Ask question

Ask question

All categories

3 years ago

14

You can feel the heat waves from a small electric heater on your hands and face even though the heater is on the other side of t

he room. Which is this is an example of?

1 answer:

NeX [460]3 years ago

8 0

That's an example of heat transfer by radiation.

You might be interested in

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down

Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

$d=ut + \frac{1}{2}at^2$

If we substitute all the values listed in part (a), we find

$d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m$

8 0

3 years ago

Two common terms for a decrease in velocity are

Colt1911 [192]

deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)

6 0

3 years ago

Read 2 more answers

1. The gravitational force acting on a falling body and its weight is constant. But the law of universal gravitation tells us th

vagabundo [1.1K]

It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.

3 0

3 years ago

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables

Juli2301 [7.4K]

They made me do it I don’t even know what to say I’m so sorry

7 0

3 years ago