80.0)\frac{7.37*10^{-4}m^2}{6.00*10^{-3}m}\\\\Q=8.69*10^{-11}C[/tex]
The charges on each plate is 8.69*10^-11 C
The force per meter between the two wires of a jumper cable being used to start a stalled car is 0.335 N/m.
Required:
a. What is the current in the wires, given they are separated by 2.90 cm?
b. Is the force attractive or repulsive?
1. The force is repulsive because the currents are in opposite directions.
2. The force is repulsive because the currents are in the same direction.
3. The force is attractive because the currents are in the same direction.
4. The force is attractive because the currents are in opposite directions.
Answer:
a) I = 1.29 10⁻⁴ A , b) the current has the opposite direction in each cable, therefore the correct answer must be 1
Explanation:
a) Let's analyze this exercise, when a conductor carries a current the elctoenes generate a magnetic field around the cable and this field can interact with the electrons that carry a current in a nearby cable, the expression for force is
F₁ = I₁ L B₂
where F₁ is the force on a cable through the field created by elotorcable.
The field sent by the other cable is
B₂ = μ₀ I² / 2πa
we substitute
F₁ = μ₀ I₁ I₂ / 2πa L
suppose the current in the two cables is the same
I₁ = I₂ = I
F₁ / L = μ₀ I₂ / 2πa
i = √ [(F₁ / L) 2πa / μ₀]
let's calculate
I = √ [0.335 2 π / 4π 10⁻⁷]
I = √ [0.335 / 2 10⁷]
I = 1.29 10⁻⁴ A
b) if the two cables have the current in the same direction the force is attractive and if it has the current in the opposite direction it is negative
the two possible answers are 1 and 2
In general, to start a car, one of the cables has a current from the battery at start-up and the other cable comes from the start to the battery, therefore the current has the opposite direction in each cable, therefore the correct answer must be 1
An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.8 times the mass of the other.
Requried:
a. If 7230 J were released in the explosion, how much kinetic energy did the heavier piece acquire?
b. How much kinetic energy did the lighter piece acquire?
Answer:
a) The heavier piece has a translational kinetic energy of 4647.857 joules, b) The lighter piece has a translational kinetic energy of 2582.143 joules.
Explanation:
a) The object breaking can be described by means of the Principle of Energy Conservation, knowing that heavier piece has 1.8 times the mass of the lighter ([tex]m_{h} = 1.8\cdot m_{l}[/tex]), both are modelled as particle due to the absence of rotation and that energy liberated by explosion is transform into kinetic energy, the equation that describes the phenomenon is:
[tex]E_{ex} = K_{h} + K_{l}[/tex]
Where:
[tex]E_{ex}[/tex] - Energy liberated by the explosion, measured in joules.
[tex]K_{h}[/tex], [tex]K_{l}[/tex] - Translational kinetic energies of the heavier and lighter piece, respectively.
This expression is expanded by using the definition of translational kinetic energy and supposing the both parts are liberated at the same initial speed ([tex]v_{o}[/tex]). Then:
[tex]E_{ex} = \frac{1}{2}\cdot (m_{h} + m_{l})\cdot v_{o}^{2}[/tex]
As can be seen, the energy liberated by expression is directly proportional to the mass of the system. Hence, the kinetic energy can be estimated by simple rule of three:
[tex]K_{h} = \frac{m_{h}}{m_{h}+m_{l}}\times E_{ex}[/tex]
If [tex]m_{h} = 1.8\cdot m_{l}[/tex] and [tex]E_{ex} = 7230\,J[/tex], then:
[tex]K_{h} =\frac{1.8\cdot m_{l}}{2.8\cdot m_{l}}\times E_{ex}[/tex]
[tex]K_{h} = \frac{9}{14}\cdot (7230\,J)[/tex]
[tex]K_{h} = 4647.857\,J[/tex]
The heavier piece has a translational kinetic energy of 4647.857 joules.
b) The translational kinetic energy of the lighter piece is calculated by using the equation derived from the Principle of Energy Conservation:
[tex]K_{l} = E_{ex} - K_{h}[/tex]
Given that [tex]E_{ex} = 7230\,J[/tex] and [tex]K_{h} = 4647.857\,J[/tex], the translational kinetic energy of the lighter piece is:
[tex]K_{l} = 7230\,J - 4647.857\,J[/tex]
[tex]K_{l} = 2582.143\,J[/tex]
The lighter piece has a translational kinetic energy of 2582.143 joules.
