Answer:
17640J
Explanation:
Give mass m= 60kg
Height h =30m
Since the student is standing on the edge of a cliff , acceleration due to gravity g is 9.8m/s^2
The student’s gravitational potential energy can be found by the formula
mgh
That’s 60 x 9.8 x 30
= 17640J
I believe it's C. The wagon does not move move because the force you apply to the wagon is equal to the force it applies to you. If the maximum friction force is greater than that force the wagon will not move.
Answer:
v = 3.24 m/s
Explanation:
Since we don't have time, we can use the formula;
(Final distance - initial distance)/time = (initial velocity + final velocity)/2
Thus;
(x_f - x_i)/t = ½(v_xi + v_xf)
We are given;
x_i = 0 m
x_f = 5 m
v_xi = 0 m/s
v_xf = 5 m/s
Thus, plugging in the relevant values;
(5 - 0)/t = (0 + 5)/2
5/t = 5/2
t = 2 s
Using Newton's first law of motion, we can find the acceleration.
v = u + at
Applying to this question;
5 = 0 + a(2)
5 = 2a
a = 5/2
a = 2.5 m/s²
To get the speed, in m/s, of the block when it had traveled only 2.1 m from the top, we will use the formula;
v² = u² + 2as
v² = 0² + 2(2.5 × 2.1)
v² = 10.5
v = √10.5
v = 3.24 m/s
Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 = 
cos 30 = 
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m
(2)
a) we substitute in 1
T = 
T = 
T = 2.26 N
b) from equation 2
v² = 
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² = 
v² = 
For the problem let us take L = 1 m
let's calculate
v = 
v = 1.68 m / s
Answer:
say a lighter car is going 80mph and a heavier car is going the same speed who do you think will have more damage i will say the heavier car
Explanation:
