Answer:
The magnitude of the static frictional force is 1200 N
Explanation:
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x 
= τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N
Answer:
352,088.37888Joules
Explanation:
Complete question;
A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.
A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)
Chane in potential energy is expressed as;
ΔGPH = mgΔH
m is the mass of the hiker
g is the acceleration due to gravity;
ΔH is the change in height
Given
m = 53kg
g = 9.8m/s²
ΔH = 2574-350 = 2224ft
since 1ft = 0.3048m
2224ft = (2224*0.3048)m = 677.8752m
Required
Gravitational potential energy
Substitute the values into the formula;
ΔGPH = mgΔH
ΔGPH = 53(9.8)(677.8752)
ΔGPH = 352,088.37888Joules
Hence the gravitational potential energy is 352,088.37888Joules
Answer:
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Explanation:
Given;
Initial length L1 = 1.2m
Initial temperature T1 = 27°C
Final temperature T2 = 0.0°C
Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C
The change i length ∆L;
∆L = L2 - L1
L2 = L1 + ∆L ...........1
∆L = xL1(∆T)
∆L = xL1(T2 - T1) ......2
Substituting the given values into equation 2;
∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)
∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)
∆L = -6.156 × 10^-4 m
From equation 1;
L2 = L1 + ∆L
Substituting the values;
L2 = 1.2 m + (- 6.156 × 10^-4 m)
L2 = 1.2 m - 6.156 × 10^-4 m
L2 = 1.1993844 m
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Yeah, like a stop sign. We know it is most likely a stop sign since it has 8 sides or is an octagon.
I hope I helped!
Answer:
35.3 N
Explanation:
U = 0, V = 0.61 m/s, s = 0.39 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0.61 × 0.61 = 0 + 2 × a × 0.39
a = 0.477 m/s^2
Force = mass × acceleration
F = 74 × 0.477 = 35.3 N
