Answer:

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 :

So, the average acceleration of badminton shuttlecock is
.
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.
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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
Answer: B (on the rough side)
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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 = - 
so we find here di by given equation that is
..................1
di = 4.65 cm
so that magnification by object will be
magnification = - 
magnification = - 
magnification = - 30
and
overall magnification will be
overall magnification = magnification by objective × magnification by eyepiece ........................2
overall magnification = -30 × 8
overall magnification = -240