Physics Problem Desperately Need Help

Physics problems- desperately need help?

I need all the help I can get!

1. A .20kg billiard ball traveling at a speed of 15 m/s strikes the side rail of a pool table at an angle of 60 degrees. If the ball rebounds at the same speed and angle, what is the change in its momentum?

2. A 2.0kg mud ball drops from rest at a height of 15m. If the impact between the ball and the ground lasts .50s, what is the average net force exerted by the ball on the ground?

3. A 1.0kg ball is through horizontally with a velocity of 15 m/s against a wall. If the ball rebounds horizontally with a velocity of 13 m/s and the contact time is .020s, what is the force exerted on the ball by the wall?

4. If the billiard ball from question 1 is in contact with the rail for .010s, what is the magnitude of the average for exerted on the ball?

5. To get off a frozen, frictionless lake, a 70.0kg person taes off a .150kg shoe and throws is horizontally, directly away from the shore with a speed of 2.00 m/s. If the person is 5.00m from the shore, how long does it take for him to reach it?

6. A projectile that is fired from a gun has an initial velocity of 90.0km/h at an agle of 60.0 degrees above the horizontal. When the projectile is at the top of its trajectory, an internal explosion causes it to separate into two fragements of equal mass. One of the fragments falls straight downward as though it had been released from rest. How far from the gun does the other fragment land?

7. In a pool game, a cue ball traveling at .75m/s hits the stationary eight ball. The eight ball moves off with a velocity of .25m/s at an angle of 37 degrees relative to the cue ball's initial direction. Assuming that the collision is inelastic, at what angle will the cue ball be deflected and what will be its speed?

8. Show that the fraction of kinetic energy lost in a ballistic-pendulum collison is equal to M/(m+M).

Please, I desperately need help with this physics problem. I have been stuck for an hour already....?

Merry-Go-Round. Two identical twins, Jena and Jackie, are playing one December on a large merry-go-round (a disk mounted parallel to the ground on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has a mass of m. The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting a distance r_1 from the center of the merry-go-round, must hold on to one of the metal posts attached to the merry-go-round with a horizontal force of F to keep from sliding off. Jackie is sitting at the edge, a distance r_2 from the center.


1)With what horizontal force must Jackie hold on to keep from falling off?

2)If Jackie falls off, what will be her horizontal velocity when she becomes airborne?

ok for the first one i put (m*v^2)/r_2 but its wrong because u cant use m and v variables. You haave to use r_1 and r_2 same for the second one..

Very urgent physics problems I desperately need help with!?

1. A lake contains approximately 2.8 × 105
kg of water.
What is its heat capacity? The specific heat of water is 1000 cal/kg ·◦ C.
Answer in units of cal/degC.

2.The specific heat of a metal (similar to copper) is 0.092 cal/g ·◦ C. The latent heat of vaporization of a liquid (similar to liquid nitrogen) is 48 cal/g. A 2 kg block of the metal at 18◦ C is dropped into a large vessel of the liquid at 75 K, which is the boiling point of the liquid.
How many kilograms of the liquid boil away by the time the metal reaches 75 K?
Answer in units of kg

3. If the molecular weight of air is 28.9,
what is the density of air at atmospheric
pressure and a temperature of 335.1 K?
1 atm = 1.013 × 105 N/m ,the mass of
a proton is 1.67262 × 10−27 kg, and k = 1.38065 × 10−23 N · m/K .
Answer in units of kg/m

4. Compressing gases requires work and the resulting energy is usually converted to heat; if this heat does not escape, the gas’s temperature will rise. This effect is used in diesel engines: The compressed air gets so hot that when the fuel is injected, it ignites without
any spark plugs.
As an example, consider a cylinder in a
diesel engine in which air is compressed to one twentieth of its original volume while the pressure rises from 1 atm to 62 atm (absolute, not gauge). Note that because the air heats up while being compressed, its pressure rises more than twenty-fold.
If the air is taken into the cylinder at 21◦C,
how hot does it get after being compressed? Answer in units of ◦C.

Physics problem? Please I desperately need help?

You are in a space shuttle above earth, when you find out a 20,000 kg asteroid 12,000 km from earth’s surface is heading straight for a collision with the planet at 500 m/s. Your only chance to save earth is to fly your ; 5000 kg shuttle into the asteroid to knock it off course so it misses earth. If you were to impact the asteroid 45 degrees off of a head-on collision, how fast do you need to be going to be able to save earth? If you can figure this out, you’ll surely go down in flames, but at least you’ll go down in the history books too! (hint: remember that a tangent line touches a circle at a right angle)

PHYSICS MIDTERM EXAM TOMORROW!!! I desperately need help with a position/time problem.?