Un bloque de 10 kg se encuentra sobre un plano rugoso inclinado 37º respecto a la horizontal, sobre él actúa una fuerza constante, horizontal, de módulo 50 N. Si el bloque desciende sobre el plano 5 m, lentamente, determine la cantidad de trabajo que realiza la fuerza de rozamiento (considere g = 10 m/s2).
Answer:
El trabajo realizado por la fuerza de rozamiento sobre el bloque tras recorrer este último una distancia de 5 metros sobre el plano es de 500.566 joules.
Explanation:
El fenómeno alrededor del bloque puede ser modelado por el Principio de Conservación de la Energía y el Teorema del Trabajo y la Energía. Al descender lentamente, significa que la aceleración neta experimentada por el bloque es aproximadamente cero. El diagrama de cuerpo libre del bloque se presenta a continuación como archivo adjunto. Las ecuaciones de equilbrio del sistema son:
[tex]\Sigma F_{x'} = P\cdot \cos \theta + m\cdot g \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y'} = N + P\cdot \sin \theta -m\cdot g\cdot \cos \theta = 0[/tex]
Donde:
[tex]P[/tex] - Fuerza externa aplicada a la caja, medida en newtons.
[tex]m[/tex] - Masa del bloque, medida en kilogramos.
[tex]g[/tex] - Aceleración gravitacional, medidas en metros sobre segundo al cuadrado.
[tex]f[/tex] - Fuerza de rozamiento, medida en newtons.
[tex]N[/tex] - Fuerza normal del plano sobre la caja, medida en newtons.
[tex]\theta[/tex] - Ángulo de inclinación del plano, medido en grados sexagesimales.
Dado que todas las fuerzas son constantes, se puede emplear la definición de trabajo como el producto de la fuerza paralela a la dirección del movimiento y la magnitud de distancia recorrida en el movimiento, entonces la primera ecuación de equilibrio queda así al multiplicar cada lado por la distancia recorrida:
[tex]P\cdot \Delta s \cdot \cos \theta + m\cdot g \cdot \Delta s \cdot \sin \theta - W_{f} = 0[/tex]
Ahora, la cantidad de trabajo realizado por la fuerza de rozamiento es:
[tex]W_{f} = (P\cdot \cos \theta+m\cdot g\cdot \sin \theta)\cdot \Delta s[/tex]
Si [tex]P = 50\,N[/tex], [tex]m = 10\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]\theta = 37^{\circ}[/tex] and [tex]\Delta s = 5\,m[/tex], entonces el trabajo realizado por la fuerza de rozamiento es:
[tex]W_{f} = \left[(50\,N)\cdot \cos 37^{\circ}+(10\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot \sin 37^{\circ}\right]\cdot (5\,m)[/tex]
[tex]W_{f} = 500.566\,J[/tex]
El trabajo realizado por la fuerza de rozamiento sobre el bloque tras recorrer este último una distancia de 5 metros sobre el plano es de 500.566 joules.
Good day can I get some help please?
Answer:
432 J
Explanation:
When moving linearly:
Kinetic Energy = (1/2)mV^2
So here you have:
KE=(1/2)(6)(12^2)=(1/2)(6)(144)=432
The unit for energy is Joules (J), so your answer would be 432 J.
A horizontal force of 14.0N is applied to a box of m=32.5kg with Vo=0. Ignoring friction, how far does the crate travel in 10.0s?
how many electrons does a neutral atom of sudium-25 have
Answer:
Option A. 11
Explanation:
The atomic number of an element does not change.
Recall:
Atomic number = proton number
If the atom is neutral, then,
Proton number = electron number
Since the element is sodium, then, the atomic number of sodium–25 is 11.
Also, we were told to obtain the electrons of a neutral atom of sodium–25
Therefore,
Atomic number = proton number = 11
Since the atom is neutral,
Proton number = electron number = 11
Answer:
A. 11
Explanation:
A neutral atom of sodium-25 has the same number of protons and electrons. Since it has 11 protons, it also must have 11 electrons!
At a certain instant, coil A is in a 10-T external magnetic field and coil B is in a 1-T external magnetic field. Both coils have the same area and are oriented at right angles to the field. Which coil will have a greater emf induced in it
Answer:
Impossible to know without more information about the fields.
Explanation:
Changing the magnetic field induces the external magnetic field, but the information regarding magnetic field variation is not provided. We need to required more information for this
Therefore according to the above explanation the correct option is Impossible to know without more information about the fields.
Hence, the b option is correct
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It strikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± [tex]\frac{1}{2} at^2[/tex] ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± [tex]\frac{1}{2} at^2[/tex] ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
From the question;
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt - [tex]\frac{1}{2} at^2[/tex]
h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her acceleration when her speed is 30.0 m/s
Answer:
6.22²
Explanation:
Given that
Mass of the skydiver, m = 80 kg
Terminal speed of the skydiver, v(f) = 50 m/s
Speed of the skydiver, v(i) = 30 m/s
Acceleration of the skydiver, a = ?
