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photoshop1234 [79]
4 years ago
11

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

Physics
1 answer:
zhenek [66]4 years ago
6 0

acceleration times time falling

25*2

50 m*s^-1

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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
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