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3 years ago

What factor affects weight( what force causes weight to change)

Physics

1 answer:

Alex Ar [27]3 years ago

3 0

Answer:

Gravitational force affects weight, weight changes with change in gravity

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igor_vitrenko [27]

Answer:

$a=40\ m/s^2$

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2$

So, the average acceleration of badminton shuttlecock is $40\ m/s^2$ .

3 0

3 years ago

How is the pressure of a gas related to its concentration of particles?

Nat2105 [25]

Increasing the pressure of gas is like exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume.
Hope this helps!

3 0

2 years ago

Read 2 more answers

A large raindrop-the type that lands with a definite splat-has a mass of 0.0014 g and hits your roof at a speed of 8.1 m/s. a. W

DENIUS [597]

Answer

given,

mass of the drop, m = 0.0014 g

speed of the drop, u = 8.1 m/s

a) Change in momentum is equal to impulse

final velocity of the drop, v = 0 m/s

J = m ( v - u )

J = 0.0014 x 10⁻³ x ( 0 - 8.1 )

J = -1.134 x 10⁻⁵ kg.m/s

impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s

b) time, t = 0.37 m s

impact of force = ?

we know

J = F x t

1.134 x 10⁻⁵ = F x 0.37 x 10⁻³

F = 0.031 N

the magnitude of the force of the impact is equal to F = 0.031 N

5 0

3 years ago

When does air become turbulent around a thrown ball?

kherson [118]

Answer: B (on the rough side)

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8 0

3 years ago

27. (a) What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being vi

denpristay [2]

Answer:

magnification = - 30

overall magnification = -240

Explanation:

given data

Focal length of microscope objective f = 0.150 cm

Object distance from microscope objective do = 0.155 cm

magnification by eyepiece = 8 ×

to find out

What magnification is produced and overall magnification

solution

we consider here Image distance from microscope objective is = di

so that

Magnification produced by objective will be = - $\frac{di}{do}$

so we find here di by given equation that is

$\frac{1}{do} +\frac{1}{di} = \frac{1}{f}$ ..................1

$\frac{1}{di} = \frac{1}{0.150} - \frac{1}{0.155}$

di = 4.65 cm

so that magnification by object will be

magnification = - $\frac{di}{do}$

magnification = - $\frac{4.65}{0.155}$

magnification = - 30

and

overall magnification will be

overall magnification = magnification by objective × magnification by eyepiece ........................2

overall magnification = -30 × 8

overall magnification = -240

7 0

3 years ago