A mass of 10kg is a point a on a table is moved to a point B.if the line joining A and is horizantal what is the done on the object by the work done on the object by the gravitational force!?

!!!...explain your answer...!!!​

Answer: 590 MW

Water flows over a section of Niagara Falls at the rate of 1.2 × 106 kg/s and falls 50 m. How much power is generated by the falling water? = 5.9 × 108 W = 590 MW , where 'MW' represents megawatts.

Explanation:

Answer:

I literally just learned this last week and if I remember correctly it is Faraday's Law of Induction.

Explanation: Hope this helps also I hope you have/had an amazing day today<3

The angular acceleration of the object can be obtained from the solution as  432 cm^2/sec from the mean position.

We know that the simple harmonic motion is the kind of motion that the acceleration of the motion is such that the motion is directed towards the center of the motion. The implication of this is that the motion of the object would always tend towards the center of the motion of the object.

Then we would have to apply the formula that involves the angular velocity and the distance that is covered and thus we have to write the formula as follows;

α =  ω^2y

α = acceleration

ω = angular velocity

y = distance covered

Then we have;

α =  (12)^2 * 3

= 432 cm^2/sec

Thus the acceleration directed to the center is  432 cm^2/sec.

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Answer:

Density=10.50kg/m³

Explanation:

[tex]Solution,\\Height(h)=0.5m\\Radius(r)=0.4m\\Now, \\Volume=\pi r^{2} h\\\\Volume=\pi *(0.4m)^{2} *0.5m\\\\Volume=0.251327m^{3} \\\\Again,\\\\Density=\frac{mass}{volume} \\\\Density=\frac{2.640kg}{0.251327} \\\\Density=10.50kg/m^{3}[/tex]

The density of the cylindrical rod is 10.516 kg/m³.

To calculate the density of the cylindrical rod, we can use the formula:

[tex]\rm \[\text{{Density}} = \frac{{\text{{Mass}}}}{{\text{{Volume}}}}\][/tex]

The mass of the rod is given as 2.640 kg. To find the volume, we can use the formula for the volume of a cylinder:

[tex]\rm \[\text{{Volume}} = \pi \times \text{{radius}}^2 \times \text{{height}}\][/tex]

Substituting the given values:

[tex]\[\text{{Volume}} = \pi \times (0.4 \, \text{{m}})^2 \times 0.5 \, \text{{m}}\][/tex]

Calculating the volume:

[tex]\[\text{{Volume}} = \pi \times 0.16 \, \text{{m}}^2 \times 0.5 \, \text{{m}}\]\\\\text{{Volume}} = 0.251 \, \text{{m}}^3\][/tex]

Now we can substitute the mass and volume values into the density formula:

[tex]\[\text{{Density}} = \frac{{2.640 \, \text{{kg}}}}{{0.251 \, \text{{m}}^3}}\][/tex]

Calculating the density:

[tex]\[\text{{Density}} = 10.516 \, \text{{kg/m}}^3\][/tex]

Therefore, the density of the cylindrical rod is 10.516 kg/m³.

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Answer:

True

Explanation:

In direct current (DC), the electric charge (current) only flows in one direction

So this true

Answer:

5,878,625,370,000 miles or 5.87 Trillion miles

Explanation:

The result: One light-year equals 5,878,625,370,000 miles (9.5 trillion km).

One light-year equals 5,878,625,370,000 miles (9.5 trillion km)

explanation: To find the distance of a light-year, you multiply this speed by the number of hours in a year (8,766).

Answer:

If the charge on an electron be 1.6X10^-19 C, find the approximate number of electrons in 1 C. ANSWER: - As it is given that   1.6 X 10-19charge is of 1 electron, So 1 C charge is of = 1/1.6X10-19electron Now number of electrons = 1019/1.6 = (100 X 1018) / 16

Statement:

Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job.

To find out:

The displacement of the window when she pulled it.

Solution:

Answer:

She pulls the window by 0.83 m.

Hope you could understand.

If you have any query, feel free to ask.

Answer: uhhhhhhhhhhhhhhhhhhhh im responding for points

Explanation:

lololololol sorry

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

Answer:

hydrosphere

Explanation:

the frozen part of the hydrosphere has its own name,the cryosphere

The horizontal and vertical component of vector A is -3 and 0 respectively.

The horizontal and vertical component of vector B is 0 and 3 respectively.

The horizontal and vertical component of vector C is 0 and 6 respectively.

The horizontal and vertical component of vector D is 4 and 0 respectively.

The horizontal and vertical component of vector E is 3.54 and 3.54 respectively.

The horizontal and vertical component of vector F is -3.54 and -3.54 respectively.

