All categories

3 years ago

• A cable is suspended over a pulley. A 5.0 kg mass is attached to

Physics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

Tension= $\frac{20g}{3}$ (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

$10g-T = 10a$

$T-5g=5a$

Now adding them we get,

$5g=15a$

$a=\frac{g}{3}$

Substituting them back in the equation we get,

$10g-10(\frac{g}{3} ) = T$

$T= \frac{20g}{3}$

You might be interested in

In typical game play situation, (with no overtime), when is a game over?

Harman [31]

C) When the time runs out (usually in sports such as soccer, but i don’t know what sport you’re referring to)

3 0

3 years ago

When an ice cube is placed in a glass of warm water, how are the signs and values q for the ice and the warm water related, assu

Valentin [98]

Answer:

q(ice) = -q(warm water)

Explanation:

Ice removes energy if ice is put in warm water and ice retains the very same amount of energy. Thus the water temperature heat sign is negative, and the ice heat sign is positive.

3 0

3 years ago

The field inside a charged parallel-plate capacitor is __________.

Bess [88]

<span>Uniform. A parallel plate capacitor is a simple arrangement of electrodes and dielectric to form a capacitor. Here two parallel conductive plates are used as electrodes with a medium or dielectric in between them. Charge separation in a parallel-plate capacitor causes an internal electric field, which is uniform.</span>

5 0

4 years ago

Read 2 more answers

A mercury thermometer is constructed as

Tamiku [17]

Answer:

The change in height of the mercury is approximately 2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

$\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C$

from which we solved for the change in volume $\Delta V$ due to a given change in temperature $\Delta T$

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

$V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3$

Therefore, the change in volume with a change in temperature of 36°C becomes:

$\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3$

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

$V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm$

8 0

3 years ago

An electronics technician wishes to construct a parallelplate capacitor using rutile (k=100) as the dielectric. The area of the

My name is Ann [436]

Given k=100, Area if plate A = 1.00 cm2

$= 1 x (10-2)2 m^{2}$

Thickness of rutile $ds = t = 1.00 mm$

$= 1 x 10-3 cm$

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd

When a dielectric is placed between capacitors plates then capacitance c1 = KoAd

Here $d = t = 1 x 10-1 cm$ , $o = 8.854 x 10-12 c^{2} m^{-2} N$

Now $c1 = 100 X 8.859 X 10^{-12} X 1 X (10-2)^{21} X 10^{-2}$

$= 8.859 X 10-1910-3= 8.85 X 10-11 F= 8.85 X 10-12 F$

$C1 = 88.5 X 10-12 F$

$C1 = 88.5 PF$

Correct Option: A

What purposes serve parallel plate capacitors?

The following are some uses for parallel plate capacitors:

This kind of capacitor is utilized in rechargeable energy systems such as batteries.
These capacitors are used in systems for dynamic digital memory.
Such capacitors are used in pulsed laser and radar circuits.

To learn more about Parallel plate capacitors, visit:

brainly.com/question/12733413

#SPJ4

5 0

2 years ago