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Jet001 [13]
4 years ago
12

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

? (Use any variable or symbol stated above as necessary.)'
Physics
1 answer:
Margarita [4]4 years ago
4 0

Answer:

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

Explanation:

  • We have to make a hollow sphere of inner  radius r_1 and outer radius r_2.

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

V_t=\frac{4}{3} \pi.r_2^3

And the volume of the hollow space in the sphere:

V_h=\frac{4}{3} \pi.r_1^3

Therefore the net volume of material required to make the sphere:

V=V_t-V_h

V=\frac{4}{3} \pi(r_2^3-r_1^3)

  • Now let the density of the of the material be \rho.

<u>Then the mass of the material used is:</u>

m=\rho.V

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

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2 years ago
As a bullet shot vertically upward rises, the kinetic energy of the bullet
algol [13]

The kenitc energy of the bullet lowers as it keeps going up.

Because gravity is pushing the bullet down as the bullet goes up.

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3 0
2 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
A car is traveling at a speed of 54 km/h. Breaks are applied so as to produce a uniform acceleration of -0.5 m/s2. Find how far
Lena [83]

Explanation:

u=54 km/h

54*5/18=15 m/s

v=0m/s

t=?

acceleration=-0.5m/s^2

we know that a=v-u/t

so,

t=v-u/a

t=15-0/0.5

=15/0.5

=30

therefore, the time is 30 second

Hope this answer helps you..

8 0
3 years ago
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