Answer:
7.0 m
Explanation:
Step 1: Given data
Initial speed of the ball (u): 1.8 m/s
Acceleration (a): 6.1 m/s²
Final speed of the ball (v): 9.4 m/s
Step 2: Calculate the displacement (s) of the ball
The ball is moving with a uniformly accelerated rectilinear motion. We can calculate the displacement using the following suvat equation.
v² = u² + 2 × a × s
s = (v² - u²)/2 × a
s = [(9.4 m/s)² - (1.8 m/s)²]/2 × 6.1 m/s²
s = 7.0 m
Answer:
16.25 m
Explanation:
we know that the equation pf parabola
from bellow figure the coordinate of parabola is (600,65) that is y=600 and x=65
putting the the value of y and x in the equation of parabola
k=0.0001805
now the equation is
we have to find the value of y at x=300m
so 
y=16.25 m
Answer: 12) 1.07 m/s (right) 13) 4.05 m/s 14) 73 m/s 15) 10.9 m/s
Explanation:
12) Conservation of momentum. Momentum is the produce of mass and velocity.
13(2) + 15(-5) = 13(-5) + 15v
v = 1.06666... ≈ 1.07 m/s (right)
13) 18(9) + 22(0) = 18v + 22v
v = 18(9)/40 = 4.05 m/s
14) 0.65(35) + 0.08(0) = 0.65(26) + 0.08v
v = 73.125
15) This is a bit trickier. Let's ASSUME you jump off at 7 m/s relative to the truck. Doing this, we can assume that the reference frame is moving along with the truck at 10 m/s
the conservation of momentum equation becomes
600(0) + 80(0) = 600v + 80(-7)
v = 0.9333333... m/s
adding back the velocity of the reference frame means the truck is now traveling.
10.9333333... ≈ 10.9 m/s
To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
<span>The function of an electric motor is to change electrical energy
into
kinetic energy.
Even if you want to use the electric motor to give a heavy weight
a lot of potential energy, by lifting it up onto, say, the roof of a house,
you still have to use the motor to MOVE the weight. So you're still
giving it kinetic energy as an intermediate step..
</span>
