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3 years ago

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A satellites are placed in a circular orbit that is 2.44 × 107 m above the surface of the earth. What is the magnitude of the ac

celeration due to gravity at this distance?

1 answer:

Vikki [24]3 years ago

7 0

2.440
X 107. The magnitude
------------ Is 261.080 plz
17.080. Make the
00.000. Brainliest
244.000
----------------
261.080

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Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum

soldi70 [24.7K]

We Know, P = m*v
Here, m = 30 Kg
v = 5 m/s

Substitute it into the expression,
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps!

3 0

3 years ago

Read 2 more answers

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

2 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

a)

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

b)

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago

Which will remain the same for two identical books, one lying flat and the other standing on an end?

Fiesta28 [93]

Weight will remain the same for two identical books, one lying flat and the other standing on an end.

The strain at a factor internal a liquid is at once proportional to the intensity of the factor. When an item is submerged in a liquid, the intensity of its backside from the floor of the liquid is extra than that of some other a part of the item.

Archimedes' precept is the declaration that the buoyant pressure on an item is identical to the load of the fluid displaced with the aid of using the item.

Ensure your scale is on a flat, strong and stage floor. Do now no longer use your scale on carpet. When taking measurements, stand nevertheless withinside the middle of the platform till all measurements are displayed, and if feasible do now no longer circulate your scale in-among measurements.

Learn more about weight here brainly.com/question/229459

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5 months ago

A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?

posledela

Answer:

$118\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of $v= 295\; \rm m\cdot s^{-1}$ . In other words, the wave would have traveled $295\; \rm m$ in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that $295\; \rm m$ ? The wavelength of this wave $\lambda = 2.50\; \rm m$ gives the length of one wave cycle. Therefore:

$\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118$ .

That is: there are $118$ wave cycles in $295\; \rm m$ of this wave.

On the other hand, Because that $295\; \rm m$ of this wave goes through that point in each second, that $118$ wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

$f = 118\; \rm s^{-1} = 118\; \rm Hz$ .

The calculations above can be expressed with the formula:

$\displaystyle f = \frac{v}{\lambda}$ ,

where

$v$ represents the speed of this wave, and
$\lambda$ represents the wavelength of this wave.

6 0

2 years ago