I'm pretty sure it's D. The stars don't influence the moon's phases.
Answer:
6.32m/s
Explanation:
note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²
okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!
when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)
now enough of the talking let solve....
so the ball was dropped .ie from rest to the ground through a distance of 2m,
the formula for calculating the distance if a body moving in a straight line is given by:
S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.
from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)
solving ....
we get t to be 0.632s
to find the speed we substitute t in the equation below:
V=u+at ,but since u=0
V=at =10•0.632=6.32m/s
therefore the speed the body uses to strike the ground is 6.32m/s
Answer:
c. 48 cm/s/s
Explanation:
Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with a mass of 2M when acted upon by a net force of 2F?
from newtons second law of motion ,
which states that change in momentum is directly proportional to the force applied.
we can say that
f=m(v-u)/t
a=acceleration
t=time
v=final velocity
u=initial velocity
since a=(v-u)/t
f=m*a
force applied is F
m =mass of the object involved
a is the acceleration of the object involved
f=m*48.........................1
in the second case ;a mass of 2M when acted upon by a net force of 2F
f=ma
a=2F/2M
substituting equation 1
a=2(M*48)/2M
a=. 48 cm/s/s
Answer:
The appropriate response is "
". A further explanation is described below.
Explanation:
The torque (
) produced by the force on the dam will be:
⇒ 
On applying integration both sides, we get
⇒ 
⇒ 
⇒ ![=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]](https://tex.z-dn.net/?f=%3DpgL%5B%5Cfrac%7Bh%5E3%7D%7B2%7D%20-%5Cfrac%7Bh%5E3%7D%7B3%7D%20%5D)
⇒ 
consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
