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3 years ago

1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0

Physics

1 answer:

igor_vitrenko [27]3 years ago

6 0

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0 m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

$\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2$

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m$

So, the plane slide a distance of 229.5 m.

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sergeinik [125]

Answer:

30ms

Explanation:

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3 years ago

(100 pts)<br>Why can't you use distance divide by time to calculate the instantaneous<br>speed?

Fiesta28 [93]

Answer:

Instantaneous speed means speed at any instant

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4 0

4 years ago

Read 2 more answers

A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?

Lena [83]

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω = 11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

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3 years ago

A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$