An object is falling downward at a rate of 25 m/s. Two seconds later, what is its acceleration?

A locomotive is pulling 15 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (w

VMariaS [17]

Answer:

298,220 N

Explanation:

Let the force on car three is T_23-T_34

Since net force= ma

from newton's second law we have

T_23-T_34 = ma

therefore,

T_23-T_34 = 37000×0.62

T_23= 22940+T_34

now, we need to calculate

T_34

Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2

F= ma

So, F= 12×37000×0.62= 22940×12= 275280 N

T_23 =22940+T_34= 22940+ 275280= 298,220 N

therefore, the tension in the coupling between the second and third cars

= 298,220 N

3 0

3 years ago

A 1.0-kg cue ball traveling at 15 m/s strikes a stationary billiard ball of mass 1.5 kg. After the collision, the cue ball remai

Igoryamba

Answer:

Other ball's velocity is 10 m/s

Explanation:

We can use conservation of momentum:

$P_{1i} +P_{2i} =P_{1i} +P_{2i}\\1 * 15 + 1.5 * 0 = 1 * 0 + 1.5 * v\\15 = 1.5 * v\\v=\frac{15}{1.5} \,\frac{m}{s} \\v=10 \,\frac{m}{s}$

8 0

3 years ago

A blue car with mass mc = 427 kg is moving east with a speed of vc = 20 m/s and collides with a purple truck with mass mt = 1282

polet [3.4K]

Answer:8540 kg-g/s

Explanation:

Given

mass of blue car $m_c=427 kg$

velocity of blue car $v_c=20 m/s$

mass of the truck $m_t=1282 kg$

speed of truck $v_t=13 m/s$

After collision they stick and lock together

Let v be the velocity of combined system at angle \theta from vertical

Conserving momentum in east direction

$m_c\times v_c=(m_c+m_t)v\cos \theta$

$427\times 20 =1709\times v\cos \theta$ ------1

Conserving Momentum in Y direction

$m_t\times v_t=(m_c+m_t)v\sin \theta$

$1282\times 13 =1709\times v\sin \theta$ -------2

squaring and then adding 1 & 2 we get

$(8540)^2+(16666)^2=(1709)^2\cdot v^2$

v=10.95 m/s

initial momentum of car $=427\times 20=8540 kg-m/s$

6 0

4 years ago

What is the magnetic field strength required to make a proton with a speed of 5.0 x 10^(5) m/s follow a circular path of radius

Nuetrik [128]

Answer:

Magnetic field strength required for this is 0.25 T

Explanation:

As we know that the proton moves in circular path in uniform magnetic field

so the radius of the path of the circle is given as

$R = \frac{mv}{qB}$

here we know that

$q = 1.6 \times 10^{-19}C$

$m = 1.6 \times 10^{-27} kg$

$R = 0.0200 m$

$v = 5 \times 10^5 m/s$

now we have

$B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}$

so we have

$B = 0.25 T$

8 0

4 years ago

Why is it that an object can accelerate while

posledela

For the same reason that you can skate around a curve at constant speed but not with constant velocity.

The DIRECTION you're going is part of your velocity, but it's not part of your speed.

If the DIRECTION changes, that's a change of velocity.

The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.

And any change of velocity is called acceleration.

3 0

3 years ago