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2 years ago

10

a girl pushes a cart to the left with a 100-N force. A boy pushes it to the right with a 50-N force. what is the net force exert

ed by the girl and the boy?

1 answer:

NNADVOKAT [17]2 years ago

3 0

Answer:

50 to the left

Explanation:

150-100: 50 to the left

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An object has a mass of 20 kg and is placed 2 m above the ground. What is its gravitational potential energy? (PEg)

Deffense [45]

Answer:

<h2>392 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 20 × 9.8 × 2

We have the final answer as

<h3>392 J</h3>

Hope this helps you

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3 years ago

The equation for free fall at the surface of a celestial body in outer space (s in meters, t in seconds) is sequals10.04tsqua

Sindrei [870]

Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

$s=\frac{1}{2}at^2$

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

$s=10.04 t^2$

this means that

$\frac{1}{2}g = 10.04$

so the acceleration of gravity on the body is

$g=2\cdot 10.04 = 20.08 m/s^2$

The velocity of an object in free fall starting from rest is given by

$v=gt$

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

is

$t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s$

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3 years ago

What is a scale on a map?

Oksi-84 [34.3K]

A map is almost always smaller than the place it describes. If a map of the US were drawn in its actual size, it would be 3,000 miles wide, and very difficult to fold. ... The scale of the map is the ratio of a distance on the map to the same distance on the real thing. ... If the map scale is 1 : 50000, then 1 foot on the map shows things that are actually spread over 50000 feet in the real city or field.

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4 years ago

Two transverse waves travel along the same taut string inopposite directions. the waves are described by following equations use

Umnica [9.8K]

Answer: y'=2Asin(kx)cos(wt)

Explanation:

Let y1=A sin (kx + wt) be the first wave

y2=A sin (kx - wt) be the second wave in the opposite direction (which we showed by putting a negative sign between the terms kx and wt)

Please do note that both wave have the same attributes (that's Amplitude, wave number and angular frequency) because they are formed on the same medium by the same source just that their directions are opposite.

By super imposing these 2 waves, we have a resulting singular wave representing both wave (law of superimposition) with a resulting value of vertical displacement y'.

Thus y' = y1 + y2.

Let us do the math.

y'=A sin (kx + wt) + A sin (kx - wt)

By factoring A out, we have that

y' = A [ sin (kx + wt) + sin (kx - wt)]

For simplicity let us use the substitution

Let (kx + wt) = a and (kx - wt) =b

Hence we have that

y' = A [sin a + sin b].

From trigonometric ratio

sin a + sin b = 2sin[(a+b)/2] * cos [(a - b)/2]

By recalling that (kx + wt) = a and (kx - wt) =b

sin a + sin b = 2sin [(kx +wt +kx-wt) /2] * cos [(kx +wt - (kx-wt))/2]

Thus we have that

sin a + sin b = 2sin [(kx+wt+kx-wt)/2] * cos[(kx+wt-kx+wt)/2]

By collecting like terms in the bracket we have that

sin a + sin b = 2sin[2kx/2] * cos [2wt/2]

By dividing

sin a + sin b = 2sin(kx) cos(wt)

Now let us get the final resultant vertical displacement (y')

Recall that

y' = A [sin a + sin b]. and we already deduced that

sin a + sin b = 2sin(kx) cos(wt)

Finally,

y' = A [2sin(kx) cos(wt)] which is

y'=2Asin(kx)cos(wt)...... Final answer

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4 years ago

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

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2 years ago