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3 years ago

A roller coaster has a mass of 300 kg. It drops from rest at the top of a hill

1 answer:

7 0

To determine the velocity of the roller coaster as it moves down, we use the kinematic equation which is expressed as 2gy = vf^2 - v0^2 where g is the gravitational acceleration, y is the elevation of the roller coaster, vf and vo are the final and initial velocity. We calculate as follows:

2gy = vf^2 - v0^2

Since it starts at rest, v0 is zero.

2gy = vf^2

vf = √2gy

vf = √2(9.8)(101)

vf = 44.5 m/s

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Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve

dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

8 0

3 years ago

alexgriva [62]

Answer:

Get your answer for $1 from www.gotit-pro.com

Explanation:

6 0

3 years ago

If you double the frequency of a vibrating object, its period:

Furkat [3]

Answer:

becomes halved.

Explanation:

trust

8 0

3 years ago

Read 2 more answers

In a typical lightning strike, 2.0 c flows from cloud to ground in 0.24 ms. what is the current during the strike?

lyudmila [28]

Current = charge/time = (2 c)/(0.00024 sec)= 8,333 Amps !

4 0

4 years ago

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

3 0

3 years ago