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wel
3 years ago
12

Find the distance covered by a car in 10 second, if it is moving at the speed of 20m/s​

Physics
1 answer:
labwork [276]3 years ago
8 0

Answer:

solution

we know that .

s=ut+½at²

now, putting the values in the second equation of motion ,

we get,

S=20×10+1/2×(10)²

S=200+1×100

S=200+100

S=300m

the distance covered by given body or object in 10 seconds is 300 m.

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A car with a mass of 1500 kg is traveling down the interstate at 25 m/s. What is the kinetic energy of the car, measured in Joul
iren2701 [21]

Answer:

The answer is

<h2>468 , 750 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}

where

m is the mass of the object

v is the speed or velocity

From the question

m = 1500 kg

v = 25 m/s

The kinetic energy of the object is

KE =   \frac{1}{2}  \times 1500 \times  {25}^{2}  \\  = 750 \times 625

We have the final answer

<h3>468 , 750 J</h3>

Hope this helps you

5 0
3 years ago
Help plz I don't understand ​
Vika [28.1K]

#78

Airplane start with initial speed

v_i = 0

now the takeoff velocity is given as

v_f = 300 km/h

v_f = 83.33 m/s

acceelration is given as

a = 1 m/s^2

now we have

v_f - v_i = at

from above equation we have

83.33 - 0 = 1(t)

t = 83.33 s

#79

Airplane start with initial speed

v_i = 0

now the takeoff velocity is given as

v_f = 300 km/h

v_f = 83.33 m/s

acceelration is given as

a = 2 m/s^2

now we have

v_f - v_i = at

from above equation we have

83.33 - 0 = 2(t)

t = 41.7 s

6 0
4 years ago
A 70 kg person walks 200 m in 4 seconds. What is their momentum?
Gemiola [76]

Answer:

z65td8ztu4a7a5857a57a75477282u2hshshsh7e7eyeyw82773888898 idhw

8 0
3 years ago
Which environmental changes occur faster?
irga5000 [103]
The answer is B. Oil Spill
6 0
4 years ago
Read 2 more answers
A track consists spring launcher on one end. A spring which is compressed 0.5 m has a
patriot [66]

(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.

(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.

<h3>Speed of the block when pushed by the spring</h3>

The speed of the block when pushed by the spring is calculated as follows;

K.E = Ux

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v² = (25 x 0.5²)/0.05

v² = 125

v = 11.18 m/s

<h3>Final velocity of the two balls after the collision</h3>

The velocity of the two balls after the collision is calculated as follows;

Pi = Pf

where;

  • Pi is initial momentum
  • Pf is final momentum

m1u1 + m2u2 = v(m1 + m2)

0.05(11.18) + 0.05(0) = v(0.05 + 0.05)

0.559 = 0.1v

v = 5.59 m/s

Learn more about velocity here: brainly.com/question/4931057

#SPJ1

4 0
2 years ago
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