To solve this, we use the formula
W - k v² = ma, where
W = weight of the skydiver
k = constant
v = speed of the skydiver
m = mass of the skydiver
So, if we substitute the values into it we have
W = mg = 80 * 9.8 = 784 N
784 - k 50² = 80 *0
784 - 2500k = 0
784 = 2500k
k = 0.3136
Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s
784 - 0.3136 * 30² = 80 * a
784 - 0.3136 * 900 = 80a
784 - 282.24 = 80a
497.76 = 80a
a = 497.76 / 80
a = 6.22 m/s²
Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²
A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:_________
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller or equal than that observed inside the bus.
e. is greator than that observed inside the bus.
Answer:
d. may be either greater, smaller, or equal to that observed inside the bus.
Explanation:
The bus is moving at a constant speed. The ball tossed and received by the ball is inside the bus at a speed equal to the speed of the ball. Therefore the speed of the bus becomes zero with respect to the observer inside the bus. Now the observer inside the bus noticed the ball from the inside of the bus, so he threw the ball back and forth from the ball with the speed v relative to the observer. Now the observer outside the bus could see the bus moving at speed relative to its reference point and also throwing the ball from front to back. The speed of the ball to the observer outside the bus The speed of the bus to the observer outside the bus is minus the speed of the ball to the observer inside the bus. Therefore, the ball speed = (u-v) relative to the observer outside the bus.A transformer has 480 primary turns and 7.8 secondary turns. (a) If Vp is 120 V (rms), what is Vs with an open circuit? If the secondary now has a resistive load of 17 Ω, what is the current in the (b) primary and (c) secondary?
Answer:
a) 1.95 V
b) 1.87 mA
c) 0.115 A
Explanation:
Given that
Number of primary turns, N(p) = 480
Number of secondary turns, N(s) = 7.8
Velocity of primary turns, V(p) = 120 V
Velocity of secondary turns, V(s) = ?
Current in the primary, I(p) = ?
Current in the secondary, I(s) ?
To solve this question, we would be using the formula
V(s)/V(p) = N(s)/N(s), now substituting the values, we have
V(s) / 120 = 7.8 / 480
V(s) = (7.8 * 120) / 480
V(s) = 936 / 480
V(s) = 1.95 V
To find the current in the primary, remember ohms law?
I = V/R
I(s) = V(s) / R(s)
I(s) = 1.95 / 17
I(s) = 0.115 A
Now, remember the relationship between current and voltage
I(p)/I(s) = V(s)/V(p)
I(p) / 0.115 = 1.95 / 120
I(p) = (1.95 * 0.115) / 120
I(p) = 0.22425 / 120
I(p) = 0.00187 A
I(p) = 1.87 mA
During a certain time interval, the angular position of a swinging door is described by θ = 4.91 + 9.7t + 2.06t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.
(a) t = 0
θ = rad
ω = rad/s
α = rad/s2
(b) t = 2.94 s
θ = rad
ω = rad/s
α = rad/s2
Explanation:
a)
θ = 4.91 + 9.7t + 2.06t² when t = 0
θ = 4.91 rad
θ = 4.91 + 9.7t + 2.06t²
ω = dθ/dt = 9.7 + 2.06t, when t =0
ω = dθ/dt = 9.7 + 0
ω = 9.7 rad/s
α = d²θ/dt² = 2.06
α= 2.06 rad/s²
b) please use same method above for t = 2.94 s
At zero seconds, the angular position is 4.91 rad, the angular velocity is 9.7 radian/second and the angular acceleration is 2.06 rad/s².
What is Angular acceleration?When the angular velocity change in relation to time, then it results in angular acceleration. It comes with direction, so it is a vector quantity.
Given: Angular position equation (θ) = 4.91 + 9.7t + 2.06[tex]t^{2}[/tex]
If the angular velocity is ω, angular acceleration is α, and t is the time then
(A) At time t = 0 second
θ = 4.91 + 9.7 × 0 + 2.06 × [tex]0^{2}[/tex]
θ = 4.91 rad
Angular velocity (ω) = dθ/dt
ω = dθ/dt,
ω = 9.7 + 2.06t (at t = 0)
ω = 9.7 radian/second
Angular acceleration (α) = d²θ/dt²
α= 2.06 rad/s²
Similarly, at t= 2.94 sec
θ = 51.23 rad
ω = 15.76 rad/s
α = 2.06 rad/s²
To know more about Angular acceleration:
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Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.1 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 4.0 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
Explanation:
Given;
moment of inertia of a skater with arms out, [tex]I_{arms \ out}[/tex] = 3.1 kg.m²
moment of inertia of a skater with arms in, [tex]I_{arms \ in}[/tex] = 0.9 kg.m²
inward angular speed, [tex]\omega _{in}[/tex] = 4 rev/s
The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.
