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Vitek1552 [10]
3 years ago
11

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass MMM and radius RRR about an axis perpendicular to th

e hoop's plane at an edge.
Physics
1 answer:
slega [8]3 years ago
3 0

Answer:

I = sum m *r^2    where m represents the (small) individual masses and r is the distance of that mass from center of rotation

Note:  sum m = M

For the hoop given all masses are at a distance RRR from the center of rotation

I = MMM * RRR^2

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A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s
Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

3 0
3 years ago
A published hypothesis:
kow [346]

Answer:

should be tested by the scientific community

4 0
3 years ago
A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

6 0
3 years ago
Scientific knowledge
emmasim [6.3K]

Answer:

D

Explanation:

As we progress and learn more, scientists make new discoveries that can contradict earlier discoveries, and since are technology is better today we are able to discover a lot more to add to or change our previous theories.

6 0
4 years ago
Read 2 more answers
A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta
vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

N_1 = 160

N_2 = 1400

\Delta V_{1rms} = 62V

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}

So the voltage for transformer 2 would be given by,

\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}

PART B) To express the number value we proceed to replace with the previously given values, that is to say

\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}

\Delta V_{2rms} = \frac{1400}{160} 62V

\Delta V_{2rms} = 1446.66V

7 0
3 years ago
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