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3 years ago

A moving train car collides with another train car that is motionless on the track. The train cars are the same

Physics

1 answer:

balandron [24]3 years ago

7 0

Answer:

A. In my own context, I say the momentum is doubled

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An object of mass 10 kg is in a circular orbit of radius 10 m at a

Paladinen [302]

A :-) F = mv^2 by t
Given - m = 10 kg
r = 10 m
v = 10 m/s
Solution -
F = mv^2 by t
F = 10 x 10^2 by 10
F = 10 x 100 by 10
( cut 10 and 10 because 10 x 1 = 10 )
F = 100 N

.:. The centripetal force is 100 N

8 0

3 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

4 years ago

Please help me on this position-time graph assignment!!!

saw5 [17]

$\text{Hello there! :)}$

$\text{The velocity is essentially the slope of the graph in a position vs. time graph, so:}$

$\text{Given: From 0 - 8 sec, velocity of 10 m/s.}\\\\\text{From 8 - 18 sec, velocity of -5 m/s. (In opposite direction because of the negative sign)}\\\\\\\text{From 18 - 22 sec, velocity of 0 m/s. (Object at rest)}\\\\\text{From 22 - 24 sec, velocity of 5 m/s.}\\\\\text{Graph the position vs. time graph using the velocities as the slope of the line }$

6 0

4 years ago

The free-body diagram below represents the force of several vehicles driving across a bridge. Assume that the bridge is in stati

krek1111 [17]

In order to calculate the unknown reaction force, we need to know that the sum of forces pointing down is equal the sum of the forces pointing up, so all forces will be in equilibrium.

So we have:

$\begin{gathered} 17800+150000+13200=121000+x \\ 181000=121000+x \\ x=181000-121000 \\ x=60000 \end{gathered}$

7 0

1 year ago

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings

ozzi

Answer:

$0.3165\ \text{rad/s}$

Explanation:

m = Mass of person = 65 kg

d = Diameter of round table = 6.5 m

r = Radius = $\dfrac{d}{2}=3.25\ \text{m}$

v = Velocity of person running = 3.8 m/s

$I_t$ = Moment of inertia of turntable = $1850\ \text{kg m}^2$

Moment of inertia of the system is

$I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2$

As the angular momentum of the system is conserved we have

$L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}$

The angular velocity of the turntable is $0.3165\ \text{rad/s}$ .

3 0

3 years ago