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AnnyKZ [126]
2 years ago
5

How does the radius of a string affect centripetal force.

Physics
1 answer:
KiRa [710]2 years ago
5 0

Answer:

because a raduis is half of 25% of a cicrle.

Explanation:

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If you double the frequency of a vibrating object, its period:
Furkat [3]

Answer:

becomes halved.

Explanation:

trust

8 0
3 years ago
Read 2 more answers
A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro
julia-pushkina [17]

Answer:

a) s \approx 6.676\,m

Explanation:

a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:

s = \frac{\pi}{4}\cdot R

s=\frac{\pi}{4}\cdot (8.5\,m)

s \approx 6.676\,m

The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

U_{g,A} = K_{B} + W_{fr}

m\cdot g \cdot h_{A} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + f\cdot s

The average frictional force is cleared within the expression:

f = \frac{m\cdot (g\cdot h_{A}-\frac{1}{2}\cdot v_{B}^{2})}{s}

f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}

f = 34.684\,N

4 0
3 years ago
How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


4 0
3 years ago
A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W
labwork [276]

Answer:

<h2>650W/m²</h2>

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

4 0
3 years ago
What is an example of strong nuclear force
Gnom [1K]
Dropping a nuke on another country.
3 0
3 years ago
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