Answer:
The angle is [tex]\theta = 15.48^o[/tex]
Explanation:
From the question we are told that
The distance of the dartboard from the dart is [tex]d = 3.66 \ m[/tex]
The time taken is [tex]t = 0.455 \ s[/tex]
The horizontal component of the speed of the dart is mathematically represented as
[tex]u_x = ucos \theta[/tex]
where u is the the velocity at dart is lunched
so
[tex]distance = velocity \ in \ the\ x-direction * time[/tex]
substituting values
[tex]3.66 = ucos \theta * (0.455)[/tex]
=> [tex]ucos \theta = 8.04 \ m/s[/tex]
From projectile kinematics the time taken by the dart can be mathematically represented as
[tex]t = \frac{2usin \theta }{g}[/tex]
=> [tex]usin \theta = \frac{g * t}{2 }[/tex]
[tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]
[tex]usin \theta = 2.23[/tex]
=> [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]
[tex]\theta = tan^{-1} [0.277][/tex]
[tex]\theta = 15.48^o[/tex]
Consider 4 charges placed at the corners of a square with side 1.25m as shown. What are the magnitude and direction of the electrostatic force on Q1 resulring from other three charges?
(Note: Please put your final answer in 4 decimal places.). pls answer:)
Answer:
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
F = 3.0226 10⁻² N , θ = 21.16º
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r₁₂²
We also use that the force is a vector magnitude, so we must calculate each component of the force , see the adjoint for the direction of the vectors
X axis
Fₓ = -F₁₄ + F₁₃ₓ
Y axis
[tex]F_{y}[/tex] = F₁₂ -F_{13y}
let's look for the expression for each force
where the side of the square is a = 1.25 m
F₁₂ = k Q₁Q₂ / a²
F₁₄ = k Q₁Q₄ / a²
the distance between 1 and 3 is
d = √(a² + a²) = a √2
F₁₃ = k Q₁Q₃ / d²
let's use trigonometry to find the components
cos 45 = F₁₃ₓ / F₁₃
F₁₃ₓ = F₁₃ cos 45
F₁₃ₓ = k Q₁Q₃ / 2a²
sin 45 = F_{13y} / F₁₃
F_{13y} = F₁₃ sin 45
F_{13y} = k Q₁Q₃ / 2a² sin 45
Taking all terms, we substitute in the force for each axis
X axis
Fₓ = - k Q₁Q₄ / a² + k Q₁Q₃ / 2a₂ cos 45
Fₓ = k Q₁ / a² ( -Q₄ + Q₃ /2 cos 45)
Fₓ = 9 10⁹ 1.5 10⁻⁶ / 1.25² (- 4.5 10⁻⁶ + 3.5/2 cos 45 10⁻⁶)
Fₓ = 8.64 10³ (3.2626 10⁻⁶)
Fₓ = 2.8189 10⁻² N
Y axis
F_{y} = k Q₁Q₂ / a² - k Q₁Q₃ /2a² sin 45
F_{y} = k Q₁ / a² (Q₂ - Q₃ /2 sin45)
F_{y} = 9 10⁹ 1.5 10⁻⁶/ 1.25² (2.5 10⁻⁶ - 3.5/2 sin 45 10⁻⁶)
F_{y} = 8.64 10³ (1.26256 10⁻⁶)
F_{y} = 1.0909 10⁻² N
The result can be given in two ways
1) F = Fₓ i ^ + F_{y} j ^
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
2) in the form of a module and an angle, for which we use the Pythagorean theorem and trigonometry
F = √ (Fₓ² + F_{y}²)
F = 10⁻² √ (2,8189² + 1,0909²)
F = 3.0226 10⁻² N
tan θ = F_{y} / Fx
θ = tan⁻¹ (F_{y} / Fₓ)
θ = tan⁻¹ (1.0909 / 2.8189)
θ = 21.16º
A spring balance is attached with string to the piece of aluminum in the preceding problem. What reading will the balance register when the metal is submerged
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
[tex]F = W_{x} = mgsin(\theta)[/tex]
Where:
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]
Now, we can find the work:
[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different car and before you even get to drive your car, the little brat crashes into you at a speed of 3 m/s.
