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Elodia [21]
3 years ago
7

If a weight hanging on a string of length 5 feet swings through 5^\circ on either side of the vertical, how long is the arc thro

ugh which the weight moves from one high point to the next high point?
Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0
<span>First we can find the circumference of the whole circle with a radius of 5 feet. circumference = 2 pi radius circumference = (2 pi) (5 feet) circumference = (10 pi) feet From one high point to the other high point, the string moves through an angle of 10 degrees. Since a full circle is 360 degrees, this angle is 1/36 of a full circle. Therefore, the arc length is 1/36 of the whole circumference. arc length = (1/36) (circumference) arc length = (1/36) (10 pi) feet arc length = 0.873 feet</span>
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A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
We are going to focus on the gravitational force between Earth and our Moon. The Earth has a mass of about 6 x 10/24 kg and the
Basile [38]

Answer

Gravity is what holds us down on the earth's (or moon's) surface. If you were to weigh yourself on a scale on Earth and then on the moon, the weight read on the moon would be 1/6 your earth weight

7 0
3 years ago
two circular loops of wire, each containing a single turn, have the same radius of 4.0 cm and a common center. the planes of the
REY [17]

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.

We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as

B = μ₀ . I / 2r

where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.

From the question above, we know that:

r = 4 cm = 0.04 m

I = 1.7 A

By substituting the parameter, we get

B = μ₀ . I / 2r

B = 4π×10¯⁷ . 1.7 / (2.0.04)

B = 2.67 x 10¯⁵ T

Due to the perpendicular plane of loops, the total magnetic field at center will be

Btotal = √(2(B²))

Btotal = √(2(2.67 x 10¯⁵²))

Btotal = 3.78 x 10¯⁵ T

Find more on magnetic field at: brainly.com/question/7802337

#SPJ4

3 0
1 year ago
Which statement best explains acceleration?
Alexxx [7]
Acceleration is the ratio of a change in velocity  to the time over which the change happend
5 0
3 years ago
Read 2 more answers
I need help on this.​
lana [24]

Answer:

The speed change during the 45-minute trip is 20[mph]

Explanation:

When we see the speed at the 45 minutes this is 20 [mph] and at the 0 minutes the speed is 0 [mph].

Therefore the change is (20 - 0) = 20 [mph]

In the attached image we can see the different figures. In fig 1 we can see the bicycle's speed after 10 minutes when the speed becames constant.

In the fig. 2 we can find the graph when the biker stopped at 30 minutes and took a 15-minute break.

Figures 3 and 4, show the differences when a horizontal line is traced on a position vs time graph, and when the horizontal line is traced in a speed vs time graph.

For fig 3 we can conclude that the body is not moving therefore there is no velocity or acceleration. And for the fig 4, we can realize that the area under the horizontal line represents a displacement during the respective interval of time.

3 0
3 years ago
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