Answer:
the expression is False
Explanation:
From the kinematics equations we can find the speed of a body in a clean fall
v = v₀ - g t
v² = V₀² - 2 g y
If the body starts from rest, the initial speed is zero (vo = 0)
v= √ (2g y)
In the first equation it gives us the relationship between speed and time.
With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.
Therefore the expression is False
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.
Wiley Coyote has missed the elusive road runner once again. This time, he leaves the edge of the cliff at 52.4 m/s with a purely horizontal initial trajectory. If the canyon is 141 m deep, how far from the base of the cliff does the coyote land
Answer:
280.86 mExplanation:
Range is defined as the distance covered in the horizontal direction. In projectile, range is expressed as x = vt where;
x is the range
v is the velocity of the runner
t is the time taken
Before we can get the range though, we need to find the time taken t using the relationship S = ut + 1/2gt²
if u = 0
S = 1/2gt²
2S = gt²
t² = 2S/g
t = √2S/g
t = √2(141)/9.8
t = √282/9.8
t = 5.36secs
The range x = 52.4*5.36
x =280.86 m
Hence, the coyort lies approximately 280.86 m from the base of the cliff
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.
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
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 person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle
Answer:
T = 2689.6N
Explanation:
Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree
while both weight are acting perpendicular to the length. Hence we have :
Torque ( clockwise) = Torque ( anticlockwise)
m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1
Where m1 = 36kg
m2 = 20kg
g = 9.81m/s^2
Theta = 12
Substituting into equation 1
36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)
353.16L/2+196.2L = T ×0.2079(2L/3)
176.58L+196.2L = T × 0.1386L
372.78L = 0.1386LT
T = 372.78L/0.1386L
T = 2689.6N
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
if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Answer:
neither will happen
Explanation:
cause the water is already defreezed
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!
Suppose a tank filled with water has a liquid column with a height of 19 meter. If the area is 2 square meters 2m squared, what’s the force of gravity acting on the column of water?
Answer:
372,400 N
Explanation:
The volume of the column is ...
V = Bh = (2 m^2)(19 m) = 38 m^3
If we assume the density is 1000 kg/m^3, then the mass of the water is ...
M = ρV = (1000 kg/m^3)(38 m^3) = 38,000 kg
The force of gravity on that mass is ...
F = Mg = (38,000 kg)(9.8 m/s^2) = 372,400 N
A plane progressive
the expression
in time, ys
where you
progressivo ware is no presented by
(At + A
y- 5 sin
in metre, t es in time the doplicensel
Calculate
the amplitude of the wave.
Answer:
Amplitude, A = 5 m
Explanation:
Let a progressive wave is given by equation :
[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)
The general equation of a progressive wave is given by :
[tex]y=A\sin (\omega t-kx)[/tex] ....(2)
Here,
A is the amplitude of the wave
[tex]\omega[/tex] is the angular frequency
k is propagation constant
We need to find the amplitude of the wave.
If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.
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.
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
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to (a) (1/3), (b) (1/10)
Answer:
35.3°
18.4°
Explanation:
a.
The first polariser polarises the unpolarised light reducing its intensity from I0 to I0/2. We have to reduce the intensity from I0/2 to I0/3.
Using to Law of Malus, I=I0cos²θ
cos²θ=I/I0=(I0/3)/I0/2 ,
cosθ=√2/3−−√=0.6667−−−−−√=0.8165
θ=cos−1(0.8165)=35.3∘
B.