Let

Xa = distance that car A has travelled (from t = 0.50 hr) when car B passes it

Xb = total distance that car B will travel before passing car A

T = time elapsed (after t = 0.5) when car B will pass car A

For car A, Xa = 31T --- call this Equation 1

For car B, Xb = 53T

Xb = 48 + 0.5(31) + Xa

and the above becomes

63.5 + Xa = 53T

Solving for Xa,

Xa = 53T - 63.5 and since this is equal to Equation 1,

53T - 63.5 = 31T

53T - 31T = 63.5

22T = 63.5

T = 2.9

Reading on the watch when car B passes car A = 0.50 + 2.9 = 3.4

Hope this helps.

Desperately need help with this physics problem!!! i will give high ratings!!! Please help!!?

OK, give "best answer".

The moment of inertia of a particle with respect to an axis is just the product

I=mXr^2

where r is the distance to the axis. It is also important to know that moments of inertia are additive.

The distances (squared) to the origin for each mass are (use Pythagorean thm):

r1²=1+1.5²=3.25
r2²=9+2.25=11.25
r3²=9+1=10
r4²=.25+.25=.5

so the corresponding I's are:

I1=18X3.25=58.5
I2=27X11.25=303.75
I3=9X10=90
I4=37X.5=18.5

now add all to get

I=58.5+303.75+90+18.5=470.75 this is answer to (a).
======================================...
Now (b) To find the position with respect to point P all you have to do is to substract
the point (-2,2) from each point. The distance squared is going to be the sum of the squares of each number, i.e
(1,1.5)-(-2,2)=(3,.5) so distance squared to P for M1 is just:
9+(.5)²=9.25.

You do the same for the other trhee points, multiply by the masses and add.

Desperately need help with this physics problem please!?

normal force Fn = weight - vertical component of rope tension
Fn = 28.5kg * 9.8m/s² - T*sin22.1º = 279N - 0.376T
Since it moves at constant speed,
horizontal component of rope tension = friction force, or
Tcos22.1º = µFn = 0.3 * (279N - 0.376T)
0.927T = 83.7N - 0.113T
1.04T = 83.7N
T = 80.5 N

Tcos22.1º = 74.6 N → Force in direction of motion
Work done = F * d = 74.6N * 30.8m = 2300 J = 2.3 kJ ←

Physics Question. Desperately need help.?

This is how I worked it out:
1. From W = V²/R, find the equivalent resistance of network.

Req = 45²/44 = 46.0227 Ω.

2. Reduce the network to its equivalent resistance:
Knowing that two equal resistors R in parallel offer an equivalent resistance of ½ R, resulting resistance when 2 R is connected in parallel with this, is ½ R × 2 R/(2 R + ½ R) = R/2.5 = 0.4 R.
Total equivalent resistance is then 1.4 R.

3. Equate both results, then solve for R.
R = 46.0227 Ω/1.4 = 32.8734 Ω.

As a check, you may compute total equivalent resistance again, in terms of its numerical value. This should yield 46.0227 Ω.

Physics Elastic Collision Question - Desperately need help?

good news, this is a classic plug-n-chug physics problem (as you'll find 99% of physics problems are, the trick is just remembering what the equations go to and the more you understand the equations the more connections with other values given ultimately allowing you to solve more with less).

(a) anyways, this is considered a 1-D elastic collision because the balls' paths lie on the same line (if you lay a grid on the table neither ball will leave the x-axis whereas 2-D collisions there is are angles in their reactions). with the information given i believe you can assume

m1v1f=0 (v1f=0)
m2v2i=0 (v2i=0)
we can assume the unknowns to be zero because this is an elastic collision and not an inelastic collision (perfectly elastic collisions mean that no kinetic energy is lost in the system, when the collision occurs all the energy [for now] is transferred to the other ball -- in inelastic collisions energy is transferred into a different form of energy) so with those values labeled as zero our KE1=KE2 equation comes out as m1vi1=m2v2f . here is a trick for the velocities -- we are given that (0.5)v2=v1 so we can mark them as (0.5)m/s=v2 and 1m/s=v1. You could choose any numbers, as long as they maintain the same proportion (go ahead and plug in different values, you'll say that you'll get the same answer as they have the same proportion). so 0.34kg*1m/s=m2*(0.5m/s). isolate and solve for m2.

(b) this one is just playing with the equation. deltaKE is the change in KE, and for our purposes we will consider dKE=(m2vf2-m1vi1). i say for our purposes because after you get through this unit you will see dKE = (1/2)m(vf^2-vi^2). and now that i read it fully i see it's just asking for KE1/KE2. (i left the above text because it may be useful for you to know what you'll probably see next week). so anyways KE1=m1v1i and KE2=m2v2f . plug the numbers into KE1/KE2 and you'll get 1. Like i said earlier, this is a perfectly elastic collision where no kinetic energy is converted to other forms of energy, so the fraction will be 0.34/0.34 or 1/1 or 100% for 100% transfer.

sorry if that was long winded, i wanted to give you the answer, not just the numbers.