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2 years ago

6

How much mercury most placed inside the glass vessel of capacity 500 cc so that the volume of space unoccupied by mercury always

remains constant?

1 answer:

Tema [17]2 years ago

4 0

Answer: 500cc of mercury should be placed

Explanation:

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Your employer asks you to build a 18-cmcm-long solenoid with an interior field of 5.8 mTmT . The specifications call for a singl

andrezito [222]

Answer:

Explanation:

The magnetic field in a solenoid is

B = μ₀ N / L I

Where N is the number of turns, L the solenoid length and I the current

N = B L / μ₀ I

Let's calculate

N = 5.8 10⁻³ 0.18 / 4 π 10⁻⁷ 1

N = 8.3 102 laps

N = 831 laps

Let's find the solenoid length

For this we use a rule of proportions

L_solenoid = Turns * wire diameter

L_ solenoid = 831 * 0.41 10--3

L_solenoid = 0.3407 m

We see that two turns are needed in the wire to have a length of 0.18 m

5 0

2 years ago

⦁ A speed skater increases her speed form 10 m/s to 12.5 m/s over a period of 3 seconds while coming out of a curve of 20 m radi

Thepotemich [5.8K]

Answer:

7.85 m/s^2

Explanation:

linear or tangential acceleration= dv/dt

⇒ $a_t= \frac{12.5-10}{3}$

=0.83 m/s^2

radial acceleration is given by = $\frac{v^2}{r}$

⇒ $a_r =\frac{12.5^2}{20}$

= 7.81 m/s^2

total acceleration

$a_T= \sqrt{a_t^2+a_r^2}$

putting values we get

$a_T= \sqrt{0.83^2+7.81^2}$

= 7.85 m/s^2

3 0

2 years ago

Read 2 more answers

A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de

Tresset [83]

Answer: 7.734 m/s

Explanation:

We have the following data:

$\theta=35\°$ The angle at which the water ballon was thrown

$x=10 m$ The horizontal distance of the water ballon

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find the initial velocity $V_{o}$ at which the water ballon was thrown, and we can find it by the following equation:

$x=V_{o}cos \theta T$ (1)

Where $T=2t$ is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time $T$ it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (2)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (3)

Remembering $T=2t$ :

$T=2\frac{-V_{o}}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (2\frac{-V_{o}}{g})$ (5)

Isolating $V_{o}$ :

$V_{o}=\sqrt{\frac{x g}{-2 cos \theta}}$ (6)

$V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}}$ (7)

Finally:

$V_{o}=7.734 m/s$

4 0

2 years ago

Generalized global air circulation and precipitation patterns are caused by

kotegsom [21]

<span> Rising, warm, moist air masses cool and release precipitation as they rise and then at high altitude, cool
and sink back to the surface as dry air masses after moving north or south of the tropics.

</span>

5 0

2 years ago

Read 2 more answers

A stack of books rests on a level frictionless surface. A force F acts on the stack, and it accelerates at 3.0 m/s2. A 1.0 kg bo

Oxana [17]

Answer:

m1 = 2 kg

m2 = 3kg

Explanation:

The force can be getting by

F = m * a

F1 = F2

a1 = 3.0 m/s^2

a2 = 2.0 m/s^2

The force F1=F2 because the force is applied so get the a2 acceleration

m1 * a1 = m2 * a2

m2 = m1 + 1kg

m1 *(3.0 m/s^2) = m2* (2.0 m/s^2)

m1 *(3.0 m/s^2) = (m1 + 1kg) * (2.0 m/s^2)

m1*(3.0m/s^2-2.0m/s^2)=2 kg*m/s^2

Solve to find the mass

m1 m/s^2= 2 kg*m/s^2

m1 = 2 kg

m2 = 3kg

4 0

2 years ago