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tresset_1 [31]
3 years ago
8

What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.085 m/s2 for 7.5 min, start

ing with an initial velocity of 3.4 m/s
Physics
1 answer:
Llana [10]3 years ago
4 0

Answer:

v_f=41.65\frac{m}{s}

Explanation:

The final velocity is given by the following kinematic equation:

v_f=v_0+at

Here, v_0 is the initial velocity, a is the body's acceleration and t is the motion time. We have to convert the time to seconds:

7.5min*\frac{60s}{1min}=450s

Now, we calculate the final velocity:

v_f=3.4\frac{m}{s}+(0.085\frac{m}{s^2}(450s))\\v_f=41.65\frac{m}{s}

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3 years ago
Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow
Helen [10]

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0
3 years ago
Which of the following changes would make a heat engine waste more energy as heat
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3 years ago
While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the
BartSMP [9]

In the experiment of free fall bob released a bag of mass 1 lb

so here we can say that initial speed of the bag is Zero

time taken by the bag to free fall is given as

t = 1.5 s

also the acceleration of free fall is given as

a = 9.8 m/s^2

now we will use kinematics equation here for finding the distance of free fall

d = v_i * t + \frac{1}{2} at^2

d = 0 + \frac{1}{2}*9.8* 1.5^2

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d = 11.025 m

so the bag will fall down by total distance of 11.025 m from its initial released position.


3 0
3 years ago
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