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dimaraw [331]
3 years ago
8

Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the d

epth of the thumbtack in the bulletin board is x centimeters is given by F(x) = 1000 (1 + 2x)2 for 0 ≤ x ≤ 1 5 . Find the work W done by pushing the thumbtack all the way into the board.
Physics
1 answer:
mario62 [17]3 years ago
3 0

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

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At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la
konstantin123 [22]

Answer:

Explanation:

Velocity of sound in air at 20 degree = 343 m/s

Velocity of sound in water at 20 degree = 1470 m/s

Time taken in to and fro movement in air

=( 2 x 10) / 343 = 0.0583 s

Rest of the time  is

.171 - .0583 = .1127 s

This time is taken to cover distance in water. If d be the depth of lake

2d / velocity = time taken

2 d / 1470 = .1127

d = 82.83 m

5 0
3 years ago
The answer and how to do it
Arturiano [62]
Current = charge per second
2 Coulombs per second = 2 Amperes

Potential difference = (current)x(resistance) in volts.

That's (2 Amperes) x (2 ohms).

That's how to do it.
I think you can find the answer now.
8 0
3 years ago
Q12. How big is a Moon? How big is a Mars? What is therefore the weight of the person from Q11 on the Moon? What is the person's
shutvik [7]

Answer:

The moon is 1,079.4 mi.

Mars is 2,106.1 mi

Multiply your weight by the moon's gravity relative to earth's, which is 0.165. Solve the equation. In the example, you would obtain the product 22.28 lbs. So a person weighing 135 pounds on Earth would weigh just over 22 pounds on the moon

Being that Mars has a gravitational force of 3.711m/s2, we multiply the object's mass by this quanitity to calculate an object's weight on mars. So an object or person on Mars would weigh 37.83% its weight on earth.

Explanation:

~Hope this helps

7 0
3 years ago
A boat produced water waves of frequency 4 Hz and wavelength 2m, if it reached the shore after 2 seconds how far is the boat fro
kap26 [50]

Answer:

S = 16 m

Explanation:

Given that

The frequency of the water waves, f = 4 Hz

The wavelength of the water waves, λ = 2 m

The time the waves reached the shore, t = 2 s

The relation between the velocity, wavelength, and the frequency of the wave is given by the relation,

                                           v = f λ    m/s

Substituting the given values in the above equation,

                                           v = 4 x 2

                                              = 8 m/s

The velocity of the water waves is v = 8 m/s

The distance between the shore and boat is given by

                                            s = v x t

                                               = 8 x 2

                                               = 16 m

Hence, the distance between the boat and the shore is, s = 16 m

7 0
3 years ago
While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38° with the vertical. While driving
Natali [406]

Answer:21.33^{\circ}

Explanation:

Given

velocity of driver v_1=25 m/s w.r.t ground towards north

driver observes that rain is making an angle of 38^{\circ} with vertical

While returning v_2=25 m/s w.r.t. ground towards south

suppose u_1=velocity of rain drop relative car while car is going towards north

u_2=velocity of rain drop relative car while car is going towards south

z=vector sum of u_1 & v_1

Now from graph

\tan 38=\frac{v_1+v_2}{u_2}

u_2=\frac{25+25}{\tan 38}=64 m/s

z=\vec{u_2}+\vec{v_2}

therefore magnitude of z is given by

|z|=\sqrt{u_2^2+v_2^2}

|z|=\sqrt{64^2+25^2}

|z|=68.70 m/s

\tan A=\frac{v_2}{u_2}

\tan A=\frac{25}{64}=0.3906

A=21.33^{\circ}

Thus rain drops make an angle of 21.33^{\circ} w.r.t to ground

6 0
3 years ago
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