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3 years ago

According to Newtons third law of motion when one object exerts a force on a second object what is true about the forces?

Physics

1 answer:

fomenos3 years ago

7 0

Answer:

D.They are equal in magnitude and opposite in direction

Explanation:

Newton's third law states that the action force is always associated with a reaction force.
When a body 'A' exerts a force on body 'B', it is called the force of action.
When the body 'B' in turn resist the force of 'A' is called the reaction force. It is the reactive force acted upon by the body 'B' on 'A'.
This reaction force is equal in magnitude of the action force.
If the two bodies remain in the same horizontal line, the 'A' exerts a force in the direction towards 'B' and the body 'B' exerts a reaction force in the direction towards 'A'.
Hence, the two forces that are exerted by the bodies are equal in magnitude and opposite in direction.

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A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli

Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

5 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

A jogger accelerates at a constant rate as she travels 5.0

VladimirAG [237]

Answer:

2.0s

Explanation:

...

3 0

3 years ago

How many protons are there in any te atom

Andreyy89

Answer:

Explanation:

Tellurium is the element of the periodic table with atomic number 52.

The atomic number of a chemical element represents the number of atoms contained in the nucleus of the atom: therefore, this means that an atom of tellurium contains exactly 52 protons in its nucleus.

Tellurium is classified as a metalloid, having properties in between metals and non-metals, and it appears with a silver color.

4 0

3 years ago

A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a

Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

$\sum F = ma$

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

$N-mg=ma$

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

$N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N$

So the closest answer is

D. 48.985 N

3 0

3 years ago