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2 years ago

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Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

rth, M = 5.97 10 24 kg and Radius of the Earth, R=5.97*1024 m, and G=6.67*10-11. , Hint use V=(GM*2pi/T)^1/3.

1 answer:

Hoochie [10]2 years ago

3 0

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

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A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s² , v = 42 m/s

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago

a car with a mass of 2000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. The curve h

melisa1 [442]

In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
      = [2000 * (25)^2]/80
      = (2000 * 625)/80
      = 1250000/80
      = 15625
So the centripetal force on the car is 15625 Newtons

4 0

3 years ago

Read 2 more answers

Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,

White raven [17]

The net force acting on the car is 65 N to the left

The net force acting on an object is simply defined as the resultant force acting on the object.

From the question given, we obtained the following data:

Force applied to the right (Fᵣ) = 250 N
Force applied to the left (Fₗ) = 315 N
Net force (Fₙ) =?

The net force acting on the car can be obtained as follow:

Fₙ = Fₗ – Fᵣ

Fₙ = 315 – 250

<h3>Fₙ = 65 N to the left </h3>

Therefore, the net force acting on the car is 65 N to the left

Learn more on net force: brainly.com/question/19549734

8 0

1 year ago

Read 2 more answers

Consider a particle with unit charge q, and mass m, in a constant magnetic field B directed along the positive z–axis. The parti

max2010maxim [7]

Answer:

it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

Explanation:

Magnetic field is along z axis while velocity is in x-z plane

so we will have

$F = q(\vec v \times \vec B)$

so here we can say

$F = q(u\hat i + w\hat k) \times (B \hat k)$

so we will have

$F = quB(-\hat j)$

so here the net force on the charge is perpendicular to its x directional velocity along - Y direction

So due to this component of motion it will move along a circle while other component of the motion will remain uniform always

So here it is combination of two parts

1) Uniform circular motion

2) Uniform motion

So we can say that it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

4 0

2 years ago

A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a

Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

$F=ma$ Eqn. (1)

where

$F$ : is the Net Force in Newtons ( $N$ )

$m$ : is the mass ( $kg$ )

$a$ : is the acceleration ( $m/s^2$ )

We also know that the acceleration is denoted by the velocity ( $v$ ) of an object as a function of time ( $t$ ) with

$a=\frac{v}{t}$ Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

$F=m\frac{v}{t}\\ \\Ft=mv$ Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. $Ft$ is in fact the Impulse $J$ of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ $22N$ and $t=4 sec$ (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

$J=Ft\\J=(22N)(4sec)\\J=88 kg m/s$

So the impulse of the cart is $J=88kg m/s$

Then, we know that the cart is moving at $v=10 m/s$ . Plugging in the values in Eqn. (3) we have:

$(22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg$

So the mass of the cart is $m=8.8kg$ .

8 0

3 years ago