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jarptica [38.1K]
3 years ago
5

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

hout slipping, how does its linear speed compare to its rotational speed? v = Rω v = ω/R They are unrelated. v = ω
Physics
1 answer:
salantis [7]3 years ago
6 0

Answer:

v = r\omega

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r

v = r\omega

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7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr
Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m

d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m

From the continuity equations in pipes we have to

A_1V_1 = A_2 V_2

Where,

A_{1,2} = Cross sectional Area at each section

V_{1,2} = Flow Velocity at each section

Then replacing we have,

(\pi r_1^2) v_1 = (\pi r_2^2) v_2

(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2

v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1

From Bernoulli equation we have that the change in the pressure is

\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)

7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)

7300= 8919.01 v_1^2

v_1 = 0.9m/s

Therefore the speed of flow in the first tube is 0.9m/s

6 0
4 years ago
A pendulum is swinging. It swings 85 complete swings and this takes 102 seconfs. What is the frequency? A. 0.83 Hz B.17Hz C.187
Serga [27]

Answer:

A: 0.83 Hz

Explanation:

Frequency can be calculated in a multitude of ways. The one way that is going to help you solve this problem is (# of times/seconds)

so you would divide 85 swings/102 seconds

= 5/6 or 0.8333 Hz

so your answer is A

3 0
3 years ago
D) A tank with dimensions: 3m x 2m x 2m contains water upto its half
lorasvet [3.4K]

Answer:

P = 9800 [Pa]

Explanation:

In order to calculate the pressure at the bottom, we must use the following formula.

P = Ro*g*h

where:

P = pressure [Pa] (units of pascals)

Ro = density of the water = 1000 [kg/m³]

g = gravity acceleration = 9.8 [m/s²]

h = height = 1 [m] (because its half of the portion, the full height is 2 m)

P = 1000*9.8*1

P = 9800 [Pa]

5 0
3 years ago
PLEASE HELP I WILL GOVE BRAINLIEST TO FIRST CORRECT ANSWER!!!!!
Dmitrij [34]
Answer is C is the correct answer
8 0
4 years ago
A wheel of diameter 78 cm has an axle of diameter 14.8 cm. A force 150 N is exerted along the rim of the wheel.What force should
bezimeni [28]

Answer:

F = 263.51 N

Explanation:

given,

diameter of wheel = 78 cm

diameter of axle = 14.8 cm

Force exerted on the rim of wheel = 150 N

Force applied outside the axle = ?

To prevent rotation wheel from rotating the Force 'F' should be applied outside of the axle.

Net momentum about the center of mass should be zero

now,

Moment of about center due to 150 N = moment about center due to F on axle

50\times \dfrac{78}{2}=F\times \dfrac{14.8}{2}

           7.4 F = 1950

                F = 263.51 N

Hence, Force exerted outside of the axle in order to prevent the wheel from rotating is equal to 263.51 N.

6 0
4 years ago
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