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Oksi-84 [34.3K]
2 years ago
9

A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that

the ratio of the string tension at the top of the circle to that at the bottom of the circle is FtopT/FbotT = 0.5. Derive an expression for the speed v.
Physics
1 answer:
adell [148]2 years ago
7 0

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

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The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0
2 years ago
In a charge-free region of space, a closed container is placed in an electric field. Which of the following is a requirement for
galina1969 [7]

Answer:

D. The requirement does not exist -the total electric flux is zero no matter what.

Explanation:

According to Gauss's law , total electric flux over a closed surface is equal to 1 / ε₀ times charge inside.

If charge inside is zero , total electric flux over a closed surface is equal to

zero . It has nothing to do with whether external field is uniform or not. For any external field , lines entering surface will be equal to flux going out.

8 0
2 years ago
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming
Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

tan \ \theta = \dfrac{12}{100} \\ \\  tan \ \theta = 0.12 \\ \\  \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0

The drag force D = Wsinθ

\dfrac{1}{2}C_v \rho AV^2 = W sin \theta

where;

\rho = 1.23 \ kg/m^3

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}

C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}

C_D =0.9378

8 0
2 years ago
An object of mass 100 kg is initially at rest on a horizontal frictionless surface. At time t = 0, a horizontal force of 10 N is
satela [25.4K]

Answer:

(D) It is moving at a constant speed

Explanation:

Before t = 1s. Due to the force, albeit small, acting on the object, since there's no static friction stopping the object from moving, this mass object would have a constant acceleration and it's velocity would be increasing.

According to Newton's 1st law, an object will stay at a constant speed if the net force acting on it is 0. After t = 1s, horizontally speaking there's no other force exerting on the mass object. There is no friction force at play here as the surface is frictionless.

Therefore the correct statement is (D) It is moving at a constant speed

8 0
2 years ago
A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. What is the acceleration of the ball in the vertical
blagie [28]

The acceleration of the ball in the vertical direction is 9.8 m/s^2, downward.

Explanation:

This is a typical example of projectile motion, which consists of two independent motions:

- A uniform horizontal motion at constant velocity (since there are no forces in the horizontal direction)

- A vertical accelerated motion at constant acceleration (due to the presence of the force of gravity)

We are considering now the vertical motion only. There is only one force acting on the ball in this direction: the force of gravity, of magnitude

F = mg

where m is the mass of the ball and g=9.8 m/s^2 the acceleration due to gravity, downward. This means that the acceleration of the ball in this direction is (using Newton's second law)

a=\frac{F}{m}=\frac{mg}{m}=g = 9.8 m/s^2

Therefore, the acceleration of the ball in the vertical direction is 9.8 m/s^2, downward.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
1 year ago
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