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KonstantinChe [14]

2 years ago

8

A boy weighing 445 N swings on a 2-m long swing. If his horizontal speed at the lowest point is 3 m/s, what total force must the

ropes holding the swing be able to withstand?

1 answer:

levacccp [35]2 years ago

8 0

Explanation:

We need to calculate the centripetal force:

Fc = W + F

With Fc being the centripetal force, W the weight of the boy, F the centrifugal force (apparent).

We know that we can calculate the apparent centrifugal force thank to the formula:

F = (m·v²)/r = 204N

So we can write:

Fc = W + F = 445N + 204N = 649N

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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that

pav-90 [236]

Answer:

The permittivity of rubber is $\epsilon = 8.703 *10^{-11}$

Explanation:

From the question we are told that

The magnitude of the point charge is $q_1 = 70 \ nC = 70 *10^{-9} \ C$

The diameter of the rubber shell is $d = 32 \ cm = 0.32 \ m$

The Electric field inside the rubber shell is $E = 2500 \ N/ C$

The radius of the rubber is mathematically evaluated as

$r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m$

Generally the electric field for a point is in an insulator(rubber) is mathematically represented as

$E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}$

Where $\epsilon$ is the permittivity of rubber

=> $E * \epsilon * 4 * \pi * r^2 = Q$

=> $\epsilon = \frac{Q}{E * 4 * \pi * r^2}$

substituting values

$\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}$

$\epsilon = 8.703 *10^{-11}$

7 0

3 years ago

Practice 3: Label the correct phase that would result from the Moon and Earth in these positions.

Anna71 [15]

Answer:

both position I think in nor

5 0

3 years ago

Read 2 more answers

Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

antoniya [11.8K]

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

$\dfrac{dT}{dt}=c(T-T_{0})$

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Where, $dT=T_{1}-T_{2}$

Here, $T =\dfrac{T_{1}+T_{2}}{2}$

$T_{0}$ = 25°C (surrounding temperature)

dt = 1 minute

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Put the value into the formula

$\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)$

$c=\dfrac{75}{17.5}$

$c=4.285\ cubic\ meter/minute$

Hence, This is the required answer.

3 0

3 years ago

What happens when the voltage increases and the resistance stays the same in a electrical circuit?

Orlov [11]

Answer:

The current in the circuit increases

Explanation:

The ohm's law states that the potential across a circuit is proportional to the current in the circuit.

V ∝ I

Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.

The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes

V = IR

According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.

So, when there is an increase in the voltage, the current on the circuit increases.

4 0

3 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

3 years ago