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viva [34]
2 years ago
5

Write the mathemetical relation between work force and displacement​

Physics
1 answer:
il63 [147K]2 years ago
8 0

Answer:

Work is measured as the product of force and the displacement in the direction of the force. Work = force × displacement in the direction of the force.

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A sanding disk with rotational inertia 1.2 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of 10 N·m a
tensa zangetsu [6.8K]

Answer:

Angular momentum = 0.7 kg.m²/s

Angular velocity = 583.3 rad/s

Explanation:

1. The torque τ  is related to the angular momentum L by the relation

τ = ΔL/Δt

ΔL = τΔt

τ = 10 N. m

Δt = 70 ms = 70 × 10⁻³s

ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s

2. The rotational inertia I relates the angular momentum L to the angular velocity w

L = Iw

w = L/I

L =  0.7 kg.m²/s

I = 1.2 × 10⁻³ kg.m²

w = (0.7 kg.m²/s)/(1.2 × 10⁻³ kg.m²) = 583.3 rad/s

4 0
3 years ago
Ok so I am taking a test I need help please
Aloiza [94]

Answer:

i know its definetly either clockwise or counter clockwise

Explanation:

6 0
3 years ago
How many feet are in a mile
klemol [59]

Hello There!

There are 5,280 feet in 1 mile.

Have A Great Day!

7 0
3 years ago
Read 2 more answers
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
What current is used to power the United States power grid?
alexdok [17]
The answer is Alternating Current
3 0
3 years ago
Read 2 more answers
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