All categories

3 years ago

Name and explain all the laws made by Newton on motion

Physics

2 answers:

Yuki888 [10]3 years ago

6 0

Answer:

first; An object will continue in it's State of rest or uniform motion unless being acted upon by an external Force

second; The force acting on a body is directly proportional to the product of the mass and acceleration.

third; all actions will have an equal and corresponding reaction.

Explanation:

are you okay with that?

Maurinko [17]3 years ago

5 0

Newton's First Law:-

It states that Every Body continues to be in it states of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.

Newton's second law:-

The rate of change of momentum of a body is directly proportional to the applied force and it takes place in the direction in which the force acts .

Mathematically

$\\ \ast\sf\hookrightarrow F=ma$

Newton's Third Law:-

To every action there is always an equal and opposite reaction.

$\\ \ast\sf\hookrightarrow F_{AB}=-F_{BA}$

You might be interested in

As an object moves from point A to point B only two forces act onit: one force is nonconservative and does -30 J of work, the ot

Romashka-Z-Leto [24]

To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

This leads to the direct conclusion that the resulting energy is 20J.

The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.

Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.

7 0

3 years ago

What is a trip valve

AnnyKZ [126]

Pressure‐sensing trip valves are for control applications where a specific valve/actuator action is required when supply pressure falls below a specific point.

7 0

4 years ago

Transition

Snezhnost [94]

Frozen water has move volume than water in liquid form

8 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

3. A Japanese bullet train slows from a speed of 50 m/sec to a speed of 40 m/sec in 500m.

Triss [41]

Answer:

-0.9 m/s²

Explanation:

Given:

Δx = 500 m

v₀ = 50 m/s

v = 40 m/s

Find: a

v² = v₀² + 2aΔx

(40 m/s)² = (50 m/s)² + 2a (500 m)

a = -0.9 m/s²

6 0

4 years ago

Name and explain all the laws made by Newton on motion ​

Name and explain all the laws made by Newton on motion