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3 years ago

The following force diagram represents Newton’s Third Law of Motion:

Physics

1 answer:

Julli [10]3 years ago

4 0

Answer:

FALSE.

Explanation:

Newton's third law states that :

Every action has equal and opposite reaction
That is , the magnitude is the same but the directions are opposite
The action reaction forces DONOT operate on the same body.

For example ,

If a block is kept on the ground , the action force is the normal force acting on it due to the ground. BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !

Thus,

we conclude that action and reaction forces donot act on the same body and therefore , this case has the answer : FALSE

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A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

2 years ago

Read 2 more answers

The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe

Crank

Answer:

The particle which completes the given equation is : $_{54}^{138}\textrm{Xe}$

Explanation:

The given reaction is of a fission reaction:

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_Z^A\textrm{X}+2_0^1\textrm{n}$

Total mass on the reactant side is equal to the total mass on the product side:

239 + 1 = 100 +A+ 2

A = 138

Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:

94 + 1(0) = 40 + Z + 2(0)

Z = 54

So atomic number 54 id of Xenon.

The particle which completes the given equation is :

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_{54}^{138}\textrm{Xe}+2_0^1\textrm{n}$

3 0

3 years ago

A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?

patriot [66]

Objects in free fall are weightless.

6 0

2 years ago

Read 2 more answers

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.The

Galina-37 [17]

The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s

7 0

3 years ago

If a 7 kg bowling ball is lifted 2 m into the air and dropped, what speed will it strike the ground? (

Naily [24]

Answer:

6.32m/s

Explanation:

note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²

okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!

when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)

now enough of the talking let solve....

so the ball was dropped .ie from rest to the ground through a distance of 2m,

the formula for calculating the distance if a body moving in a straight line is given by:

S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.

from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)

solving ....

we get t to be 0.632s

to find the speed we substitute t in the equation below:

V=u+at ,but since u=0

V=at =10•0.632=6.32m/s

therefore the speed the body uses to strike the ground is 6.32m/s

4 0

2 years ago