A. Knowing that the bumper cars each weigh 80 kg, while you and your brother weigh 60 and 30 kg,respectively, write down the equations you need to use to figure out how fast you and your brother are moving after the collision.
B. After the collision, your little brother reverses direction and moves at 0.36 m/s. How fast are you moving after the collision?
C. Assuming the collision lasted 0.05 seconds, what is the average force exerted on you during the collision?
D. Who undergoes the larger acceleration, you or your brother? Explain.
Answer:
a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) Your velocity after collision is 2.64 m/s
c) The force you felt is 7392 N
d) you and your brother undergo an equal amount of acceleration
Explanation:
Your mass [tex]m_{y}[/tex] = 60 kg
your brother's mass [tex]m_{b}[/tex] = 30 kg
mass of the car [tex]m_{c}[/tex] = 80 kg
your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)
your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s
your final speed [tex]v_{y}[/tex] after collision = ?
your brother's final speed [tex]v_{b}[/tex] after collision = ?
a) equations you need to use to figure out how fast you and your brother are moving after the collision is
[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
but [tex]u_{y}[/tex] = 0 m/s
the equation reduces to
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) if your little brother reverses with velocity of 0.36 m/s it means
[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)
then, imputing values into the equation, we'll have
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)
330 = 140[tex]v_{y}[/tex] - 39.6
369.6 = 140[tex]v_{y}[/tex]
[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s
This means you will also reverse with a velocity of 2.64 m/s
c) your initial momentum = 0 since you started from rest
your final momentum = (total mass) x (final velocity)
==> (60 + 80) x 2.64 = 369.6 kg-m/s
If the collision lasted for 0.05 s,
then force exerted on you = (change in momentum) ÷ (time collision lasted)
force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N
d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s
your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2
your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s
his deceleration is (3 - 0.36)/0.05 = 52.8 m/s
you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost
An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is 11) held in place, how fast will the electron be moving when it is 3.00 cm from the proton?
Answer:
Vf = 1.43 m/s
Explanation:
From Coulomb's Law, the electrostatic force between electron and proton is given as:
F = kq₁q₂/r²
F = Electrostatic force = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
q₂ = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = distance between electron and proton = 9 cm = 0.09 m
Therefore,
F = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(0.09 m)²
F = 2.84 x 10⁻²⁶ N
but, from Newton's second law:
F = 2.84 x 10⁻²⁶ N = ma
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
a = acceleration of electron = ?
Therefore,
2.84 x 10⁻²⁶ N = (1.67 x 10⁻²⁷ kg)(a)
a = 2.84 x 10⁻²⁶ N/1.67 x 10⁻²⁷ kg
a = 17.03 m/s²
Now, we apply 3rd equation of motion to the motion of electron from a distance of 9 cm to 3 cm near to the proton:
2as = Vf² - Vi²
where,
s = distance traveled = 9 cm - 3 cm = 6 cm = 0.06 m
Vf = speed of electron when it is 3 cm from proton = ?
Vi = Initial speed of electron = 0 m/s
Therefore,
2(17.03 m/s²)(0.06 m) = Vf² - (0 m/s)²
Vf = √2.04 m²/s²
Vf = 1.43 m/s
An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=
Answer
0.2067m or 0.2067m
Explanation;
Let lenght of spring= Lo= 11cm=0.110m
It is hang from a mass of
3.05-kg having a length of L1= 12.40cm= 0.124m
Force required to stretch the spring= Fkx
But weight of mass mg= kx then K= Mg/x
K= 3.05-kg× 9.8)/(0.124m-.110m)
K=2135N
But potential Energy U= 0.5Kx
X=√ 2U/k
√(2*10)/2135
X=0.0967m
The required new length= L2= L0 ±x
=
.110m ± 0.0967m
X= 0.2067m or 0.2067m hence the total lenghth
You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
An object has an acceleration of 12.0 m/s/s. If the net force was doubled and the mass were tripled, then the new acceleration would be _____ m/s/s.
