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Alecsey [184]
2 years ago
13

A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i

n 10 s. Find (a) the total work done on the car
Physics
1 answer:
attashe74 [19]2 years ago
4 0

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

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Using examples, explain why the first and second Newton laws of motion are significant for living organisms.
Triss [41]

Answer:

1) Newton's first law of motion states an object will remain at rest or in uniform will be in uniform motion in a straight line unless a force acts on it

2) Newton's second law states the acceleration of an object is directly proportional to the applied force acting on an object and inversely proportional to the mass of the object

Explanation:

1) With Newton's first law, we are able arrange things within a space and schedule meetings in time knowing that they will remain in place unless an external force changes their positions

2) An example of Newton's second law of motion is that small objects such as a ball are easily accelerated and can be given appreciable acceleration for flight by single, one time contact (such as kicking the ball) while larger objects such as a rock require sustained force application to change their location.

6 0
3 years ago
Why does the candle light has no shadow when light fall on it?​
andrezito [222]
Shadows are the absence of light, they are created when an object blocks light. In other words, shadows are the product of light particles, known as photons. These particles “bounce off” of the object without reaching the other side. Therefore light by itself will not form a shadow.
6 0
3 years ago
A hiker hikes south 0.7 km. she then hikes east 2.4 km. what is the magnitude and unit of her displacement from her initial posi
Usimov [2.4K]
If you illustrate the problem, you will somewhat come up with the figure shown. The missing value is the hypotenuse of the right triangle. Using the pythagorean theorems, the value is determined to be

x = √(0.7^2 + 2.4^2)
x = 2.5 km

2.5 km is the magnitude of the distance. If you want to incorporate the displacement, the answer is reported as 2.5 km, southeast. The direction is determined from the starting point to the endpoint.

7 0
4 years ago
If a truck has the mass of 2,000 kilometres and a velocity of 35 m/s, what is the momentum
alexira [117]

Answer:

<h2>70,000 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 2000 × 35

We have the final answer as

<h3>70,000 kg.m/s</h3>

Hope this helps you

5 0
3 years ago
A law enforcement officer in an intergalactic "police car" turns on a red flashing light and sees it generate a flash every 1.2
butalik [34]

Answer:

The velocity of the police car relative to earth is v_{rel} = 2.51\times 10^{8} m/s

Given:

time for flash generation of the inter galactic police car, t = 1.2 s

time between flashes as measured from earth, t' = 2.2 s

Solution:

Utilising Einstein's equation for time dilation to calculate the velocity of the police car, the equation is given by:

t' = \frac{t}{\sqrt {1 - \frac{v^{2}}{c^{2}}}}                                (1)

where, c = speed of light in vacuum = c = 3\times 10^{8}

re arranging eqn (1) for velocity, v:

v_{rel} = c\times \sqrt {1 - (\frac{t}{t'})^{2}}                               (2)

Now, from eqn (2)

v_{rel} = 3\times 10^{8}( \sqrt {1 - (\frac{1.2}{2.2})^{2}})

v_{rel} = 3\times 10^{8}\times 0.838

v_{rel} = 2.51\times 10^{8} m/s

3 0
3 years ago
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