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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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Answer:

graph A

Explanation:

the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)

but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)

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The three forces shown act on a particle. what is the direction of the resultant of these three forces?
melisa1 [442]
Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N
in the positive direction.

y-axis) The resultant on the y-axis is
F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
\tan \alpha =  \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7
from which we find 
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7 0
3 years ago
A water pump is a positive displacement-type pump true or false
Kay [80]

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

6 0
3 years ago
An object with an acceleration of –10 m/s² is _____.
Sidana [21]

The answer might be B because the value is negative, and negative could mean slowing down.

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
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