The horizontal and vertical component of each vector is calculated as follows;

[tex]A_x = Acos (\theta) = 3 \times cos(180) = -3\\\\A_y = A sin(\theta) = 3 \times sin(180) = 0[/tex]

[tex]B_x = 3 \times cos(90) = 0\\\\B_y = 3 \times sin(90) = 3[/tex]

[tex]C_x = 6 \times cos(90) = 0\\\\C_y = 6 \times sin(90) = 6[/tex]

[tex]D_x = 4 \times cos(0) = 4\\\\D_y = 4 \times sin(0) = 0[/tex]

[tex]E _x = 5 \times cos (45) = 3.54 \\\\E_y = 5 \times sin(45) = 3.54[/tex]

[tex]F_x = 5 \times cos(225) = -3.54\\\\F_y = 5 \times sin(225) = -3.54[/tex]

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a. Between the time the ball is first dropped and the first time it hits the ground, the ball's vertical velocity v at time t is given by

v = -gt

and its height y is

y = 2.00 m - 1/2 gt²

Solve the second equation for time t when the height is y = 0 (corresponding to the moment the ball first hits the ground) :

0 = 2.00 m - 1/2 gt²

1/2 gt² = 2.00 m

gt² = 4.00 m

t² ≈ 0.408163 s²

t ≈ 0.638877 s ≈ 0.639 s

Solve for v in the first equation at time this t :

v = -gt

v ≈ -6.26099 m/s ≈ -6.26 m/s

(That is, the ball hits the ground with a downward velocity of 6.26 m/s.)

b. The ball loses some energy after hitting the ground, and when it bounces back up, so it does so with a smaller upward initial velocity v' ("v prime").

At its maximum height h, the ball has zero vertical velocity. It's in freefall, so acceleration is constant. Then

0² - (v')² = -2gh

Solve for v' using the fact that h = 1.10 m after the first bounce:

(v')² = 2g (1.10 m)

(v')² = 21.56 m²/s²

v' ≈ 4.64327 m/s ≈ 4.64 m/s

We have that the Force  is mathematically given as

F=170.833N

Question Parameters:

Generally the equation for the Force   is mathematically given as

[tex]F=\frac{T}{d}[/tex]

Therefore

F=41/0.24

F=170.833N

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Answer: Hope this helps!!!!

Explanation:

Answer:

D. is the correct answer.

Explanation:

We can use sin cos tan method to solve Mx:

[tex]\frac{Mx}{5}=cos(37)[/tex]

[tex]Mx= cos(37) * 5[/tex]

[tex]Mx = 4[/tex]

So Mx should be 4 and head to right. as its positive.

We can use sin cos tan method to solve My:

[tex]\frac{My}{5} = sin(37)[/tex]

[tex]My = sin(37) * 5[/tex]

[tex]My = 3[/tex]

So My should be 3 and head to upwards. as its positive.

Final conclusion: D. is the correct answer.

According to Ptolemy's model, he, too, believed in a geocentric Universe and that the planets and stars were perfect spheres, though Earth itself was not.

He further thought that the movements of the planets and stars must be circular since they were perfect and, if the motions were circular, then they could go on forever.

Shooting stars are very different from comets, although the two can be related. A Comet is a ball of ice and dirt, orbiting the Sun (usually millions of miles from Earth). ... A shooting star on the other hand, is a grain of dust or rock (see where this is going) that burns up as it enters the Earth's atmosphere.

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We have that the Velocities is mathematically given as

a)v=2.8e5m/s

b)v_2=3.11e5m/s

From the question we are told

Generally the equation for the workdone   is mathematically given as

a)

[tex]-qEr=0.5m(u/2)^2\\\\Therefore\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31*v^2/4[/tex]

v=2.8e5m/s

b)

applying work

[tex]-qEr=0.5m(u/2)^2\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31(v_2^2-7.75e10)[/tex]

v_2=3.11e5m/s

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Answer:

laser pointer does not make use of total internal reflection

Explanation:

some car crashes only produce minor injuries because the car crash isn't that bad.

The distance the ball would have traveled is 1,066.72 m.

The relative velocity of two objects moving in the same or opposite direction can be used to determined the time of motion when the two objects move to a certain distance.

[tex](V_a + V_b) t = d\\\\(3 + 3 ) t =400\\\\t = \frac{400}{6} \\\\t = 66.67 \ s[/tex]

[tex]d = Vt\\\\d = 16\ m/s \ \ \times \ \ 66.67 \ s\\\\d = 1,066.72 \ m[/tex]

Thus, the distance the ball would have traveled is 1,066.72 m.

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Answer:

Approximately [tex]50.5\; {\rm g}[/tex] (rounded to three significant figures.)

