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d1i1m1o1n [39]
3 years ago
15

A rough surface does not shine why?​

Physics
2 answers:
ANTONII [103]3 years ago
8 0

Explanation:

Smooth surfaces could reflect light, and rough surfaces do not. Because of the nature of the substances. Things with a rough surface could scatter light, or reflect it in all directions. This makes them look dull and doesn't shine.

bonufazy [111]3 years ago
7 0
This is the answer to the question
I hope this help you
Thank you for your question.!

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A1.200 kg car is sliding down an icy
Triss [41]

Answer:

1585.67N

Explanation:

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5 0
3 years ago
A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
1. You get hit with a force of 45 N by a ball with a mass of 0.75 kg. How fast was the
dlinn [17]

Answer:

60 {ms}^{ - 2}

Explanation:

As we know,

=》Force = Mass × Acceleration

=》45 N = 0.75 × Acceleration

=》Acceleration = 45 ÷ 0.75

=》Acceleration = 60

hence, the Acceleration of the ball would be. 60 meters per second square

60m {s}^{ - 2}

6 0
3 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
ArbitrLikvidat [17]

Answer:

The mass of the solid cylinder is m  =  1612.5  \  kg

Explanation:

From the question we are told that

   The radius of the grinding wheel is R =  0.330 \ m

   The  tangential force is F_t =  250 \ N

    The angular acceleration is  \alpha  =  0.940 \ rad/s^2

The torque experienced by the wheel is mathematically represented as

     \tau  =  I  *  \alpha

Where  I  is the moment of inertia

The torque experienced by the wheel can also be  mathematically represented as

       \tau  =  F_t  * r

substituting values

       \tau  =  250 * 0.330

      \tau  = 82.5  \ N\cdot m

So

   82.5  =  I *  \alpha

    82.5  =  I *  0.940

So

   I  =  87.8 \ kg \cdot m^2

This moment of inertia can be mathematically evaluated as

     I  =  \frac{1}{2} * m* r^2

substituting values

  87.8  =  \frac{1}{2} * m* (0.330)^2

=>   m  =  1612.5  \  kg

5 0
3 years ago
Isostatic rebound
svp [43]
I think it is B the rise of the land
4 0
3 years ago
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