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KonstantinChe [14]
3 years ago
12

An open train car, with a mass of 2130 kg, coasts along a horizontal track at the speed 2.93 m/s. The car passes under a loading

chute and, as it does so, gravel falls vertically into it for 3.03 s at the rate of 465 kg/s. What is the car's speed v f after the loading is completed
Physics
1 answer:
crimeas [40]3 years ago
4 0

Answer:

The final speed after the loading is completed is 1.764 m/s

Explanation:

Given;

mass of open train car, m₁ = 2130 kg

initial speed of the car, u₁ = 2.93 m/s

loading rate of gravel = 465 kg/s

time of loading = 3.03 s

mass of the gravel, m₂ = 465 kg/s x 3.03 s = 1408.95 kg

From the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final speed after the loading is completed

2130 x 2.93 + 0 = v (2130 + 1408.95)

6240.9 = v (3538.95)

v = 6240.9/3538.95

v = 1.764 m/s

Therefore, the final speed after the loading is completed is 1.764 m/s

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3 years ago
A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic
Ilia_Sergeevich [38]

Answer:

\mu_k=0.27

Explanation:

According to the free body diagram, in this case, we have:

\sum F_x:-F_k=ma\\\sum F_y:N=mg

Recall that the force of friction is given by:

F_k=\mu_k N

Replacing and solving for the coefficient of kinetic friction:

-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}

Finally, we calculate \mu_k:

\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27

4 0
3 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

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3 years ago
On the asteroid Cere, the acceleration due to gravity is about 0.27 m/s^2. How long would it take a ball to fall to the ground i
KiRa [710]

Answer:

F. 25.82 s

Explanation:

Given:

Δy = 90 m

v₀ = 0 m/s

a = 0.27 m/s²

Find: t

Δy = v₀ t + ½ at²

90 m = (0 m/s) t + ½ (0.27 m/s²) t²

t = 25.82 s

7 0
3 years ago
7. Two bikes travelling in the same direction move at a speed of 30 km/hr. The bikes are separated by a distance of 5 km. What w
Fantom [35]

Answer:

Explanation:

Call the bike on the right A

Call the bike on the left B

The car begins it's time when it passes A

4 minutes later, it passes B.

But B has moved in 4 minutes and that is the key to the problem.

How far has B moved.

t = 4 minutes = 4/60 hours = 1/15 of an hour.

d = ?

rate = 30 km / hr

d = r * t

d = 30 km/hr * 1/15 hours = 2 km

The distance between the bikes is 5 km.

So the car has traveled 5 - 2 = 3 km

d = 3 km

r = ?

t = 4 minutes = 1/15 hour

r = d/t = 3/(1/15)= 3 / 0.066666666  = 45 km/hr.

6 0
3 years ago
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