Ask question

Ask question

All categories

3 years ago

11

What do all the materials that interact with the magnets have in common?

1 answer:

zepelin [54]3 years ago

4 0

They both have a magnetic field

You might be interested in

Two balls move down a incline. One spins, while the other slides. Which one is going faster at the bottom?

natta225 [31]

Answer: the one spinnin

Explanation:

7 0

3 years ago

Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running

tangare [24]

Answer:

Since both start with the same vertical velocity from the same position with the same acceleration they will reach the lake at the same time.

6 0

3 years ago

Which of the following supplies the heat for the hot reservoir in a car's engine?

nata0808 [166]

Out of the choices given, igniting the gas-air mixture supplies the heat for the hot reservoir in a car's engine. The correct answer is C.

8 0

3 years ago

Read 2 more answers

What is the acceleration of a 1000kg car subject to a 550N net force?

igomit [66]

Answer:

a=550÷1000

a=0.55m/s²

5 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago