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Ad libitum [116K]
3 years ago
15

[OSS.01]Which of these statements is most likely correct about Newton's law on gravity?

Physics
2 answers:
irga5000 [103]3 years ago
6 0

Answer:

It does not explain why objects exert gravitational force.

Explanation:

As per the Newton's law of gravitation he said that two objects of different masses are placed at some distance from each other then the gravitational attraction force between two masses is

(1) proportional to the product of the two masses

(2) inversely proportional to the square of the distance between two masses

so it is given by

F = \frac{Gm_1m_2}{r^2}

so here as per Universal law of gravitation Newton gives an experimental result about the gravitational force while the theoretical  explanation why the gravitational force exist is not explained by Newton's law of gravitation

So correct answer will be

It does not explain why objects exert gravitational force.

inessss [21]3 years ago
4 0
I think the first one is the correct answer

note: are not always 100% true
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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:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \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 car travels at an average speed of 60km/h for 15 minutes. How far does the car travel in this time?
jok3333 [9.3K]

Answer:

In 15 minutes the car travels a distance of 15 km.

Explanation:

4 0
2 years ago
Read 2 more answers
One event occurs at the origin at t equal to zero, and a second events occurs at the point x=5m along the x-axis at time with ct
Anastasy [175]

Answer:

  • The separation will be spacelike.
  • The first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

Explanation:

We can define the separation between two events (using the + - - - signature)  as :

(\Delta s )^2  = (ct_2 - c t_1 )^2 - (x_2 - x_1)^2

where the separation will be lightlike if is equal to zero, timelike if is positive and spacelike if is negative.

For our problem

c t_1 = 0

x_1 = 0

ct_2 = 4 \ m

x_2 = 5 \ m

(\Delta s )^2  = (4 \ m - 0 )^2 - ( 5 \ m - 0)^2

(\Delta s )^2  = (4 \ m )^2 - ( 5 \ m 0)^2

(\Delta s )^2  = 16 \ m^ 2 - 25 \ m^2

(\Delta s )^2  = - 9\ m^2

So the separation will be spacelike, and the first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

8 0
3 years ago
An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os
aniked [119]

Answer:

(a)  T = 0.412s

(b)  f = 2.42Hz

(c)  w = 15.25 rad/s

(d)  k = 86.75N/m

(e)  vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz

The frequency is 2.42 Hz

(c) The angular frequency is:

\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}

The spring constant is 86.75N/m

(e) The maximum speed is:

v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

F=kA=(86.75N/m)(0.2m)=17.35N

The maximum force that the spring exerts on the block is 17.35N

3 0
3 years ago
What is position?<br> the definition
inn [45]

Answer:

a place where someone or something is located or has been put.

4 0
2 years ago
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