consider the right direction as positive and left direction as negative.
M = mass of the ball = 5 kg
m = mass of stone = 1.50 kg
= initial velocity of the ball before collision = 0 m/s
= initial velocity of the stone before collision = 12 m/s
= final velocity of the ball after collision = ?
= final velocity of the stone after collision = - 8.50 m/s
using conservation of momentum
M + m = M + m
(5) (0) + (1.5) (12) = 5 + (1.50) (- 8.50)
= 6.15 m/s
h = height gained by the ball
using conservation of energy
Potential energy gained by ball at Top = kinetic energy at the bottom
Mgh = (0.5) M
(9.8) h = (0.5) (6.15)²
h = 1.93 m
Answer:
Explanation:
According to the exercise we know the angle which the bait was released and its maximum height
To find the initial y-component of velocity we need to do the following steps:
At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0
Since the bait is released at 25º
Well, <span>v = u + a×t is the equation.</span>
<span>
v: final velocity, which is 23 m/s in this equation.</span>
<span>u: initialo velocity = 13 m/s </span>
<span>a: acceleration = ? </span>
<span>t: time = 30s
</span>
Your equation would be...
<span>23 = 13 + a×30 </span>
<span>a = (23 - 13) / 30 </span>
<span>a = 1 / 3 </span>
<span>a = 0.333 m/s</span>
#1
In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.
So here we can say let say sea lion chase the fish in time "t"
then here we have
here
d1 = distance covered by sea lion in time t
d2 = distance covered by fish in the same time t
L = distance between fish and sea lion initially = 60 m
So it will take 9 s to chase the fish by sea lion
# 2
velocity of truck on road = 25 m/s along North
velocity of dog inside the truck = 1.75 m/s at 35 degree East of North
we can write the relative velocity as
now plug in the velocity of truck in this
so it is given as
direction will be given as
so with respect to ground dog velocity is 26.44 m/s towards 2.2 degree East of North
Answer:
A torque of 102.5375 Nm must be exerted by the fireman
Explanation:
Given:
The rate of water flow = 6.31 kg/s
The speed of nozzle = 12.5 m/s
Now, from the Newton's second law we have
The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction
thus we have,
F = 6.31 kg/s x 12.5m/s
or
F = 78.875 N
Now,
The torque (T) exerted by water force about the fireman's will be
T = (F x d)
or
T = 78.875 N x 1.30 m
T = 102.5375 Nm
hence,
<u>A torque of 102.5375 Nm must be exerted by the fireman</u>
