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Nataly_w [17]
2 years ago
10

If you push a chair across the floor at a constant velocity, how does the force of friction compare with the force you exert? ex

plain.
Physics
1 answer:
patriot [66]2 years ago
6 0
If the velocity of the chair is constant, then the net force acting on it is zero.

The force you exert to keep it going is equal and opposite to the force of friction.
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Which sector is not part of the circular flow model of the economy ?
morpeh [17]

Answer:

a. government sector

Explanation:

7 0
3 years ago
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An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
How does friction make it possible for you to walk across the floor?
Tema [17]

if we are walking on a perfectly smooth ground which has no friction our force would simply cancel out the force reverted by the ground and we would fall.

We need it to help push out feet off the ground

Hope those helps :)

5 0
3 years ago
Ideally, the resistance of an ammeter should be:
pav-90 [236]
Ideally the resistance should be ZERO
7 0
3 years ago
The dispatcher of courier service receives a message from truck A that reports a position of +5 after a displacement of +2. What
Romashka [77]

The formula we need to use is displacement.

d=\Delta{x}=x_f-x_i, where xf is final position and xi is initial position.

We report the final position of 5 and the displacement of 2 so the formula is now:

2=5-x_i\Longrightarrow x_i=3.

So the initial position of truck A is 3.

Hope this helps.

r3t40

8 0
2 years ago
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