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irina1246 [14]
3 years ago
10

the apparent weight of a body wholly immersed in water is 32N and its weight in 96N and calculate volume of the body

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

0.0065 m³

Explanation:

Apparent weight = weight − buoyancy

32 N = 96 N − (1000 kg/m³) (9.8 m/s²) V

V = 0.0065 m³

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Alondra applies 24 N of force to the end of a stick (a type of lever) to open a can of paint.
Assoli18 [71]

Answer:

1/4

Explanation:

Mechanical Advantage = Load/Effort

Given

Effort applied = 24N

Load = 6N

Substitute

MA = 6/24

MA = 1/4

Hence the mechanical advantage is 1/4

8 0
3 years ago
Read 2 more answers
10. During 4th period we put Klaudia in a box because she was talking too much. We still heard her voice through the box so we d
Alex787 [66]

Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

Explanation:

Given that;

friction force of ground box = 68 N

student of 7th grade = n

Whitmore can apply a force of 25 N

every other 7th grade student can apply a force of 6 N.

now

friction force = forced applied by whitmore + total force ny 7th grade student

we substitute

68 = 25 + 6n

6n = 68 - 25

6n = 43

n = 43/6

n = 7.17

Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

6 0
3 years ago
Use the diagram below modeling a football kicked from a horizontal surface B
djyliett [7]
B is the correct one
5 0
3 years ago
Read 2 more answers
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0
3 years ago
How much time will it take to do 33j of work with 11 w of power
Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s

 

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}

8 0
3 years ago
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