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max2010maxim [7]
3 years ago
11

A 4 kg bowling boll sliding to the right at 8 m/s has elastic head-on collision with another 4K bowling ball initially at rest.

The first bus stops after collision
A. Find the velocity of the second ball after the collision

B. Verifier answered by calculating the total kinetic energy before and after the collision
Physics
1 answer:
MAXImum [283]3 years ago
6 0
The energy would be transferred from the objects. Also do not forget, add direction to your answer.
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Noah Formula is in an airplane which is flying at constant speed in a circular course of radius 6500 m, circling O’Hare‘s airpla
algol [13]

Answer: this one looks hard

Explanation:

3 0
4 years ago
A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of
WINSTONCH [101]

The correct answer to the question is : A) The velocity of the cart after it hits the wall.

EXPLANATION:

Before answering this question, first we have to understand impulse.

Impulse of a body is defined as the change in momentum or the product of force with time.

Mathematically impulse = m ( v- u ).

Here, v is the final momentum and u is the initial momentum.

Hence, we need the velocity of the cart after it hits the wall in order to calculate the impulse of the lab cart.

4 0
3 years ago
Read 2 more answers
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
elena-s [515]

Answer:

g=13.42\frac{m}{s^2}

Explanation:

1) Notation and info given

\rho_{center}=13000 \frac{kg}{m^3} represent the density at the center of the planet

\rho_{surface}=2100 \frac{kg}{m^3} represent the densisty at the surface of the planet

r represent the radius

r_{earth}=6.371x10^{6}m represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of  the radius

r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}

r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}

So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

So then our linear model would be

\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be dV=r^2 sin\theta d\phi d\theta dr.

And the total mass would be given by the following integral

M=\int \rho (r) dV

Replacing dV we have the following result:

M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

And replacing the values we got:

M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

And now that for any shape the gravitational acceleration is given by:

g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}

4 0
3 years ago
A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g
mr Goodwill [35]

The distance an object falls from rest through gravity is

                         D  =  (1/2) (g) (t²)

            Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if  ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.

                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t² 

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)


Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes: 

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
                                             and smaller 'g' ==> longer 't' .

-- They don't change by the same factor, because  1/g  is inside
the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value
of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
4 0
4 years ago
Read 2 more answers
Please someone answer this ASAP❗️❗️‼️‼️
kogti [31]
Starting making jokes and rapping
5 0
3 years ago
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