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Alexandra [31]
3 years ago
15

A ball weighing 2.0 N falls through a distance of 4 m.

Physics
1 answer:
Naddik [55]3 years ago
4 0

Answer:

This question is incomplete, a possible question would be to calculate the work done by the ball.

Work done = 8J

Explanation:

Work done by an object is calculated my multiplying the force by the distance.

The weight of this ball is 2.0N, which also represents the force exerted by the ball. The ball travels through a distance of 4m

Hence, work done = F × d

Work done = 2N × 4m

Work done = 8Nm or 8J

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A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net for
Dmitriy789 [7]

Answer:

The answer  is The acceleration is double its original value.

Explanation:

<h2><u>It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.</u></h2><h2><u></u></h2>

Hope this helps....

Have a nice day!!!!

6 0
3 years ago
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What is produced as the result of unequal warming of the earth's surface?
Soloha48 [4]
Unusual precipitation patterns 
7 0
3 years ago
A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i
Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2    ;  μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0
3 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

Distance covered in the slowing down phase = 250 m

v^2-u^2=2as

a= \frac {(v^2-u^2)}{2s}

a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2

v=u+at

t= \frac {(v-u)}{a}

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

The car is in the pit stop for 5s t_2=5 s

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

v=u+at

t= \frac{(v-u)}{a}

t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0
3 years ago
3. A cat pushes a 0.25-kg toy with a net force of 8 N. According to Newton's second
jek_recluse [69]
  • Mass=0.25kg
  • Force=8N

\\ \sf{:}\!\implies F=ma

\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}

\\ \sf{:}\!\implies Acceleration=\dfrac{8}{0.25}

\\ \sf{:}\!\implies Acceleration=32m/s^2

5 0
3 years ago
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