Cos²θ=I/Io =Io/10/Io9
Cosθ= √9/10= 0.9487
= cos−10.9487
=18.4°
(a) The angle of polaroid such that intensity reduces by 1/3 is 35.26°
(b) The angle of polaroid such that intensity reduces by 1/10 is 63.43°
Angle of polarisation:According to the Malus Law: The intensity of light when passing through a polarizer is given by:
I = I₀cos²θ
where θ is the angle of the polarizer axis with the direction of polarization of the light
I₀ is the initial intensity
When an unpolarised light passes through a polarizer, θ varies from 0 to 2π, so the intensity after passing the first polarizer is :
I = I₀<cos²θ> { average of cos²θ, for 0<θ<2π}
I = I₀/2
Now, this emerging light passes through a second polarizer such that:
(a) the intensity is I' = I₀/3
From Malus Law:
I' = Icos²θ
I₀/3 = (I₀/2)cos²θ
cos²θ = 2/3
θ = 35.26°
(b) the intensity is I' = I₀/10
From Malus Law:
I' = Icos²θ
I₀/10 = (I₀/2)cos²θ
cos²θ = 1/5
θ = 63.43°
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A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle \theta\:=\:θ = 35.4 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?
Answer:
The velocity is [tex]v = 2.84 1 \ m/s[/tex]
Explanation:
The diagram showing this set up is shown on the first uploaded image (reference Physics website )
From the question we are told that
The mass is m = 4 kg
The length of the string is [tex]L = 2.0 \ m[/tex]
The constant angle is [tex]\theta = 35.4 ^o[/tex]
Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as
[tex]Tcos (\theta ) - mg = 0[/tex]
=> [tex]mg = Tcos (\theta )[/tex]
Now let the force acting on mass horizontally be k so from SOHCAHTOA rule
[tex]sin (\theta ) = \frac{k }{T}[/tex]
=> [tex]k = T sin \theta[/tex]
Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as
[tex]F_v = \frac{m v^2}{r}[/tex]
So
[tex]k = F_v[/tex]
Which
=> [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]
So
[tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]
=> [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]
=> [tex]v = \sqrt{r * g * tan (\theta )}[/tex]
Now the radius is evaluated using SOHCAHTOA rule as
[tex]sin (\theta) = \frac{ r}{L}[/tex]
=> [tex]r = L sin (\theta)[/tex]
substituting values
[tex]r = 2 sin ( 35.4 )[/tex]
[tex]r = 1.1586 \ m[/tex]
So
[tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]
[tex]v = 2.84 1 \ m/s[/tex]
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.
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an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?
Answer:
Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.
If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.
A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from the star and is moving at 55 km/s. What is the semimajor axis of the planet's orbit
Answer:
32
Explanation:
The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?
Answer:
Torque = 8.38Nm
Explanation:
Time= 8.00s
angular speed (w) =400 rpm
Moment of inertia (I)= 1.60kg.m2 about its rotation axis
We need to convert the angular speed from rpm to rad/ sec for consistency
2PI/60*n = 0.1047*409 = 41.8876 rad/sec
What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?
Then we need to use the formula below for our torque calculation
from basic equation T = J*dω/dt ...we get
Where : t= time in seconds
W= angular velocity
T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm
Therefore, constant torque that is required is 8.38 Nm
Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.
Given-
Inertia of the flywheel is 1.60 kg m squared.
Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,
[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]
[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]
[tex]\omega = 41.89[/tex]
Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.
When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,
[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]
[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]
[tex]\tau=8.38[/tex]
Hence, the required constant torque is 8.38 N-m.
to know more about the torque, follow the link below-
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radiation transfers energy through___. a metal. b liquid. c touch. d waves.
Answer:
Radiation is transferred through electromagnetic waves so D.
Explanation:
Answer:
D. Waves
Explanation:
a and b don't make much sense, conduction is transfer of energy through touch
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]
You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will have a value that is
Answer:
The equivalent is 6R/11Explanation:
We know that the equivalent resistance of resistors connected in parallel is expressed as
[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]
the L.C.M is 6R
[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]
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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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 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
A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
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]
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-half its original value, and the charge of B to one-tenth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
An object has an acceleration of 6.0 m/s/s. If the net force was tripled and the mass were doubled, then the new acceleration would be _____ m/s/s.
Answer:
The new acceleration would be 9 m/s².
Explanation:
Acceleration of an object is 6 m/s²
Net force is equal to the product of mass and acceleration i.e.
F = ma
[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]
If the net force was tripled and the mass were doubled, it means,
F' = 3F
m' = 2m
Let a' is new acceleration. So,
[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]
So, the new acceleration would be 9 m/s².