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marishachu [46]
3 years ago
15

Which force acts between two surfaces to prevent or slow motion

Physics
1 answer:
vredina [299]3 years ago
4 0
The answer is friction
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Compare the signs of ƒ for lenses and mirrors.
STALIN [3.7K]

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

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3 years ago
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What type of heat does not require matter?
Lana71 [14]
It would be Thermal Radiation
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3 years ago
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Which of the following statements is accurate?
cupoosta [38]

Answer: A and B

Explanation:

A

The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Because wavelength is the distance between the two successful crest or trough.

B) 

Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.

Because amplitude is the measure of maximum displacement from the original position

4 0
3 years ago
What is the index of refraction of a material in which the speed of light is 7.50% slower than the speed of light in vacuum?
Inessa [10]

Answer:

The value is n = 1.081

Explanation:

From the question we are told that

   The speed of in a vacuum is c = 3.0 *10^{8} \  m/s

    The speed of light in the material is  v  =  c -0.075c = 0.925 c =  0.925 * 3.0*10^{8}  = 2.775 *10^{8} \ m/s

   Generally the reflection of the material is mathematically represented as

             n = \frac{c}{v}

=>    n = \frac{3.0*10^{8}}{2.775 *10^{8}}

=>    n = 1.081

8 0
3 years ago
g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0
3 years ago
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