Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
Answer:
The velocity of the ball after 5 seconds will be 49 m/s
Explanation:
<em>v = final velocity</em>
<em>u = initial velocity</em>
<em>g = acceleration due to gravity</em>
<em>t = time</em>
Initial velocity of the ball = 0 (As the ball is dropped from rest )
Acceleration due to gravity = 9.8 m/s
Time taken = 5 sec
As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question
Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v
Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.
Velocity of the ball after 5 seconds
Distance covered by the ball in 5 sec
To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,
The electric field is
Here,
k = Coulomb's Constant
Q = Charge
R = Distance (At this case from the center of mass of the earth to the surface)
Rearranging to find the charge,
Replacing,
Since the electric field is directed towards the center of earth, the charge is negative.
PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore
Replacing,
Solving for q,
PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is
PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.
M, mass=84 kg
height, h=3.9m
gravity, g= 9.8m/s2
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 84•9.8•3.9J
W= 3210.48
Therefore the answer will be 3210.48J.
Answer:
Explanation:
We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.
