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lesantik [10]
2 years ago
15

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Physics
1 answer:
Vesnalui [34]2 years ago
7 0

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=\rho

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

Density=\frac{mass}{volume}

Mass=Density\times volume=Constant

Therefore,\rho_1 V_1=\rho_2V_2

Volume of sphere=\frac{4}{3}\pi r^3

Using the formula

\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3

\rho R^3=\rho_2\times \frac{R^3}{8}

\rho_2=8\rho

Hence, the density of  the compressed sphere=8\rho

Option a is correct.

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Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

v_x=149\ m/s

In the y axis

v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}

The resultant velocity is given by

v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

4 0
3 years ago
The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril
STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

                         T = 5252*P / N

                         T = 5252*5.85 / 1900

                         T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

                          T = F*r

Where,   r = d / 2

                          F = 2T / d

                          F = 2*16.171 / 0.98

                          F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

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Answer:

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Answer:

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