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Vitek1552 [10]
3 years ago
8

An athlete does one push-up period in the process, she moves half of her body weight, 250 N, a distance of 20 cm, this distance

is the distance her gravity moves when she fully extend her arms how much work did you do after one push-up
Physics
1 answer:
ioda3 years ago
3 0
In the push-up, the athlete's weight  (or force) of 250 N moves through a distance of 20 cm.

By definition,
Work = force * distance.

Therefore, work done in one push-up is
W=(250 \, N)*(20 \, cm)*(10^{-2} \,  \frac{cm}{m}) = 50 \, J

Note that 1 J = 1  N-m.

Answer:  1 J
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A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors
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Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, F_1 = 140 N

let the second force = F_2

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N

F_2 = 280 \ East

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N

F_2 = 560 \ West

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a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe
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Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, m=1\ kg

Extension length of the spring is, x=10\ cm=0.1\ m

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

T=2\pi\sqrt{\frac{m}{k}}

Plug in the given values and solve for period 'T'. This gives,

T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s

Therefore, the period of oscillation is 0.635 s.

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