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2 years ago

What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter

Physics

1 answer:

Kobotan [32]2 years ago

7 0

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

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express each of the following aa indicated a. 2 dm expressed in millimeter b. 2h 10min expessed in sec c. 16g expressed in centi

Alex787 [66]

I thinks or know it is A or B cuz magnets and metal attract

4 0

3 years ago

Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long

nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

$R = \dfrac{\rho L }{A}$

where;

A = πr²

$R = \dfrac{\rho L }{\pi r ^2}$

For the shorter cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{ 4 \rho L }{\pi \ D ^2}$

For the longer cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{32\rho L }{\pi \ (4 D) ^2}$

$R = \dfrac{2\rho L }{\pi \ (D) ^2}$

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

$\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}$

$\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}$

$\dfrac{R_s}{R_L} =2$

${R_s}=2{R_L}$

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0

3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

Harman [31]

When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.

Given:
a = 86 m/s^2
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m

vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m

Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m

5 0

3 years ago

An unknown mass of each substance, initially at 25.0 ∘C, absorbs 1920 J of heat. The final temperature is recorded. Find the mas

zhannawk [14.2K]

Answer: mass for Pyrex glass 84.21g

mass for sand 61.6g

mass for ethanol 41.32g

mass for water 62.07g

Explanation

By definition specific heat is the amount of heat required to change the temperature of 1 kg mas by 1°C

Q=mcΔT is formula for specific heat

Q is heat transfer

m is mass

ΔT is change in temperature

c is specific heat

c of Pyrex glass= 0.75 j/g°C

c of sand = 0.84 j/g°C

c of ethanol= 2.42 j/g°C

c of water = 4.18 j/g°C

now we will make M(mass) the subject, so equation becomes

m=Q/cΔT

for

pyrex glass Tf=55.4°C

m=1920/(55.4-25)*0.75

m=84.21g {after cutting J(joules) and °C we are left with g(grams)}

for

sand Tf=62.1°C

m=1920/(62.1-25)*0.84

m=61.6g {after cutting J(joules) and °C we are left with g(grams)}

for

ethanol Tf=44.2°C

m=1920/(44.2-25)*2.42

m=41.32g {after cutting J(joules) and °C we are left with g(grams)}

for

water Tf=32.4°

m=1920/(32.4-25)*4.18

m=62.07g {after cutting J(joules) and °C we are left with g(grams)}

i hope you understand the solution, thank you.

7 0

3 years ago

A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal

Natalija [7]

Answer : The specific heat of unknown sample is, $8748.78J/kg^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-[q_2+q_3]$

$m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]$

where,

$c_1$ = specific heat of unknown sample = ?

$c_2$ = specific heat of water = $4186J/kg^oC$

$c_3$ = specific heat of copper = $390J/kg^oC$

$m_1$ = mass of unknown sample = 72.0 g = 0.072 kg

$m_2$ = mass of water = 203 g = 0.203 kg

$m_2$ = mass of copper = 187 g = 0.187 kg

$T_f$ = final temperature of calorimeter = $39.4^oC$

$T_1$ = initial temperature of unknown sample = $80.0^oC$

$T_2$ = initial temperature of water and copper = $11.0^oC$

Now put all the given values in the above formula, we get

$0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]$

$c_1=8748.78J/kg^oC$

Therefore, the specific heat of unknown sample is, $8748.78J/kg^oC$

7 0

3 years ago