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oee [108]
3 years ago
13

A sanding disk with rotational inertia 1.2 × 10−3 kg ⋅ m2 is attached to an electric drill whose motor delivers a torque of magn

itude 16 N ⋅ m about the central axis of the disk. About that axis and with the torque applied for 33 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?
Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

L = 0.528kgm²/s

ω = 440 s^-1

Explanation:

given

Inertia, I = 1.2 * 10^-3kgm²

Torque, τ = 16Nm

Time, t = 33ms

Angular momentum is gotten using the relation

τ = ΔL/ΔT

Where τ = Torque

ΔL = change in angular momentum

ΔT = change in time taken

ΔL = τΔT

ΔL = 16Nm * 33*10^-3s

ΔL = 0.528kgm²/s

Angular velocity of the disk will be gotten by using the formula

L = Iω where

L = angular momentum

I = rotational inertia

ω = angular velocity

ω = L/I

ω = 0.538/1.2*10^-3

ω = 440 s^-1

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Basketball thrown over the shoulder to make a basket: Is this a projectile? why or why not
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Answer:

An object that is thrown, kicked or otherwise launched through the air is called a projectile.

Explanation:

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2 years ago
HELP PLZ
Anon25 [30]

Answer:

A = 11.77 m/s²

Explanation:

Mass of box (m) = 66kg

Force (f) = 777n

Acceleration (a) = ??

F = ma

777 = 66 · a

A = \frac{777}{66} = 11.77m/s²

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2 years ago
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lozanna [386]
1.022 km/s
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2 years ago
Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit aro
Sergeu [11.5K]

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

          T² = (\frac{4\pi ^2}{ G M_s} a³ = K_s a³

           

Let's apply this equation to our case

          a = \sqrt[3]{ \frac{T^2}{K_s} }

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

             10² = K_s  a_{w}^3

             a_w = \sqrt[3]{ \frac{100}{ K_s} }

             a_w = \frac{1}{ \sqrt[3]{K_s} }  4.64

Planet X

             a_x = \sqrt[3]{ \frac{640^3}{K_s} }

             a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

              a_y = \sqrt[3]{ \frac{80^2}{K_s}  }

              a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

              a_z = \sqrt[3]{ \frac{270^2}{K_s} }

              a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

3 0
3 years ago
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