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timama [110]
3 years ago
5

Does the area of contact effect the frictional force . explain?​

Physics
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

The force due to friction is generally independent of the contact area between the two surfaces. This means that even if you have two heavy objects of the same mass, where one is half as long and twice as high as the other one, they still experience the same frictional force when you drag them over the ground.

Plz mark 5 star, thanks, and brainliest

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A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have i
GarryVolchara [31]

Answer:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

The ball momentum is

\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s

As there is a similar momentum

So,

\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

5 0
3 years ago
For a glacier to melt, it must be
nata0808 [166]

Answer:

warmed should be ur answer

Explanation:

hope it helps , pls mark me as brainliest

7 0
4 years ago
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52. How did people spend Life in the<br>hunting age-​
kondor19780726 [428]

Answer:

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4 years ago
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A Michelson interferometer uses light with a wavelength of 452.446 nm. Mirror M2 is slowly moved a distance x, causing exactly 2
I am Lyosha [343]

Answer:

Option b. Dark spot

Explanation:

Michelson interferometer gives the path difference of the light as path difference, d as twice the distance moved or covered by the mirror M_{2}, x and is given as:

d= 2x

Since, its an even multiple, we obtain a bright fringe

now, when the move half the distance, i.e., \frac{x}{2}

Therefore, the path difference for half the distance \frac{x}{2} is:

d = x

As it is clear that its an odd multiple which correspond to dark spot as the final image

3 0
4 years ago
What is the final velocity of a car that is originally traveling 12 m/s and then undergoes an acceleration of 2.3m/s squared for
Katarina [22]

To solve this problem we use the general kinetic equations.

We need to know the time it takes for the car to reach 130 meters.

In this way we have to:

x(t) = x_0 + v_0t + 0.5at ^ 2

Where

x_0 = initial position

v_0 = initial velocity

a = acceleration

t = time

x(t) = position as a function of time

130 = 0 + 12(t) + 0.5(2.3)t ^ 2

1.15t ^ 2 + 12t - 130.

We use the quadratic formula to solve the equation.

t = \frac{-12 \± \sqrt {(12) ^ 2-4(1.15)(- 130)}}{2 (1.15)}

t = 6.63 s and t = -17.1 s

We take the positive solution. This means that the car takes 6.63 s to reach 130 meters.

Then we use the following equation to find the final velocity:

v_f = v_0 + at

Where:

v_f = final speed

v_f = 12 +2.23(6.63)

The final speed of the car is 27.25 m/s

3 0
3 years ago
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