✴ Case - I
⟶ Force = F
⟶ Mass = m
⟶ Acceleration = 12m/s²
✴ Case - II
⟶ Force = 2F
⟶ Mass = 3m
To Find :➳ Acceleration in second case.
Concept :⇒ This question is completely based on the concept of newton's second law of motion.
⇒ As per this law, Force is defined as the product of mass and acceleration.
Mathematically, F = ma
Calculation :[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]
New acceleration would be 12 m/s²
Given that;
Acceleration of object = 12 m/s²
New net force = 2f
New mass = 3m
Find:
New acceleration
Computation:
[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]
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Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answer:
The inductance is [tex]L = 40\mu H[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 163 \ turns[/tex]
The diameter is [tex]D = 6.13 \ mm = 6.13 *10^{-3} \ m[/tex]
The length is [tex]l = 2.49 \ cm = 0.0249 \ m[/tex]
The radius is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{6.13 *10^{-3}}{2}[/tex]
[tex]r = 3.065 *10^{-3} \ m[/tex]
The inductance of the Tarik's solenoid is mathematically represented as
[tex]L = \frac{\mu_o * N^2 * A }{l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
A is the area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 3.065*10^{-3}]^2[/tex]
[tex]A = 2.952*10^{-5} \ m^2[/tex]
substituting values into formula for L
[tex]L = \frac{ 4\pi *10^{-7} * [163]^2 * 2.952*10^{-5} }{0.0249 }[/tex]
[tex]L = 40\mu H[/tex]
The greater the frequency of the waves, the ____________ the pitch.
Answer:
Higher.
Explanation:
The greater the frequency the bigger the amplitude gets and the greater pitch gets.
Think - more energy, bigger waves, more waves, and higher sound
A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coil has 700 turns, what is the maximum emf generated in it
Answer:
The maximum emf generated in the coil is 60527.49 V
Explanation:
Given;
area of coil, A = 0.320 m²
angular frequency, f = 100 rev/s
magnetic field, B = 0.43 T
number of turns, N = 700 turns
The maximum emf generated in the coil is calculated as,
E = NBAω
where;
ω is the angular speed = 2πf
E = NBA(2πf)
Substitute in the given values and solve for E
E = 700 x 0.43 x 0.32 x 2π x 100
E = 60527.49 V
Therefore, the maximum emf generated in the coil is 60527.49 V
lock of mass m2 is attached to a spring of force constant k and m1 . m2. If the system is released from rest, and the spring is initially not stretched or com- pressed, find an expres- sion for the maximum displacement d of m2
Answer:
The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]
Explanation:
Kinetic Energy (K) = 1/2mv²
Potential Energy (P) = mgh
Law of Conservation of energy states that total energy of the system remains constant.
i.e; Total energy before collision = Total energy after collision
This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.
[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]
d = maximum displacement of the mass m₂
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s
Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s
A screen is placed 43 cm from a single slit which is illuminated with 636 nm light. If the distance from the central maximum to the first minimum of the diffraction pattern is 3.8 mm, how wide is the slit in micrometer
Answer: The width is 1.25692 μm
Explanation:
The data that we have here is:
Distance between the single slit to the screen = L = 43cm
λ = 636 nm
Distance from the central maximum to the first minimum = Z = 3.8mm
We know that the angle for the destructive diffraction is:
θ = pλ/a
where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,
then we have:
θ = (636nm/a)
And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:
Tg(θ) = Z/L
Tan(636nm/a) = 3.8cm/43cm
First, we need to use the same units in the right side:
3.8mm = 0.38cm
Tg(636nm/a) = 0.38cm/43cm
636nm/a = Atg( 0.38cm/43cm ) = 0.506
a = 636nm/0.506 = 1,256.92 nm
1 μm = 1000nm
then:
a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm
A tornado passes in front of a building, causing the pressure to drop there by 25% in 1 second. Part A If a door on the side of the building is 8.1 feet tall and 3 feet wide, what is the net force on the closed door.
