It takes 6 sec for a stone to fall the the bottom of a mineshaft. calculate the
1 answer:
Ok so this is simple projectile motion problem.
if we have an object falling in free fall it is subject to gravity of -9.80m/s^2
so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m
so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground
hope this helps you! Thanks!!
You might be interested in
Answer:

Explanation:
Given that,
Radius, r = 2 m
Velocity, v = 1 m/s
We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

So, the magnitude of centripetal acceleration is
.
Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)
Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N
-F
towards the sister (-F) (greater force applied)
A textbook would hit the ground first
Factors:
-Textbook weighs most
-Pillow is flat and fluffy not very aerodynamic) also is very light
-Paper airplane will glide to the ground do to its wings and will hit the ground last
Answer:
The specific heat of a gas may be measured at constant pressure. - is accurate when discussing specific heat.
Explanation:
The answer to this would be B