Answer:
A = m³/s³ = [L]³/[T]³ = [L³T⁻³]
B = m³s = [L³T]
Explanation:
We have the equation:
V = At³ + B/t
where, the dimensions of each variable are as follows:
V = m³ = [L]³
t = s = [T]
substituting these in equation, we get:
m³ = A(s)³ + B/s
for the homogeneity of the equation:
A(s)³ = m³
<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>
Also,
B/s = m³
<u>B = m³s = [L³T]</u>
ANSWER:
F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.
STEP-BY-STEP EXPLANATION:
F(h) is Horizontal Force = 200 N
V is Speed = 2.4 m/s
The total weight increase by 42%
coefficient of rolling friction decrease by 19%
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
F(h) = F(f)
F(h) = mg* u
m is mass
g is gravitational acceleration = 9.8 m/s^2
200 = mg*u
Since weight increases by 42% and friction coefficient decreases by 19%
New weight = 1+0.42 = 1.42 = (1.42*m*g)
New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u
F(h) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
Answer:
4.37 * 10^-4 J
Explanation:
Energy stored :
mgΔl / 2
m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m
Δl = mgl / πr²Y
Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m
Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10
Δl = 98 / 3.5 * π * 10^6
Δl = 0.00000891267
Energy stored :
mgΔl / 2
(10 * 9.8 * 0.00000891267) / 2
= 0.00043672083 J
4.37 * 10^-4 J
An object in motion stays in motion while an object at rest stays at rest.
a = 7.8 m/s^2
Explanation:
Let Fnet = net force = ma
m = mass of the skydiver
a = acceleration caused by Fnet
W = weight = mg
f(air) = frictional force due to air resistance
Fnet = W - f(air)
= (100 kg)(9.8 m/s^2) - (200 N)
= 780 N
Therefore, the acceleration of the skydiver due to Fnet is
a = Fnet/m
= (780 N)/(100 kg)
= 7.8 m/s^2
