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fiasKO [112]
2 years ago
11

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

is doubled, what happens to the tension in the string?
Physics
1 answer:
Liono4ka [1.6K]2 years ago
8 0

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

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Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

6 0
2 years ago
Read 2 more answers
A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli
Katen [24]

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of kgm^{2 }\  / s.

  • In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
  • All the other option is incorrect because it is not follows the given scenario
3 0
1 year ago
Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 /  - 52.5 - 0 = 1 / f

f =  -52.5 cm

= -.525 m

Power P = 1 / f = -  1 / .525

= -  1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = -  52.5 cm

1 / v - 1 / u = 1 / f

  1 / v + 1 / 17.2 = -  1 / 52.5

1 / v  = - 1 / 17.2 -    1 / 52.5

= - .05813 -  .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0
2 years ago
Select the correct text in the passage. This paragraph attempts to explain the rain shadow effect, but it gets some of the facts
valentina_108 [34]

the sentence with leeward side and the sentence that has windward side both have errors.

5 0
2 years ago
Read 2 more answers
Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i
andrey2020 [161]

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field U=4\times 10^{-6}J

Value of inductance L =0.060 H

Energy stored in magnetic field is given by U=\frac{1}{2}Li^2

i=\sqrt{\frac{2U}{L}}

i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A

So the current flowing through rectangular toroid will be 0.01154 A

3 0
2 years ago
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