Evaluate the amount of thermal energy lost from a 0.5 kg glass

Determine the resistance of a 250-w lightbulb connected to its proper source voltage of 110 v.

ollegr [7]

1.P=UI
2.U=IR
so P=U^2/R
R=U^2/P=110*110/250=48.4ohm

5 0

3 years ago

Can someone help me please? I’ve been trying to solve these questions all day.

qwelly [4]

#16

If we put a resistor in circuit it will slow the speed of current

Let's check ohms law

$\\ \rm\Rrightarrow \dfrac{V}{I}=R$

So if resistance is more current is less

#17

Again use ohms law

$\\ \rm\Rrightarrow V=IR$

$\\ \rm\Rrightarrow V\propto I$

Voltage must be increased

4 0

2 years ago

What are the differences between glaciers and geysers

liq [111]

A glacier is frozen and a geyser has boiling water coming from it. A glacier is usually still but it sometimes breaks and a geyser erupts. Also geysers are mostly rare and glaciers aren't.

7 0

3 years ago

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro

Alexxx [7]

Explanation:

It is given that,

Radius of the circular orbit, $r=0.53\times 10^{-10}\ m$

Speed of the electron, $v=2.2\times 10^6\ m/s$

Mass of the electron, $m=9.1\times 10^{-31}\ kg$

(a) The force acting on the electron is centripetal force. Its formula is given by :

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

$F=\dfrac{9.1\times 10^{-31}\times (2.2\times 10^6)^2}{0.53\times 10^{-10}}$

$F=8.31\times 10^{-8}\ N$

(b) The centripetal acceleration of the electron is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(2.2\times 10^6)^2}{0.53\times 10^{-10}}$

$a=9.13\times 10^{22}\ m/s^2$

Hence, this is the required solution.

5 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago