3. How can we determine the volume of a regular object?

1. When in the past have you pushed your personal limits? Give at least one

Yuri [45]

Answer:

Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.

Explanation:

4 0

3 years ago

What is a tool that can be used to measure the size of a force

Travka [436]

The answer to this question would be: a spring scale.

The spring scale that you use to determine your body weight is actually a device that measures your body gravitational force. The force itself influenced by your body weight, that is why it can determine your body weight.
More weight means more force, more force will shrink the spring more.

7 0

3 years ago

Need Help with this physics question, just a written response

Nadusha1986 [10]

Answer:

Explanation:

According to Newton's third law, every action has an equal and opposite reaction

so it tells us that the force exerted by the earth on the spacecraft is equal to the force exerted by the spacecraft on the earth. But we do not see the earth moving towards the spacecraft because the inertia of the spacecraft is very less than the inertia of the earth.

3 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

Joan stirred a pot of hot soup with a metal spoon. after a minute or two, the handle of the spoon became too hot to hold. what k

natima [27]

When the metal spoon comes in contact with the hot soup, heat transfer from hot soup to the spoon because of difference in temperature of the two. The kind of heat transfer represented here is conduction. In conduction, heat is transferred from a region of higher temperature to a region of lower temperature when there is a physical contact between the two.

4 0

3 years ago