Answer:
F_net = 264, 26 pound
Explanation:
For this exercise we use Newton's second law
F_net = F_int - F_outside
where the force can be found from the definition of pressure
P = F / A
F = P A
we substitute
F_net = P_inside A - P_outside A
F_net = A (P_inside - P_outside)
indicate that the pressure on the outside is 25% less than the pressure on the inside
P_outside = 0.25 P_inside
The area is
A = L W
we substitute
F_net = L W P_inside (1-0.25)
let's calculate
suppose the pressure inside is atmospheric pressure
P_inside= P_atmospheric = 1,013 10⁵ Pa = 14.7 PSI
F_net = 8.1 3 14.5 0.75
F_net = 264, 26 pound
What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol suppresses the immune system and is produced when the body is under stress. B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low. C. The hormone melatonin induces sleep and its production is slowed by exposure to light. D. The hormone oxytocin promotes labor contractions of the uterus during childbirth.
B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 41 m/s (about 91.7 mph, the speed of a good major league fastball).
(a) Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
(b) Now calculate the maximum height using universal gravitation.
(c) How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a) h = 13,205.4 m
b) r_f = 2.12 106 m
c) e% = 0.68%
Explanation:
a) This is an exercise we are asked to use energy conservation,
Starting point. On the surface of Mimas
Em₀ = K = ½ m v²
Final point. Where the ball stops
[tex]Em_{f}[/tex] = U = m g h
Em₀ = Em_{f}
½ m v² = m g h
h = ½ v² / g
let's calculate
h = ½ 41² / 0.0636
h = 13,205.4 m
b) For this part we are asked to use the law of universal gravitation, write the energy
starting point. Satellite surface
Em₀ = K + U = ½ m v² - GmM / r_o
final point. Where the ball stops
[tex]Em_{f}[/tex]= U = - G mM / r_f
Em₀ = Em_{f}
½ m v² - G m M / r_o = - G mM / r_f
In this case all distances are measured from the center of the satellite
1 / rf = 1 / GM (-½ v² + G M / r_o)
let's calculate
1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)
1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)
1 / r_f = 4,714 10⁻⁷
r_f = 1 / 4,715 10⁻⁷
r_f = 2.12 106 m
to measure this distance from the satellite surface
r_f ’= r_f - r_o
r_f ’= 2.12 106 - 1.98 105
r_f ’= 1,922 106 m
c) the percentage difference is
e% = 13 205.4 / 1,922 106 100
e% = 0.68%
The estimate of part a is a little low
While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current through the resistor RRR will
Answer:
The induced current through resistor R will
b) flow from a to b
Explanation:
The image is shown below, and the full question is written down as
The two solenoids in the figure are coaxial and fairly close to each other. While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current through the resistor R will
a) Flow from b to a
b) flow from a to b
c) be zero because the rate is constant.
From the image, the current in the left hand solenoid flows from the positive terminal of the battery to the negative terminal in an anticlockwise direction by convention.
Varying a rheostat causes a change in the resistance of electricity through the solenoid, and a changing current through a solenoid will induce current to flow through another solenoid placed nearby. Therefore, the left-hand solenoid induces a current flow on the right-hand solenoid.
Since the current in the left-hand solenoid flows in an anticlockwise direction, then it will have an equivalent magnetic polarity of a north pole on a magnet.
Also remember that Lenz law states that the induce current acts in such a way as to oppose the motion, or action producing it.
In this case, the induced current in the right-hand solenoid will act as to repel the left-hand solenoid away from itself. The only way is by the right-hand solenoid also having a north pole equivalent magnetic pole on it since like poles repel each other. This means that the induced current in the right-hand solenoid will flow in an anticlockwise manner too, from a to b.
The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the electric field strength? (The value is surprisingly large, but correct.) You may assume a uniform E-field.
