Difficult Physics Question

Difficult Physics question?

Before 1960 they had long known that a coefficient greater than 1 was easily obtainable under acceleration. The effect is WEIGHT SHIFT, and has nothing to do with tyre design improvements other than the greater "stickiness" of the modern compounds + surface treatments. Without weight shift, no tyre can possibly have a coefficient greater than 1.
The "front wheels off the ground" is the indicator of weight shift to the rear wheels, short of actually flipping the car.
1/4 mile = 0.25(1609.3) = 402.325 metres.
Acceleration = (2d/t^2) = (804.65/5.16^2) = 30.221m/sec^2.
Accelerating force = (ma) = (1kg x 30.221) = 30.221N. per kg. mass.
(30.221/g) = µ of 3.08. (9.8 for g).

Difficult physics question?

you are going to have to change the numbers and calculate, but here is an answer with a readius of 100 km and a height of 4 km:

Hurricanes can involves winds in excess of 120 km/h. make crude estimate of (a) the energy, (b) angular momentum. Approximate the hurricane as a rigidly uniform cylinder of air (density 1.3 kg/m3) of radius 100 km and height 4.0 km.


The first thing to do is to estimate the mass of air involved in the hurricane.

Mass = density×volume of cylinder

= (1.3 kg/m3)(pr2h)

= (1.3 kg/m3)(p(100,000 m)2(4000 m)

= 1.63×1014 kg.

For a rigid cylinder the moment of inertia about the center is

I = ½ MR2

= ½ (1.63×1014 kg)(100,000 m)2

= 8.17×1023 kg×m2

a. The energy is in the form of rotational KE.

KEr = ½ Iw2 where w = v/r = 33.3 m/s/100,000m = 3.33×10-4 rad/s since v = 120 km/h = 33. m/s.

Thus KEr = ½ (8.17×1023 kg×m2)(3.33×10-4 rad/s)2

= 4.53×1016 J


b. The angular momentum

L = Iw = (8.17×1023 kg×m2)(3.33×10-4 rad/s)

= 2.72×1020 kg×m2/s

Note: To put the rotational KE (4.5×1016 J) in prospective, a 900 MW power plant produces ~2.8×1016 J in 1 year. Thus a hurricane has more energy than a power plant produces in a whole year. A 900 MW power plant can supply the power needs of twice the population of Tallahassee.

Difficult physics question?

This question has to be solved in two parts. The first part, we consider only the collision between M and m and we use the principle of conservation of momentum to find the velocity, u which both the masses embedded together will have immediately after the collision.

So, Total momentum before collision = Total momentum after collision
mv = (m+M)u
u = mv/(m+M)

Now we come to the second part, here we use the principle of conservation of energy. So the total initial kinetic energy of both the masses embedded togather should be equal to the potential energy of both the masses at the peak of the arc.(which will be twice length,l)

1/2(m+M)u^2 = (m+M)g2l
now we substitute the value of u from the first part into this second equation.
1/2(m+M)[mv/(m+M)]^2 = 2(m+M)gl
m^2v^2/(m+M) = 4(m+M)gl
v^2 = 4gl(m+M)^2/m^2
v= [2(m+M)√gl]/m

Difficult Physics Question?

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g=30 m/s2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.)
It will smash his face.
It will take more time to return to the point from which it was released.
It will take less time to return to the point from which it was released.
Its mass will be greater.
It will stop well short of his face.

Difficult Physics Question?

Physics: Physics and Engineer Question?
Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with person-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle θ with the vertical as shown in the figure below. Take D = 55.0 m, F(vec) = 109 N, L = 40.0 m, and θ = 50.0°.

http://www.webassign.net/pse/p8-71.gif

(a) With what minimum speed must Jane begin her swing to just make it to the other side? (Hint: First determine the potential energy associated with the wind force.) If Jane can make it across with zero initial velocity, enter 0.
???? m/s

(b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume that Tarzan has a mass of 80.0 kg.
???? m/s

Difficult Physics Question?

