solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.
Answer:
Explanation:
height of pole = 15 ft
height of man = 6 ft
Let the length of shadow is y .
According to the diagram
Let at any time the distance of man is x.
The two triangles are similar
15 y - 15 x = 6 y
9 y = 15 x
Differentiate with respect to time.
As given, dx/dt = 4 ft/s
ft/s
For these question, it has two separate equations: 2f(a) and f(2a) .
For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation
For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.
Answer: Projectiles travel with a parabolic trajectory due to the influence of gravity. There are no horizontal forces acting upon projectiles and thus no horizontal acceleration, The horizontal velocity of a projectile is constant (a never changing in value.
Explanation: I hoped that helped yall !!
Answer:
<h2>9.3kN</h2>
Explanation:
Step one:
given data
mass of bullet= 0.02kg
speed v=700m/s
time taken =1.5ms= 0.0015 seconds
Step two:
we know that from the first law
F=ma-----1 first law of motion
also, we know that
a=v/t----2
put a=v/t in equation 1 we have
F=mv/t
Step three:
substitute our given data to find force
F=0.02*700/0.0015
F=14/0.0015
F=9333.33N
F=9.3kN
<u>The average force exerted is 9.3kN</u>
