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Mama L [17]
3 years ago
9

Turn this scentence to repirted speach.i ate icecreamShe said that..........​

Physics
2 answers:
kenny6666 [7]3 years ago
7 0
Fvhjgxvbbdfhj if Dan kf xdhgxzsf
Luda [366]3 years ago
5 0

Answer:

dhfhffuththt9tr8tujtngigjtjrjrjrurur

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A 195 g glider is moving at 5.3 m/s on a frictionless air track. It then collides with a stationary 295 g glider.
lord [1]

Answer:For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart. For a perfectly inelastic collision, the final velocity of the cart system will be 1/2 the initial velocity of the moving cart.

Explanation:

3 0
3 years ago
What is the magnitude of the force acting on a spring with a spring constant of 275 N/m that is stretched 14.3 cm?
Travka [436]
15 i thank 
if u need more help just ask me

8 0
3 years ago
List the rock layers from oldest to youngest
MakcuM [25]

Answer:

I,J,h,g,f,e,d,c,b,a

Explanation:

8 0
2 years ago
Read 2 more answers
A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
3 years ago
A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

4 0
3 years ago
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