All categories

2 years ago

A compass in a magnetic field will line up __________.

Physics

1 answer:

Anastaziya [24]2 years ago

5 0

Line up in a direction parallel to the magnetic field lines<span />

You might be interested in

a coin press creates a pressure of 3.20*10^8 Pa on a nickel of radius 0.0106 m. how much force does the press exert on the coin?

Naya [18.7K]

Answer:

1.13 x 10⁵N

Explanation:

Given parameters:

Pressure of the coin press = 3.2 x 10⁸ Pa

radius of the nickel coin = 0.0106m

Unknown:

Force of the press on coil = ?

Solution:

Our knowledge of pressure will help us solve this problem.

Pressure is defined as the force applied per unit area on a body.

Pressure = $\frac{force}{area}$

Force = Pressure x Area

Since the pressure is known;

Area of the coin = Area of a circle = π r²

where r is the radius of the coin;

Area of the coin = π x 0.0106² = 3.53 x 10⁻⁴m²

Force = 3.2 x 10⁸ x 3.53 x 10⁻⁴ = 1.13 x 10⁵N

3 0

2 years ago

What is the name for family labeled #4 (Yellow)?

goldfiish [28.3K]

Answer:

transition metals im sorry if this was too late

4 0

2 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

2 years ago

Read 2 more answers

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

2 years ago

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate

tigry1 [53]

Explanation:

It is given that,

Average power per unit mass generated by Lance, $\dfrac{P}{m}=6.5\ W/kg$

$P=6.5\times 75=487.5\ W$

(a) Distance to cover race, $d = 160\ km =160\times 10^3\ m$

Average speed of the person, v = 11 m/s

If t is the time taken to cover the race.

$t=\dfrac{d}{v}$

$t=\dfrac{160\times 10^3\ m}{11\ m/s}$

t = 14545.46 s

Let W is the work done. The relation between the work done and the power is given by :

$P=\dfrac{W}{t}$

$W=P\times t$

$W=487.5\times 14545.46$

W = 7090911.75 J

(b) Since, $1\ J=2.389\times 10^{-4}\ calories$

So, in 7090911.75 J, $W=7090911.75 \times 2.389\times 10^{-4}$

W = 1694.01 J

Hence, this is the required solution.

6 0

2 years ago