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3 years ago

What is the answer for number 10

Physics

1 answer:

Finger [1]3 years ago

3 0

The vertical component is = vsinx m/s

If you know the angle, substitute the value of x.

If you know the velocity at which it is moving, substitute it for v

Hope it helps :)

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A solid cylinder of mass 12.0 kgkg and radius 0.250 mm is free to rotate without friction around its central axis. If you do 75.

dybincka [34]

Answer:

The final angular velocity is 20rad/s

Explanation:

We are given;

mass, m = 12 kg

radius, r = 0.25 m

Work done;W = 75 J

Moment of inertia of cylinder, I = (1/2) mr²

Thus,

I = (1/2) x 12 x 0.25² = 0.375 kg.m²

Now, from work energy theorem,

Work done = Change in kinetic energy

So, W = KE_f - KE_i

Now, Initial Kinetic Energy (KE_i) = 0

Final Kinetic Energy; KE_f = (1/2)Iω²

So, KE_f = (1/2) x 0.375 x ω²

KE_f = 0.1875 ω²

Now, W = 75 J

Thus,

From, W = KE_f - KE_i, we have;

75 = 0.1875 ω² - 0

75 = 0.1875 ω²

ω² = 75/0.1875

ω² = 400

ω = √400

ω = 20 rad/s

5 0

3 years ago

A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of

Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

$K_1 = K_2 + P + W$

then, the work done by the engine is:

$W = K_1 - K_2 - P$

The formulas for the Kinetic and potential energy are:

$K=\frac{1}{2}mV^2\\P=mgh$

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

$W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh$

Replacing the values:

$W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J$

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

$t=\frac{3200m}{20m/s}=160s$