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3 years ago

If the mass of an object increases, predict what will happen to the coefficient of sliding friction.

Physics

2 answers:

anyanavicka [17]3 years ago

5 0

Answer:

f the mass of an object increases, predict what will happen to the coefficient of sliding friction. The coefficient of sliding friction will increase. The coefficient of sliding friction will decrease.

Explanation:

Julli [10]3 years ago

4 0

Answer:

The coefficient of friction remains the same. The mass affects the force of friction, however the coefficient of friction depends the structural properties of the surfaces of the object and the floor. Therefore, the mass has no effect of the coefficient of friction.

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Fiesta28 [93]

alll factor is possible

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3 years ago

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. How far away is the reflecting object? (c

Assoli18 [71]

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time = 20 μs = 20 * $10^{-6}$ s

let distance to the reflecting surface be = x

total distance travelled by pulse will be = 2x

speed = 3.0 × $10^{8}$ m/s

distance = speed * time

2x = 3.0 × $10^{8}$ * 20 * $10^{-6}$

x = 3000 m

The reflecting object would be 3000 m away

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5 0

2 years ago

Question 3 A woman sitting in a stationary car notices a man cycling past her at 40 km/hr. Five seconds after he passes her car,

Naya [18.7K]

The shortest possible distance in which she catches up with the man is 129.7 m.

The given parameters:

speed of the man, v₁ = 40 km/hr

speed of the woman, v₂ = 70 km/hr

time when the woman started moving, t₁ = 5 seconds

To find:

the shortest distance when she catches up with the man

Convert the given speed in km/h to m/s

3.6 km/h = 1 m/s
40 km/h = 40/3.6 = 11.11 m/s
70 km/h = 70/3.6 = 19.44 m/s

Calculate the distance moved by the man before she started following him.

Distance moved by the man = speed x time

= 11.11 m/s x 5 s

= 55.55 m

Let the time the woman catches up with the man = t

Apply the following simple principle to calculate the time.

The man is moving forward and the woman is moving forward to close the gap between them, thus both speeds are working against each other.

$(v_2-v_1)t = 55.55 \ m$

the difference in the speed is because they are working against each other.

$(19.44 - 11.11)t = 55.55\\\\8.33t = 55.55\\\\t = \frac{55.55}{8.33} \\\\t = 6.67 \ s$

The time when the woman catches up with the man = 6.67 s

The shortest distance at which the woman catches up with the man is calculated as;

Shortest distance = woman's speed x time when she catches up

= 19.44 m/s x 6.67 s

= 129.7 m

Thus, the shortest possible distance in which she catches up with the man is 129.7 m.

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3 years ago

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi

Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

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This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

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1 year ago

What do butterflies have on their feet which allows them to sense different strengths and types of nectar?

Andru [333]

The answer is D for sure.

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3 years ago

Read 2 more answers