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Arturiano [62]
3 years ago
10

A Russian athlete lifts 300kg to a height of 2.5 m in 2 seconds. What is the work done? And what power is exerted?

Physics
1 answer:
snow_tiger [21]3 years ago
5 0
We want to find the work done and power exerted, let’s start with work first.

We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.

Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2

(The equation says x but can be used with y values)

If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.

To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.

Force = 300 kg * 1.25 m/s^2 = 375 Newtons

So, work = 500 N * 2.5 m = 1000 Joules

Power: W/t

So, Power = 1000 J / 2 seconds = 500 Watts

Hope this helps!

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You're driving in a car at 50 km/h and bump into a car ahead traveling at 48 km/h in the same direction. the speed of impact is
salantis [7]

To solve this problem, we must remember about the law of conservation of momentum. The initial momentum mist be equal to the final momentum, that is:

m1 v1 + m2 v2 = (m1 + m2) v’

where v’ is the speed of impact

Since we are not given the masses of each car m1 and m2, so let us assume that they are equal, such that:

m1 = m2 = m

Which makes the equation:

m v1 + m v2 = (2 m) v’

Cancelling m and substituting the v values:

50 + 48 = 2 v’

2 v’ = 98

v ‘ = 49 km/h

 

<span>The speed of impact is 49 km/h.</span>

6 0
3 years ago
A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m
victus00 [196]

Answer:

The acceleration of the car will be a=9600m/sec^

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion s=ut+\frac{1}{2}at^2

200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2

a=9600m/sec^

So the acceleration of the car will be a=9600m/sec^

6 0
4 years ago
A 20-kilogram child is riding on a 10-kg sled over a frictionless icy surface at 8.0 meters per second. Calculate the kinetic en
Veseljchak [2.6K]

Answer:

K = 960 J

Explanation:

Given that,

Mass of a child = 20 kg

Mass of a sled = 10 kg

Speed of child on sled = 8 m/s

We need to find the kinetic energy of the sled with the child.

The total mass of child and the sled = 20 kg + 10 kg

= 30 kg

The formula for the kinetic energy of an object is given by :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 30\times (8)^2\\\\K=960\ J

Hence, the kinetic energy of the sled with the child is 960 J.

6 0
3 years ago
A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

4 0
3 years ago
Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt
Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of  A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am  1 m away from B , the path difference will be

8 - 1 = 7 m  which is  odd multiple of 1 or λ /2

So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

8 0
4 years ago
Read 2 more answers
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