Answer:
The magnitude of the electric field intensity is [tex]E = 7.89 *10^{6} \ V/m[/tex]
Explanation:
From the question we are told that
The voltage is [tex]\epsilon = 72.7 \ mV = 72.7 *10^{-3} V[/tex]
The thickness of the membrane is [tex]t = 9.22 \ nm = 9.22 *10^{-9} \ m[/tex]
Generally the electric field intensity is mathematically represented as
[tex]E = \frac{\epsilon }{t}[/tex]
substituting values
[tex]E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}[/tex]
[tex]E = 7.89 *10^{6} \ V/m[/tex]
A truck accidentally rolls down a driveway for 8.0\,\text m8.0m8, point, 0, start text, m, end text while a person pushes against the truck with a force of 850\,\text N850N850, start text, N, end text to bring it to a stop. What is the change in kinetic energy for the truck?
Answer:
Explanation:
According to work energy theorem
change in kinetic energy of truck = work done against it
work done against it = force x displacement
= - 850 x 8 = 6800 J
change in kinetic energy of truck = - 6800 J .
energy will be reduced by 6800 J
Answer:-6800
Explanation:
A light ray in air strikes water at an angle of incidence equal to 40°. If the index of refraction for water is 1.33, what is the angle of refraction?
Answer:
The angle of refraction is 28.68°Explanation:
Given data
angle of incidence, i=40°
angle of refraction,r = ?
index of refraction u= 1.33
applying the formula
[tex]u=\frac{ sin i}{ sin r}[/tex]
According to Snell's law, the incident, normal and refracted rays all act on the same point.
Substituting and solving for r we have.
[tex]1.33= \frac{sin (40)}{sin (r)} \\\\sin(40)= 1.33* sin(r)\\\0.642= 1.33* sin(r)[/tex]
divide both sides by 1.33 we have
[tex]sin(r)= \frac{0.642}{1.33} \\\sin(r)= 0.48\\\r= sin^-^10.48\\\r= 28.68[/tex]
r= 28.68°
If you were in a smooth-riding train with no windows, could you sense the difference between uniform motion and rest or between accelerated motion and rest?
1. Both acclerated and uniform motion can be sensed.
2. Only uniform motion can be sensed.
3. Only accelerated motion can be sensed.
4. No motion can be sensed.
Answer:
3. Only accelerated motion can be sensed
Explanation:
Without windows on such a train, you'd have no frame of reference for your speed. By that I mean, without being able to see how fast you are moving past other things, it's almost as if you aren't moving at all... almost.
At rest you obviously aren't moving and in uniform motion, with a constant speed, it would feel as though you aren't moving. But during periods of acceleration you'll feel the force on your body (F=ma) and would be able to tell if you were moving in a particular direction.
You've probably felt this before. Maybe not on a windowless train but perhaps in a car or on a roller coaster. Speeding up makes you go back into your seat a bit and slowing down makes you lean forward a bit. Both speeding up and slowing down are examples of acceleration (just in different directions) and how fast you accelerate will affect how much force you experience.
So the answer would be option 3.
Side note: If the train wasn't smooth riding then there would be some amount of friction going on and you could probably tell if you were in motion by the products of that friction (like sound and vibrations) even at a constant speed.
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton box 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc
Answer:
The frequency is [tex]f = 0.221 \ Hz[/tex]
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]
Let the voltage of the capacitor when it is fully charged be [tex]V_o[/tex]
Then the voltage of the capacitor at time t is said to be [tex]V = \frac{V_o}{2}[/tex]
Now this voltage can be mathematical represented as
[tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]
Where RC is the time constant
substituting values
[tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]- \frac{0.5}{RC} = ln (0.5)[/tex]
[tex]-\frac{0.5}{RC} = -0.6931[/tex]
[tex]RC = 0.721[/tex]
Generally the cross-over frequency for a low pass filter is mathematically represented as
[tex]f = \frac{1}{2 \pi * RC }[/tex]
substituting values
[tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]
[tex]f = 0.221 \ Hz[/tex]
Dr. Stein's hypothesis is that excess sugar causes hyperactivity. He is interested in doing research.
Which research method would be the best to use?
Answer:
The correct answer would be - dependent independent variable experiment.
Explanation:
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.
Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, the correct answer would be - dependent independent variable experiment.
The best research method to use for the research of hyperactivity, would be dependent-independent variable experiment.
