3 years ago

A sports car accelerates at a constant rate from rest to a speed of 90 km/hr in 8 s. What is its acceleration?

1 answer:

3 0

First convert 90km/hr to m/s.

Initiate velocity = 0m/s (car was at rest)

Final velocity is 25m/s (90km/hr converted)

25m/s - 0m/s / 8s = 3.125 m/s^s

Therefore the answer is option A (3.13m/s^2)

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A solid sphere is placed on a frictionless floor in a very long corridor and is given a quick push so that it begins to slide, w

prohojiy [21]

I think it’s D I hope it helps

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3 years ago

A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting

emmainna [20.7K]

Answer:

400

Explanation:

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2 years ago

A ceiling fan has three blades. The moment of inertia of a blade is 0.2kgm^2. The net torque exerted on fan blades is 8Nm. Find

olchik [2.2K]

Answer:

(A) the angular acceleration of the blades is 13.33 m/s.

Explanation:

Given;

moment of inertia of a blade, I = 0.2 kgm²

net torque exerted on fan blades, ∑τ = 8Nm

Torque is given as product of moment of inertia and angular acceleration;

τ = Iα

where;

α is the angular acceleration

Since there are three blades of the ceiling fan, the net torque is given as;

∑τ = (3I)α

∑τ = 3Iα

α = ∑τ / 3I

α = (8) / (3 x 0.2)

α = 13.33 m/s

Therefore, the angular acceleration of the blades is 13.33 m/s.

8 0

2 years ago

Gravity pulls downward on a rock with a force of 800 N. If you pull upward on the rock with a force of 400 N, what is the total

wel

The answer is
a. 400 N downward

3 0

3 years ago

Read 2 more answers

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

3 years ago