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Inessa [10]
4 years ago
6

A 5-kg object slides down a frictionless surface inclined at an angle of 30º from the horizontal. The total distance moved by th

e object along the plane is 10 meters. The work done on the object by the normal force of the surface is?
Physics
1 answer:
erma4kov [3.2K]4 years ago
7 0

Answer:

Workdone =0

Explanation:

The body is inclined at θ=30°,     s=10m,  m=5kg

Workdone by the object = FsCosθ

but the surface is a friction-less surface, this means coefficient of friction =0

F=μR

F=0*R

F=0

W=FScosθ

W=0*SCosθ

W=0

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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
Which parts of the graph would increase if all of earth's polar ice melts? Which part would decrease?
Sedbober [7]

Answer:

the answer to the question is the part that is reflected by clouds and atmosphere 26%

8 0
4 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
bogdanovich [222]

Answer:

magnitude of force on charge 2Q  = \frac{KQ^{2} }{I^{2} }

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q  i.e x-component of the net force and the y-component of the net force

║F║ = \sqrt{f_{x}^{2} + f_{y}^{2}    }  =  after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = \frac{KQ^{2} }{I^{2} }

The direction of the force on charge 2Q is calculated as

tan ∅ = \frac{f_{y} }{f_{x} } = 1.8284

therefore ∅ = tan^{-1}  1.8284

= 61⁰

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3 years ago
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If I remember correctly, it is the 3rd answer choice.
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