Please help I have until 9 to turn in.

A solenoid of length 0.250 m and radius 0.0250 m is comprised of 440 turns of wire. Determine the magnitude of the magnetic fiel

Bad White [126]

Answer: 0.02654

Explanation:

in the attachment

6 0

4 years ago

Read 2 more answers

If a spacecraft in earth orbit is pushed by a thruster what will happen?

grin007 [14]

For the orbital speed of a spacecraft we know that

$v = \sqrt{\frac{GM}{r}}$

here

M = mass of the planet around which satellite is revolving

r = orbital radius

Now when thruster is used by the spacecraft its speed will change due to which orbital speed will change.

Since here while changing the speed mass of the planet will be same

we can say the speed of the spacecraft will changed by thruster due to which its orbital radius will change

so the correct answer must be

<em>b. the spacecraft will change motion and will maintain this new orbit until the thruster is fired again.</em>

8 0

3 years ago

Suppose high tide is at midnight, the water level at midnight is 3 m, and the water level at low tide is 0.5 m. Assuming the nex

aev [14]

We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

$t=10.0hours$

From the Question we are told that

Maximum height $h_{max}=3m$

Minimum height $H_{min}=0.5m$

Time for next high tide will occur $T=12 hours =>720 min$

Generally Average Height

$h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75$

Therefore determine Amplitude to be

$A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25$

Generally, the equation for Time is mathematically given by

At t=0

$h(x)=Acos(Bx)+h_{avg}$

Where

$B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}$

Therefore

$h(t)=Acos8.73*10^{-3}(t)+h_{avg}$

Hence the Time at $T=1.125$ is

$1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75$

$-0.1249t=1.75$

$t=10.0hours$

For more information on this visit

brainly.com/question/22361343

5 0

3 years ago

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 1

astra-53 [7]

The acceleration experienced by the particle is given by
$a=113000 g=113000 \cdot 9.81 m/s^2$
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed $\omega$ of the particle and its distance r from the axis by the relationship
$a= \omega ^2 r$
In our problem, $r=6 cm=0.06 m$ , so we can solve for $\omega$ :
$\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s$
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to $( \frac{1}{2 \pi} )$ revolutions, while $1 s = \frac{1}{60} min$ . So we the conversion is $\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm$

4 0

4 years ago

A car with the mass of '800 kg is moving at a velocity of 34 m/s. What is 10 points

klio [65]

Answer:462400joules

Explanation:

Mass=m=800kg velocity=v=34m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(800x34^2)/2

Kinetic energy =(800x34x34)/2

kinetic energy =924800/2

Kinetic energy =462400joules

7 0

3 years ago