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3 years ago

If a 5000-kg is moving at a speed of 43 m/s, what is its momentum?

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

215000kgm/s

Explanation:

Given parameters:

Mass of the moving body = 5000kg

Velocity = 43m/s

Unknown:

Momentum = ?

Solution:

The momentum of a body is the amount of motion a body possess.

It is mathematically expressed as:

Momentum = mass x velocity

Now:

Momentum = 5000 x 43 = 215000kgm/s

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in

insens350 [35]

Answer:

659.01W

Explanation:

The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be

Wc = mass of the cab × acceleration due to gravity in m/s²

Wc = 1250 × 9.81 = 12262.5 N

but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N

Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N

the motor of the elevator will have to provide this in form of work

work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m

power provided by the motor of the elevator = workdone by the motor / time in seconds

Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W

5 0

3 years ago

Some people wish that we lived in a recollapsing universe that would eventually stop expanding and start contracting. For this t

Monica [59]

Answer:

(B) Dark energy does not exist and there is much more matter than current evidence suggests.

Explanation:

The repulsive force which is accelerating expansion of the universe is called as dark energy. Most of matter present in the universe is the dark matter of about eighty five percent.

So, a collapsing universe would not have the dark energy and there is more matter which is not the dark matter. This theory is rejected because expansion of the universe is observable.

4 0

3 years ago

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago

They don't have an Astronomy Section and I believe that this falls the closest to Physics

Ksenya-84 [330]

The only thing we know of so far that can shift light to longer wavelengths is the "Doppler" effect. If the source and the observer are moving apart, then the observer sees wavelengths that are longer than they should be. If the source and the observer are moving toward each other, then the observer sees wavelengths that are shorter than they should be. It works for ANY wave ... sound, light, water etc. The trick is to know what the wavelength SHOULD be. If you know that, then you can tell whether you and the source are moving together or apart, and you can even tell how fast. If the lines in a star"s spectrum are at wavelengths that are too long, then from everything we know right now, the star and Earth are moving apart.

5 0

3 years ago

Neglecting air, what speed does a rock thrown straight up have to be to reach the edge of our atmosphere: say 100 km? Still negl

Ugo [173]

To solve this problem it is necessary to apply the kinematic equations of movement description, specifically those that allow us to find speed and acceleration as a function of distance and not time.

Mathematically we have to

$v_f^2-v_i^2 = 2ax$

Where,

$v_{f,i} =$ Final velocity and Initial velocity

a = Acceleration

x = Displacement

From the description given there is no final speed (since it reaches the maximum point) but there is a required initial speed that is contingent on traveling a certain distance under the effects of gravity

$0 - v_i^2 = 2(9.8)(100*10^3)$

$v_i = 14*10^2m/s$

Therefore the speed which must a rock thrown straight up is 14*10^2m/s to reach the edge of our atmosphere.

The displacement and gravity traveled are the same, therefore the final speed will be the same but in the opposite vector direction (towards the earth), that is $14 * 10 ^ 2m / s$

5 0

3 years ago