The given problem is based on the effect of sugar on hyperactivity. Hyper activity refers to the increased movement, impulse actions and a shorter attention span.
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, we can conclude that the best research method to use, would be - dependent-independent variable experiment.
Learn more about the hyperactivity here:
https://brainly.com/question/15539672
In the figure, suppose the length L of the uniform bar is 3.2 m and its weight is 220 N. Also, let the block's weight W = 270 N and the angle θ = 45˚. The wire can withstand a maximum tension of 450 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?
Answer:
a) x = 2.46 m
b) 318.2 N
c) 177.8 N
Explanation:
Need to resolve the tension of the string at end say B.
The vertical upward force at B due to tension is 450 sin 45°.
Using Principle of Moments, with the pivot at A,
Anti clockwise moments = Clockwise moments
450 sin 45° X 3.2 = 220 X (3.2/2) + (270 X x)
x = 2.46 m
(b) The horizontal force is only due to the wire's tension, so it is
450 cos 45° = 318.2 N
(c) total downward forces = 270 + 220 = 496 N
Total upward forces = 450 sin 45° (at B) + upForce (at A)
Equating, upForce = 496 - 318.2
= 177.8 N
2. A wire 4.00 m long and 6.00 mm in diameter has a resistance of 15.0 mΩ. A potential difference of 23.0 V is applied between the end. a) What is the current in the wire? b) Calculate the resistivity of the wire material. c) Try to identify the material.
Answer:
Explanation:
a )
current in the wire = potential diff / resistance
= 23 / (15 x 10⁻³ )
= 1.533 x 10³ A .
b )
For resistance of a wire , the formula is
R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area of wire
putting the given values
15 x 10⁻³ = 4ρ / π x .003²
ρ = 106 x 10⁻⁹ ohm. m
= 10.6 x 10⁻⁸ ohm m
The metal wire appears to be platinum .
(a) The current in the wire is 1.533 x 10³ A
(b) The resistivity of the wire material is 10.6 x 10⁻⁸ Ωm
(c) The material of the wire is Platinum
Ohm's Law and resistivity:
(a) According to the Ohm's Law:
V = IR
where V is the potential difference
I is the current
and R is the resistance
So,
I = V/R
I = 23 / (15 x 10⁻³ )
I = 1.533 x 10³ A
(b) The resistance of a wire is expressed as:
R = ρ L / A
where ρ is the resistivity,
L is length
and A is the cross-sectional area
15 x 10⁻³ = 4ρ / π x .003²
ρ = 106 x 10⁻⁹ Ωm
ρ = 10.6 x 10⁻⁸ Ωm
The metal from which the wire is made is platinum.
Learn more about Ohm's Law:
https://brainly.com/question/796939?referrer=searchResults
To compensate for acidosis, the kidneys will
Answer:
Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.
In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.25 m/s2. Determine the orbital period of the satellite.
Answer:
118 minutes( 2 hours approximately )
Explanation:
Here, we are interested in calculating the orbital period of the satellite
Please check attachment for complete solution
Answer:
T = 7101 s = 118.35 mins = 1.9725 hrs
Explanation:
To solve the question, we apply the formula for gravitational acceleration
a = GM/r², where
a = acceleration due to gravity
G = gravitational constant
M = mass of the earth
r = distance between the satellite and center of the earth
Now, if we make r, subject of formula, we have
r = √(GM/a)
Recall also, that
a = v²/r, making v subject of formula
v = √ar
If we substitute the equation of r into it, we have
v =√a * √r
v =√a * √[√(GM/a)]
v = (GM/a)^¼
Again, remember that period,
T = 2πr/v, we already have v and r, allow have to do is substitute them in
T = 2π * √(GM/a) * [1 / (GM/a)^¼]
T = 2π * (GM/a³)^¼
T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼
T = 6.284 * [(3.982*10^14) / 244.140]^¼
T = 6.284 * (1.63*10^12)^¼
T = 6.284 * 1130
T = 7101 s
T = 118.35 mins
T = 1.9725 hrs