Explanation:

Let [tex]m_\text{p}[/tex] and [tex]m_\text{a}[/tex] denote the mass of the pineapple and the apple, respectively.

Let [tex]r_{\text{p}}[/tex] denote the distance between the pineapple and the pivot. Let [tex]r_{\text{a}}[/tex] denote the distance between the apple and the pivot.

Because of gravity, both the pineapple and the apple would exert a normal force on the seesaw. The magnitude of that force is equal to the weight of the fruit. Let [tex]g[/tex] denote the gravitational field strength.

Since these two forces are perpendicular to the seesaw, the magnitude of the torque exerted by the pineapple and the apple would be:

For the seesaw to maintain a rotational equilibrium, these two torques need to balance each other. Thus:

[tex]m_{\text{p}} \, g\, r_{\text{p}} = m_{\text{a}} \, g\, r_{\text{a}}[/tex].

Rewrite and simplify this equation to find an expression for the unknown mass of this apple, [tex]m_{\text{a}}[/tex]:

[tex]\begin{aligned}m_{\text{a}} &= \frac{m_{\text{p}}\, g\, r_\text{p}}{g\, r_\text{a}} \\ &= \frac{m_{\text{p}}\, r_{\text{p}}}{r_\text{a}}\end{aligned}[/tex].

Substitute in the values [tex]m_\text{p} = 121\; {\rm g}[/tex], [tex]r_\text{p} = 1.23\; {\rm m}[/tex], and [tex]r_{\text{a}} =2.95\; {\rm m}[/tex] and evaluate:

[tex]\begin{aligned}m_{\text{a}} &= \frac{m_{\text{p}}\, r_{\text{p}}}{r_\text{a}} \\ &= \frac{121\; {\rm g} \times 1.23\; {\rm m}}{2.95\; {\rm m}} \\ &\approx 50.5\; {\rm g}\end{aligned}[/tex].

Thus, the mass of the apple should be approximately [tex]50.5\; {\rm g}[/tex].

Answer:

h=4+85*time - 1/2 32*time^2

16time^2-85*time-106=0

use the quadratic formula to solve for time (notice two solutions, one going up, one going back down).

time=(85+-sqrt(85^2+4*16*106) )/32

Explanation:

Recheck my work just in case I'm not good at this type of class but I wanted to help.

a. At maximum height h, the baseball has zero vertical velocity, so that with initial velocity v we have

[tex]0^2-v^2 = -2gh \implies h = \dfrac{v^2}{2g}[/tex]

Given that [tex]v = 30 \frac{\rm m}{\rm s}[/tex], the ball reaches a height of

[tex]h = \dfrac{\left(30\frac{\rm m}{\rm s}\right)^2}{2g} \approx \boxed{46\,\mathrm m}[/tex]

b. At max height, the baseball's vertical velocity is such that

[tex]0=v-gt[/tex]

For [tex]v = 30 \frac{\rm m}{\rm s}[/tex], we find

[tex]gt=30\dfrac{\rm m}{\rm s} \implies t = \dfrac{30\frac{\rm m}{\rm s}}g \implies t \approx 3.1 s[/tex]

which is half the time it spends in the air, making the total time about 6.1 s.

c. We saw in part (a) that

[tex]h = \dfrac{v^2}{2g}[/tex]

If we halve the initial speed and replace [tex]v[/tex] with [tex]\frac v2[/tex], we get

[tex]h = \dfrac{\left(\frac v2\right)^2}{2g} = \dfrac14 \cdot \dfrac{v^2}{2g}[/tex]

which means the max height is reduced by a factor of 1/4.

In part (b), we found the time to max height is

[tex]t = \dfrac{v}{g}[/tex]

and halving the initial speed gives

[tex]t = \dfrac{\frac v2}g = \dfrac12 \cdot \dfrac{v}{g}[/tex]

That is, the time to max height is reduced by a factor of 2 when the speed is halved, and so the overall time in the air is reduced by a factor of 2.

A wooden log is displaced to a distance of 20m in 10 seconds by applying 500N effort . Calculate the workdone and power...

Solution,

displacement = 20 m

time = 10 sec

force = 500 N

work done = ?

power = ?

Now ,

work done = f × s

= 500 N × 20 m

= 10000 j

Now ,

[tex]power \: = \frac{w}{t} \\ = \frac{10000j}{10sec} \\ = 1000W[/tex]

~nightmare 5474~

The work done and the power is 10000 J and 1000 W respectively.

To calculate the work done by the wooden log, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

To calculate the power, we use the formula below.

Where:

Given:

Substitute the values above into equation 2

Hence, the work done and the power is 10000 J and 1000 W respectively.
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Answer:

Im Pretty Sure Its 15.7m/s

Explanation:

Hope This Helps tell Me if im wrong