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2 years ago

True or false. when objects collide , some momentum is lost

Physics

2 answers:

ycow [4]2 years ago

5 0

Answer:

It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

omeli [17]2 years ago

5 0

Answer: True

Explanation:

You might be interested in

Which unit is equivalent to j/s

mrs_skeptik [129]

To be honest, there's no sure way to answer that, because you haven't defined your terms and we can't be sure of what j or s might be.

Tell you what I'll do:. I'll assume definitions for j and s, and then I'll answer the question that I invented.

Assume that j stands for Joule, the unit of energy. And assume that s stands for 'second', the unit of time.

Then j/s is the rate of transferring energy or doing work.

Its unit is the Watt, equivalent to 1 Joule per second.

In your system of notation, it would be 'w' .

4 0

3 years ago

Below is an "oracle" function. An oracle function is a function presented interactively. When you type in a t value, and press t

Bingel [31]

Answer:

The rate of change of height with respect to time is -10.64 feet/sec

Explanation:

Given that,

There are three lines, so you can calculate three different values of the function at one time.

The function f(t) represents the height in feet of a ball thrown into the air, t seconds after it has been thrown.

Given table is,

Time t = 0, 1, 1,02

Function is, $F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015$

We need to calculate the initial height of ball

Using equation of motion

$h=h_{0}+ut-\dfrac{1}{2}gt^2$

Where, h₀ = initial height

u = initial velocity

t = time

g = acceleration due to gravity

At t = 0,

Put the value into the formula

$-3.053113177191196\times10^{-18}=h_{0}+0-0$

$h_{0}=-3.053113177191196\times10^{-18}$

We need to calculate the height of ball at t = 1

Using equation of motion

$h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32$

$6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}$

$u_{0}=22\ feet/s$

Velocity is the rate of change of height with respect to time

So, velocity at 1.02 sec is given

We need to calculate the height

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

On differentiating w.r.to t

$h'(t)=u-\dfrac{1}{2}g(2t)$

Put the value into the formula

$h'(t)=22-\times32\times(1.02)$

$h'(t)=-10.64\ feet/sec$

Hence, The rate of change of height with respect to time is -10.64 feet/sec.

4 0

3 years ago

What is the limitation of relative dating?

astraxan [27]

Your not really supposed to date your relative

7 0

3 years ago

A ball rolls off the edge of a table with a fairly large horizontal velocity. Which of the following statements are true? (Selec

Degger [83]

Answer:

A.The vertical velocity is constantly increasing as the ball falls.

B.The horizontal velocity does not noticeably change as the ball falls.

G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.

H.The velocity vector of the ball changes as it travels through the air.

Explanation:

As the ball is projected horizontally so here the vertical component of the velocity is zero

So the time to reach the ground is given as

$H = \frac{1}{2} gt^2$

so we will have

$t = \sqrt{\frac{2H}{g}}$

so this is the same time as the ball is dropped from H height

Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.

So overall the velocity vector will change due to net acceleration g

4 0

3 years ago

Two force vectors are oriented such that the angle between their directions is 38 degrees and they have the same magnitude. If t

Usimov [2.4K]

Answer:

Sum of the forces will be equal to 3.479 N

Explanation:

We have given two same forces are oriented at an angle of 38°

Magnitude of each force is given $F_1=F_2=1.84$

We have to find the sum of the forces

Sum of the forces will be equal to

$F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\Theta }$

So sum of the forces will be equal to $F=\sqrt{1.84^2+1.84^2+2\times 1.84\times 1.84\times cos38^{\circ} }=3.479$

So sum of the force will be equal to 3.479 N

6 0

3 years ago