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4 years ago

Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?

2 answers:

8 0

Newton’s Second Law concerns the generation of force based on an object’s mass and acceleration, as described by the equation F=ma.

Hope this helps!

Ronch [10]4 years ago

5 0

The two variables which influences the body force addressed in “Newton’s second law of motion” s mass and acceleration.

<u>Explanation:</u>

The “Newton’s second law of motion” states that rate of change of momentum is directly in proportional to force applied on body.

$\begin{array}{l}{F \propto \frac{d p}{d t}} \\ {F=m \cdot \frac{d v}{d t}} \\ {F=m \cdot a} \\ {\text { Force }=\text {Mass} \times \text {Acceleration}}\end{array}$

This is how the Second law of motion is derived in formula by the Newton which establishes the relationship between the two variables which influences the force of the body.

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Hydrometer is a devise used to measure the volume of a liquid. *<br> True<br> False

RSB [31]

Answer:

false

Explanation:

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3 years ago

Changing the length of an air column will alter its _______

bija089 [108]

Natural Frequency
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3 years ago

Read 2 more answers

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

Westkost [7]

Answer:

(I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

Explanation:

Given that,

Mass of black hole $m= 1\times10^{8} M_{sun}$

(I). We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=2.94\times10^{8}\ km$

(II). Mass of block hole $m= 6 M_{sun}$

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=17.7\ km$

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}$

$R_{g}=1.1\times10^{-7}\ km$

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}$

$R_{g}=7.4\times10^{-29}\ km$

Hence, (I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

8 0

3 years ago

An empty oﬃce chair is at rest on a ﬂoor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc

Bumek [7]

4. 1 and 2 only.

1. the downward force is the force of gravity.

<span>2. The upward force exerted is the Normal reaction from the floor.</span>

8 0

3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

3 years ago