Ask question

Ask question

All categories

3 years ago

11

An ant sits on a cd at a distance of 17 cm from the centre. If it sits there for 42 seconds, it travels a total distance of 913

cm. a. What angle has the ant turned through? (in radians) b. What speed has the ant been traveling at? (in cm/sec) c. What angular velocity has the ant been spinning at? (in radians/sec) d. What rpm is the cd turning at?

1 answer:

lys-0071 [83]3 years ago

4 0

a) 53.7 rad

b) 21.7 cm/s

c) 1.28 rad/s

d) 12.2 rpm

Explanation:

a)

The ant sits on the cd at a distance of

r = 17 cm

from the centre: this is therefore the radius of the circle covered by the ant.

Therefore, we can find the length of the circumference of the circle:

$c=2\pi r=2\pi(17)=106.8 cm$

A full circle corresponds to an angle of

$\theta = 2\pi$ rad

In this problem instead, the distance covered by the ant along the circle is

d = 913 cm

Therefore, the angle corresponding to this distance can be found by using the rule of three:

$\frac{\theta}{c}=\frac{\theta'}{d}$

And so we find:

$\theta'=\theta \frac{d}{c}=(2\pi)\frac{913}{106.8}=53.7 rad$

b)

The speed of the ant can be calculated by using

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

where

d is the distance travelled by the ant

t is the time taken to cover that distance

v is the speed

In this problem:

d = 913 cm is the distance covered by the ant

t = 42 s is the time elapsed

Therefore, the speed of the ant is:

$v=\frac{913}{42}=21.7 cm/s$

c)

The angular velocity of an object in rotation is the rate of change of the angular displacement.

The angular velocity is related to the linear speed by

$\omega=\frac{v}{r}$

where

$\omega$ is the angular velocity

v is the linear speed

r is the radius of the circle

In this problem:

v = 21.7 cm/s is the speed of the ant

r = 17 cm is the radius of the circle

So, the angular velocity of the ant is:

$\omega=\frac{21.7}{17}=1.28 rad/s$

d)

The rpm means "revolutions per minute", and it is another units used to express the angular velocity.

In order to convert the angular velocity from radians per second to rmp, we must keep in mind that:

1 revolution = $2\pi$ radians

1 minute = 60 seconds

The angular velocity of the ant in this problem is

$\omega=1.28 rad/s$

Therefore, in order to convert to rpm, we apply the two conversion factors above, and so we get:

$\omega = 1.28 \frac{rad}{s}\cdot \frac{60 s/min}{2\pi rad/rev}=12.2 rev/min$

So, the cd is turning at 12.2 rpm.

You might be interested in

Base your answer(s) to the following question(s) on the information and diagram

klio [65]

Done I don't know answer of this question or this photo is the answer can you tell me

8 0

2 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

b)

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

c)

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

3 years ago

Four pairs of objects have the masses as described below, along with the distances between

lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

The gravitational force is defined as

$F=G\frac{m_{1} m_{2} }{r^{2} }$

By definition, the gravitational force depends directly on the product of the masses and indirectly on the distance between the masses, which means the further they are, the less gravitational force would be. And, the greater the masses, the greater the gravitational force.

Among the options, the pair that would have the greatest gravitational force is Mass of 1 Kg and 2 Kg, with 1 meter between them.

Notice that the last choice includes the same masses but with a greater distance between them, that means it would be a weaker graviational force.

Therefore, the right answer is the second choice.

7 0

3 years ago

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

3 years ago