Answer:
The output power the weightlifter is 2916.67 W.
Explanation:
Given;
weight lifted, W = 700 N
height the weight is lifted, h = 2.5 m
time taken to lift the weight, t = 0.60 s
The output power the weightlifter is calculated as;
Power = Energy applied / time taken
Energy applied = weight lifted x height the weight is lifted
Energy applied = 700 x 2.5
Energy applied = 1750 J
Power = 1750 / 0.6
Power = 2916.67 J/s = 2916.67 W.
Therefore, the output power the weightlifter is 2916.67 W.
Answer: In a logical Pace forum subject to the distance
Explanation:
Answer:
The distance away the center of the earthquake is 1083.24 km.
Explanation:
Given that,
Speed of transverse wave = 9.1\ km/s
Speed of longitudinal wave = 5.7 km/s
Time = 71 sec
We need to calculate the distance of transverse wave
Using formula of distance

....(I)
The distance of longitudinal wave
....(II)
From the first equation

Put the value of t in equation (II)




Hence, The distance away the center of the earthquake is 1083.24 km.
Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
Answer:
a.After
second Mr Comer's speed

b.Distance travelled by Mr.Comer in
seconds

Explanation:
a. Lets recall our first equation of motion 
Now we know that
,
and

Plugging the values we have.




Then Mr.Comer's speed after
sec

b.
Lets find the distance and recall our third equation of motion.

So
distance covered.
Dividing both sides with 2a we have.

Plugging the values.


So Mr.Comer will travel a distance of
.