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Vanyuwa [196]
2 years ago
15

A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially

at rest, is dropped onto the car.
What will be the car's new speed?
Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
Alchen [17]2 years ago
5 0

The new speed of car is 10.9 m/s

<h3 />

According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

m_1u_1 + m_2u_2 = (m_1 + m_2) v

v =\frac{m_1u_1}{m_1+m_2}

v = \frac{7950*15}{7950+2950}

v= 10.9 m/s

Therefore, the new speed of care is 10.9 m/s

Learn more about momentum here:

brainly.com/question/22257327

#SPJ1

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A running student has half the kinetic energy that his brother has. The student speeds up by 1 m/s, at which point he has the sa
hoa [83]

Answer:

V = (√2) + 1) m/s

Explanation:

Let the mass and speed of the running student be M and V respectively.

We are told that when he speeds up by 1 m/s, he has the same kinetic energy as his brother.

Thus, his speed at which he mow has the same kinetic energy as his brother is (V + 1) m/s

Now, we are told that the mass of the student is twice as large as that of his brother. Thus, his brother's mass is; M/2

Since kinetic energy is given by the formula K.E = ½mv²

Therefore, since we want to find the original speed of both students and that the initial condition says that the running student had half the kinetic energy of the brother, we now initial condition as;

½MV²= ½(½(M/2)V²) - - - - (eq 1)

Since he has sped up by 1 m/s, and has a kinetic energy now equal to that of his brother, we have;

(½M(V + 1)²) = (½(M/2)V²) - - - - (Eq2)

Dividing eq 1 by eq 2 gives;

V²/(V + 1)²= 1/2

Taking square root of both sides gives;

V/(V + 1) = 1/√2

Cross multiply to give;

(√2)V = V + 1

(√2)V - V = 1

V((√2) - 1) = 1

V = 1/((√2) - 1)

Simplifying this using surfs gives;

V = [1/((√2) - 1)] × ((√2) + 1))/((√2) + 1))

V = ((√2) + 1))/1

V = (√2) + 1) m/s

8 0
3 years ago
a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second
lawyer [7]

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

3 0
3 years ago
A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f
meriva

Answer:

v=\sqrt{\dfrac{10g(R-r)}{7}}

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

I=\dfrac{2}{5}mr^2

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for  rolling without slipping       v= ωr

Form energy conservation

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

 v= ωr

I=\dfrac{2}{5}mr^2

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2

mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2

2mg(R-r)=mv^2+\dfrac{2}{5}mv^2

2g(R-r)=\dfrac{7}{5}v^2

v=\sqrt{\dfrac{10g(R-r)}{7}}

3 0
3 years ago
How much force is needed to accelerate a toy car of mass 8 kg at 2m/s
atroni [7]

Answer:

4N

Explanation:

f = ma

8kg x 2m/s = 4N

4 0
2 years ago
What was the first manmade satellite put into orbit
Nataly_w [17]

The Soviet Union's Sputnik 1, it was launched October 4th 1957.

5 0
3 years ago
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