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2 years ago

1. A biker travels at an average speed of 11.1 m/s along a 8900 m

Physics

1 answer:

Karolina [17]2 years ago

4 0

Answer:

Time = 801.802 secs.

Explanation:

Speed can be defined as the rate of change in the magnitude of distance with respect to time. It is a scalar quantity because it only has magnitude but no direction.

Given the following data;

Average speed = 11.1m/s

Time =?

Distance = 8900m

$Speed (S) = \frac{distance}{time}$

$S = \frac{d}{t}$

Making time (t) the subject of formula, we have;

$Time (t) = \frac{d}{s}$

$Time (t) = \frac{8900}{11.1}$

Time = 801.802 secs.

<em>Hence, the time the biker takes to cover the segment is 801.802 seconds. </em>

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g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

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2 years ago

A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

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Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

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2 years ago

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull

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Answer:

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(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E) 351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

work done = force x distance x cos θ

since the rope is parallel to the ramp the angle between the rope and

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(B) calculate the work done by gravity

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The weight of the carton = force exerted by the mass of the carton = m x g

the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

work done by gravity = 128 x 5.2 x cos 120

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(C) find the work done by the normal force acting on the ramp

work done by the normal force = force x distance x cos θ

the angle between the normal force and the ramp is 90 degrees

work done by the normal force = Fn x distance x cos θ

work done by the normal force = Fn x 5.2 x cos 90

work done by the normal force = Fn x 5.2 x 0

work done by the normal force = 0 J

(D) what is the net work done ?

The net work done is the addition of the work done by the rope, gravitational force and the normal force

net work done = 374.4 - 332.8 + 0 = 41.6 J

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