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Advocard [28]
3 years ago
14

A student is bouncing on a trampoline. at her highest point, her feet are 65 cm above the trampoline. when she lands, the trampo

line sags 15 cm before propelling her back up. part a for how long is she in contact with the trampoline?
Physics
1 answer:
Ede4ka [16]3 years ago
3 0
<span>she is in contact with the trampoline for less than one second.</span>
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An object is 27.0 cm from a concave mirror of focal length 15.0 cm. find the image distance.
pantera1 [17]

The distance of the Image will be -33.75 cm

A concave mirror has an inward-curving reflecting surface that faces away from the light source. Unlike convex mirrors, a concave mirror's image forms a variety of images based on the object's proximity to the mirror.

Given that, an object placed 27 cm from a concave mirror having the focal length of 15 cm

We have to find distance of the Image

Using Mirror Formula:

1/f = 1/v + 1/u

Where,

f = focal length

v =  Image distance from the mirror

u = object distance from the mirror (concave)

Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.

1/(-15) = 1/v + 1/(-27)

1/(-15) = 1/v - (1/27)

1/v = -0.029

v = -33.75 cm

Therefore the distance of the Image will be -33.75 cm

Learn more about concave mirror here:

brainly.com/question/9816370

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8 0
2 years ago
As the distance between two point charges is tripled, the electrostatic force between the charges will become
MatroZZZ [7]
Relation between electrostatic force and distance is inverse square i.e
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Fα ----
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Hence if r is tripled, new electrostatic force will be 1/9 times old force.          
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4 years ago
Uan created a chart to help him study for a test.
yuradex [85]

Answer:

Y: transparent objects i think

6 0
3 years ago
Read 2 more answers
15 points. give me the method.
AveGali [126]

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

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3 years ago
If the sun became cooler but kept the same size it would emit
jek_recluse [69]
If the sun would become cooler having constant size, it would emit less ultraviolet light and less visible light than what it currently gives to earth. Hope this answers the question. Have a nice day. Thank you for posting here.
4 0
3 years ago
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