What is the main function of a protein within the body?
2 answers:
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Answers:
A. A car driving at steady speed on a straight and level road.
B. A car driving at steady speed up a 10∘ incline.
Explanation:
An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:
Since F=0, a=0 as well.
Let's now analyze each case.
A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.
D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.
So the only two choices which are correct are A and B.
Answer:
Both of them reach the lake at the same time.
Explanation:
We have equation of motion s = ut + 0.5at²
Vertical motion of James : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²
Vertical motion of John : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²
So both times are same.
Both of them reach the lake at the same time.
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N