in the first case, the speed of the red car is 7.78m/s; this means it takes 43.8m/7.78m/s = 5.63s for the two cars to pass

since we know xg = 218m, we know the green car covered a distance of 218m-43.8m = 174.2 m in 5.63s, or by using the distance equation:

dist = v0 t + 1/2 a t^2

we get:

174.2 = 5.63v0 + 1/2 a (5.63)^2

in the second case the time it takes to meet is 76m/15.6m/s = 4.88s, so the green car covers a distance of (218 - 76)m in 4.88s or:

142 = 4.88v0 + 1/2 a (4.88)^2

these two equations become:

174.2 = 5.63v0 + 15.8a and
142 = 4.88v0 + 11.9a

this is a system of two equations in 2 unknowns, solving them simultaneously with standard techniques gives:

v0=16.9m/s
a = 5m/s/s

check: in the first case, the green car travels a distance

16.9m/s x 5.63s + 1/2 x 5m/s/s x (5.63s)^2 = 174.2m

in the second case:

16.9m/s x 4.88s + 1/2 x 5m/s/s x(4.88s)^2 = 142m

Difficult physics question?

John is a fairly common name. If you are the same John who asked this a couple days ago, but found my answer (which I'll paste below) confusing, let me know what, in particular, confuses you, and I'll see if I can clarify it.

"You know that the intensity (power per unit area) is the average value of the Poynting Vector

S = (1/μ_0) E × B

I = = (c/μ_0) (Erms)²

where I've used the fact that E = cB in an electromagnetic wave, and the average value of sin²θ = ½.

If you want the energy, multiply the intensity by the area and the time

I A t = A t (c/μ_0) (Erms)²
"

****************
Hmmm ... well, let's see what I get. Wait! Oh, poo. When I used E = cB, I should have gotten

I = = [1/ (c μ_0)] (Erms)²

that ought to make the value a lot smaller!

A t [1/ (c μ_0)] (Erms)² = (0.650 m²)(30 s)(2.60×10^−2 V/m)² / [ (2.998×10^8 m/s)(4π×10^-7 T∙m/A) ] = 35 μJ

which is a lot more reasonable. I apologize for my error (I should have checked my dimensions).

Difficult Physics Questions?

(1)
Let s be the the distance to the stone.
the time the sound need to travel through air to the person is
s = c_air · t_air
<=>
t_air = s/c_air
the time the sound need to travel through air to the person is
t_concrete = s/c_concrete.

The time difference is
Δt = t_air - t_concrete = s/c_air - s/c_concrete

Therefore
s = Δt / (1/c_air - 1/c_concrete)
= 1.07s / ( 1/344m/s - 1/2950m/s)
= 416.67m


(2)
Sound intensity level is defined as:
L = 10 · log( I / I₀ )
=>
I = I₀ · 10^(L/10)

The energy absorbed per unit time is
P = I · A
= I₀ · 10^(L/10) · A
= 1.0×10-12 W/m² · 10^(50/10) · 5.8×10-5 m²
= 5.8×10-12 W
= 5.8pJ/s

Difficult physics questions?

(a) An object in simple harmonic motion has amplitude 12 cm and frequency 3 Hz. At time t = 0 s, it passes through the equilibrium point moving to the right. Write the function x(t) that describes the object's position. The answer will be in the form m * cos((n*t) + p)where m, n and p are real numbers.

I have m = 12.0 cm. and n = 18.8 radians but for some reason I'm having trouble getting the phase constant, p. Any help would be appreciated.

(b) A 232.0 g mass attached to a horizontal spring oscillates at a frequency of 2.56 Hz. At t = 0 s , the mass is at x = 2.67 cm and has a velocity of -40 cm/s. Determine the position (in cm) at t = 3.81 s.

I already have the following information:
period = 0.391 s
angular frequency = 16.1 rad/s
amplitude = 3.65 cm
phase constant = 0.750 rad
max velocity = 58.7 cm/s
max acceleration = 944 cm/s/s
total energy = 0.0400 J

I just cannot determine this last part (the position at t = 3.81 s) I've tried using the formula (x(t) = A cos(omega*t + phi) where omega is angular frequency and phi is phase constant but the answer is incorrect. Any help on either of these two questions would be very nice.