What is the mass of a bicycle moving at 8ms with ke of 384j ?

Why does a rocket have such great momentum even if it is moving at a slow speed?

Artist 52 [7]

Answer:

Rockets provide a wonderful example of Momentum Conservation. As momentum in one direction is given to the rocket's exhaust gases, momentum in the other direction is given to the rocket itself.

Explanation:

First, think of two masses connected by a lightweight (massless!) compressed spring. When the two spring apart, conservation of momentum tells us the Center of Mass remains where it was (or moving as it was).

PTot,i = p1i + p2i = 0 + 0 = 0

PTot,f = p1f + p2f = PTot,i = 0

p1f + p2f = - m1 v1f + m2 v2f = 0

4 0

3 years ago

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A force of 58 newtons is applied to a crate at an angle of 32° with the horizontal. What is the value of force acting in the y–d

Mumz [18]

90 degrees would be the answer. hope this helps

5 0

3 years ago

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A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (6)

Where:

$a=-4.9 m/s^{2}$

$b=48.1 m/s$

$c=1.5 m$

Substituting the known values:

$t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}$ (7)

Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

3 0

3 years ago

Two friends, Joe and Sam, go to the gym together to strength train. They decide to start off with an exercise called flat bench

astra-53 [7]

Answer:

216.31 (the work done by gravity is -216.31) positive for going up.

Explanation:

We look at this question first by getting the right equation for work.

Which should be... W = F x D.

From this, we can do everything, we need the Force (F) first - the question tells us that Joe is lying on his back and moves his arms upward to raise the barbell. This means that he is countering the force of graving on the object.

What is the formula for the force of gravity on an object near the earth?

Right here --- $F_{grav}$ = mg

m = the mass and...

g = the acceleration due to gravity which is 9.81 m/s2

Before we plug things in though, we need to convert everything to SI units,

the weight is in kg - so we're good to go there, but the length of Joe's arms are in "cm" we need m or meters. Converting 70 cm to m = .7 m.

Now, we just put it all together - (31.5kg)(9.81m/s2)(.7m) = 216.31 J or 216.31 N m.

3 0

3 years ago

A cannon fires a 40.5kg shell toward a target and the shell moves with a velocity of 120 m/s. Calculate the shells momentum

kvv77 [185]

Answer:

4860 kg m/s

Explanation:

P = mv

P = 40.5 x 120

P = 4860 kg m/s

8 0

2 years ago