Miriam Said:

Can someone help me with these physics problems?

We Answered:

Remember your basic right-triangle trigonometry. Sine = opposite over hypotenuse and cosine = adjacent over hypotenuse. If you make a sketch of each vector (always a good idea), you'll see that

1) the hypotenuse is 120 units. The horizontal component is the adjacent leg to the 30 degree angle, so

horizontal = 120 cos(30)

The vertical component is the opposite, so

vertical = 120 sin(30)

You can use a calculator for those.

2) Same problem, except the hypotenuse is 10 units and the angle is 48 degrees.

Tracy Said:

Can someone help with a couple physics problems please?

We Answered:

Here's my try:

1993M1.-

k= 400 N/m
mC= 4.0 kg
mD= 2.0 kg
x= 0.50 m
μ= 0.40

a)
* Since Elastic Energy is a Potential Energy that is determined due to the position of masses, the Elastic Energy stored in the system by the spring is determined by the distance the spring is compressed. The equaiton for Elastic Energy is Us = (1/2)kx^2.
So:

Us= (1/2)kx^2
Us= (1/2) 400 N/m * ( 0.50 m )^2.
Us= 50 J

b)
When the spring is decompressed, al the Elastic Energy stored on it, becomes Kinetic Energy done by the mass of block C. But, since there's friction between the surfaces, there will be another Work done on the block C, the Work done by friction. By definition Work equals Force times Distance, and the Force done by Friciton equals Normal Force times the Coefficient of Kinetic Friction. This will explain it more clearly:

W= F*d
F= m*g*μ
so
W= m*g*μ*d
substituting
W= 4.0 kg * 9.8 m/s^2 * 0.40 * 0.50 m
W= 7.84 J

Then we sum both of the works to get the Total Work done (since the Work done by Friction opposes original motion, it is substracted):
Wt = 50 J - 7.84 J
Wt = 42.16 J

By knowing this would be Kinetic Energy, we can say that
42.16 J = (1/2)mv^2

Substituting values
42.16 J = (1/2) * 4.0 kg * v^2

Clearing for V we get
v= (2 * 42.16 J / 4.0 kg)^.5

v= 4.59 m/s

That's the velocity just before colliding with block D, but i still don't know why they tell you they collide, if they don't ask you anything, wierd stuff.

Cassandra Said:

How do you know when to use potential energy and when to use kinetic energy when solving physics problems?

We Answered:

potential energy = mgh ( m= mass, g= acceleration due to gravity and h = height from ground).

kinetic energy = 1/2 m v^2( m= mass, v= velocity)

if an object is in motion , it has kinetic energy.
on the other hand in free fall , it has potential energy.
any way the sum total of the energy possesed by an object an any point of time is the sum total of both KE
and PE.so u can use either formula .provided the necessary data is provided.

Ethel Said:

Please can someone help me with these physics problems?

We Answered:

The first one is an excellent, and tricky question!
So, i'll cut straight to the chase - A satellite in orbit. The earth exerts a gravitational force on the satellite, but the earth does no work on it. Thought W = FD, and a satellite does orbit (D) and have a Force due to gravity (F), work CAN NEVER been done perpendicular to the gravity well.

Second question, the PE would double. PE = mgh, if it were doubled then it would be PE = 2hmg or, to make it clearer, 2(mgh). Doubled.

Final question:
Impulse is defended as a change in momentum. Momentum can't hurt us, we can have a ton (going really fast) but we'll be fine until the momentum suddenly changes (impulse). Padded Dashboards increase the time of impact, thus, reducing the force form the resulting collision and this, reducing the change in momentum (impulse) of the collision (your face hitting the dashboard).

Gilbert Said:

Is there a good website out there to help solve physics problems?

We Answered:

can computers really do that?!
when i need physics help i just google the problem, and sometimes i can actually find the exact answer.

but if there IS a site like that, it would be totally awesome!