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Ilia_Sergeevich [38]
3 years ago
9

Newton's third law states that when an object exerts force on another object, the second object exerts an equal but opposite for

ce on the first object. Which of the following shows an example of this
A. A girl hits a volleyball. Her hand exerts a force on the ball, and the ball exerts a force on her hand.

B. A woman presses on both sides of a box with equal force.

C. A moving cart becomes more difficult to stop as more mass is added to it.
Physics
2 answers:
svlad2 [7]3 years ago
6 0
<span>A. A girl hits a volleyball. Her hand exerts a force on the ball, and the ball exerts a force on her han</span>
Tju [1.3M]3 years ago
5 0

Answer:

A. A girl hits a volleyball. Her hand exerts a force on the ball, and the ball exerts a force on her hand.

Explanation:

As rightly stated, from Newton's third law, we understand that for every action, there is an equal and opposite reaction, and if we closely examine the listed options, starting with A

A. The said girl acts by hitting the volleyball, i.e her hand exerts a force on the ball and the reaction occurs as the ball exerts a force on her hand.

B. Here, there is an action with no account of reaction, the woman presses the boxes (action), even thought with equal force on the sides, this option gives no detail of an equal and opposite reaction.

C. There is an action taking place by adding more masses to the moving cart, however, there is no explanation in this option for a reaction

Therefore, with a close look of the above listed options, it can be concluded that only option A shows an example of Newton's third law.

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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
4 years ago
If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ra
igomit [66]

Answer:

\frac{E}{E_c} =3.125

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

E_c=\frac{1}{2} mv^2

The equivalence between mass and energy established by the theory of relativity is given by:

E=mc^2

This formula states that the equivalent energy E can be calculated as the mass m multiplied by the speed of light c squared.

Where c is approximately 3\times 10^{8} m/s

Hence:

E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J

E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J

Therefore,  the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125

4 0
3 years ago
a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t
Agata [3.3K]

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

5 0
3 years ago
The distance between two successive maximaof
denpristay [2]

Answer:

v = 1.224 m/s

Explanation:

given,                          

distance between the two successive maxima = 1.70 m

number of crest = 8          

time = 11 s                            

frequency is equal to number of cycle per secod

f = \dfrac{8}{11}          

f = 0.72\ Hz                

velocity of wave

v = f x λ                

v = 0.72 x 1.70        

v = 1.224 m/s          

Hence, the wave speed is equal to v = 1.224 m/s

4 0
4 years ago
An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De
gayaneshka [121]
We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
F=T-W
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
a= \frac{F}{m}
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
F=T-W=6000 N-5000 N=1000 N
To find m, we can use the weight of the system. In fact, the weight of an object is given by
W=mg
where g=9.81 m/s^2. Solving for m, and using W=5000 N, we find
m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg

and at this point, we can calculate the acceleration of the system (elevator+people):
a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2
and the acceleration has the same direction of the resultant force, so upward.
8 0
3 years ago
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