[tex]L_{out} = L_{in}[/tex]
[tex]I_{out} \omega_{out} = I_{in} \omega_{in}\\\\\omega_{out} = \frac{ I_{in} \omega_{in} }{I_{out} } \\\\\omega_{out} = \frac{0.9*4}{3.1} \\\\\omega_{out} = 1.161 \ rev/s[/tex]
Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the object for 2.0 s, after which the object's velocity is v 2 = (16.0 i^ + 29.0 j^) m/s.
Required:
Find the work done by the force in joules.
Answer:
The work done by the force is 820.745 joules.
Explanation:
Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:
[tex]K_{1} + W_{F} = K_{2}[/tex]
Where:
[tex]W_{F}[/tex] - Work done by the external force, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Translational potential energy, measured in joules.
The work done by the external force is now cleared within:
[tex]W_{F} = K_{2} - K_{1}[/tex]
After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:
[tex]W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})[/tex]
Where:
[tex]m[/tex] - Mass of the object, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the object, measured in meters per second.
Now, each speed is the magnitude of respective velocity vector:
Initial velocity
[tex]v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}[/tex]
[tex]v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex]
Final velocity
[tex]v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}[/tex]
[tex]v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex]
Finally, if [tex]m = 3.5\,kg[/tex], [tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex] and [tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex], then the work done by the force is:
[tex]W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]W_{F} = 820.745\,J[/tex]
The work done by the force is 820.745 joules.
If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water knocks two rocks together, you'll barely hear the sound.
Match the words.
The air-water interface is an example of boundary. The( )portion of the initial wave energy is way smaller than the( )portion. This makes the( ) wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can( ) .
1. reflect more efficiently
2. transmitted
3. travel directly to your ears
4. boundary
5. reflected
6. discontinuity
Answer:
The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.
Explanation:
The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.
Answer: The air-water interface is an example of boundary. The (transmitted) portion of the initial wave is way smaller than the (reflected) portion. This makes the (transmitted) wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can (travel directly to your ears.)
Explanation:
The part of the sound wave that is transmitted across the boundary between air and water is much smaller than the part of the wave that is reflected. This is what makes it hard to hear your friend knocking two rocks together above the surface.
When you and the rocks are underwater, the sound that comes from knocking the rocks together can travel directly to your ears rather than having to be transmitted across mediums.
A piston absorbs 42 J of heat from its surroundings while being compressed from 0.0007 m3 to 0.0002 m3 at a constant pressure of 1.0 × 105 Pa. What are the correct values for heat and work for the piston?
Answer:
D
Explanation:
W = P∆V
Use the above equation and substitute, thanks
A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.
a. 3m/s
b. 30.3 m/s
c. None of the above
Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s
A positive charge moves in the direction of an electric field. Which of the following statements are true?
a. The potential energy associated with the charge decreases.
b. The electric field does positive work on the charge.
c. The electric field does negative work on the charge.
d. The potential energy associated with the charge increases.
e. The electric field does not do any work on the charge.
f. The amount of work done on the charge cannot be determined without additional information.
Answer:
The potential enwrgy associated with charge decreases.
The ele ric field does negative work on the charge.
Explanation:
Answer:
The potential energy associated with the charge decreases
The electric field does positive work on the charge.
The Earth, which has an equatorial radius of 6380 km, makes one revolution on its axis every 23.93 hours. What is the tangential speed of Fayetteville, NC, whose latitude is 35o N
Answer:
The tangential speed of Fayetteville, NC is 1372.2 km/h.
Explanation:
To find the tangential speed at the latitude 35 °N, first, we need to calculate the radius at that latitude [tex]r_{lat}[/tex]:
[tex] r_{lat} = r_{eq}*cos(\theta) [/tex]
Where:
[tex]r_{eq}[/tex]: is the equatorial radius = 6380 km
θ = 35 °
[tex] r_{lat} = r_{eq}*cos(\theta) = 6380 km*cos(35) = 5226.2 km [/tex]
Now, the tangential speed is:
[tex] v = \omega*r = \frac{2\pi}{23.93 h}*5226.2 km = 1372.2 km/h [/tex]
Therefore, the tangential speed of Fayetteville, NC is 1372.2 km/h.
I hope it helps you!
A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor
Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.