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3 years ago

Every force has one and only one 3rd law pair force.

Physics

2 answers:

viktelen [127]3 years ago

5 0

Answer: Please see below as the answer is self-explanatory.

Explanation:

The Newton's 3rd law, in simple words, explains, that for any force exerted on an object by another one, there exists a reaction from the object acted upon, which consists in another force, that is of equal value that the one is being exerted on it, but in the opposite direction, and is acting on the the other object.

So, Newton's 3rd Law says that for any pair of objects interacting, there is a pair of forces, equal and opposite, that each of them exerts on the other.

It is very important to understand, that in any free body diagram, it must be showed only one of these forces, the reaction that the other object is exerting on the object for which we are drawing the FBD, not the two forces, as the another one is external to the other object.

One example for this is the normal force, the gravity force, etc.

FromTheMoon [43]3 years ago

3 0

Answer:

Explanation:

This statement is true. According to Netwon's third Law, every force has one and only same and opposite force as reaction.

Actually, the two forces must be originated in two physical points, which act in opposite direction with equal magnitude. The best example of this the force due to gravity and the normal force, which is the unique pair reaction force of the one due gravity.

Therefore, the given statement is true.

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A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the

Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$ . What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

$= 3050.19 \ kg \ m^2$

So the moment of inertia is $3050.19 \ kg \ m^2$ .

4 0

2 years ago

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

5 0

3 years ago

Convert 5g/cm^3 into kg/m^3

Sliva [168]

Answer:

0.01135624

Explanation:

7 0

2 years ago

If you shout at a cliff wall that is 440 m away and the air temperature is at 25 °C, how long will it take before you hear your

ira [324]

The speed of sound at T=25°C is Vs=346 m/s. So the sound has to reach the cliff and return back to you so the path it needs to travel is s=2*440 m = 880 m.
Since the speed of sound is constant s=Vs*t, and t= s/Vs=880/346=2.54335 s. You will hear the echo after t=2.54335 s after you shouted.

7 0

3 years ago

How to find acceleration?

Yanka [14]

Acceleration = Change in Velocity / time

a = (v - u) / t

Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²

A proper record of the changes in velocity with the corresponding time would help find the acceleration.

4 0

3 years ago

Read 2 more answers