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3 years ago

What is the distance between two spheres, each with a charge of 2.5 x 10-6 c, when the force between them is 0.50 n?

1 answer:

7 0

In this case, Coulomb's Law applies:

F = 1/(4πε₀) · (Q₁Q₂/r²)

You can solve it for r:

r = √[1/(4πε₀) · (Q₁Q₂/F)]

Plugging in numbers:

r = √[1/(4π·8.85×10⁻¹²) · (2.5×10⁻⁶)²/0.50]
= 0.335m

The correct answer is: the two charges are 0.335m apart.

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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.

Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' = 19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0

3 years ago

Question 7 (2 points)

DiKsa [7]

This has to do with independent and dependent variables. The independent variable is what affects the dependant variable, so the question is, does a students mark on a test affect their hours of sleep or does their hours of sleep affect their test results? Which one makes more sense? I would say the latter.

6 0

3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

lara [203]

Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

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4 years ago

Canada is the country (dash)of the united States

lana66690 [7]

I believe it is north.

5 0

3 years ago

What property is used to distinguish the layers of the atmosphere?

Margarita [4]

There are many porperties. You can use Altitude, Temperature, Pressure and Density, but the best one is temperature. The resaon for that is that based on the temperature changes then the athmosphere can be broken into four major layers. Remember that the layers are the following: the troposphere,the stratosphere, the mesosphere, and thethermosphere.

3 0

3 years ago

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