Answer:
Distance 5 km, Displacement 3 km east
Explanation:
Neither they’re both moving the same . 150/3 = 50km
100/2 = 50km
therefor they’re traveling at the same rate.
Answer:
Explanation:
From work energy theorem
Work done by all forces = Change in kinetic energy
Lets take
m= mass of object
h=height from the ground surface
initial velocity of object = 0 m/s
The final velocity of object is v
Work done by gravitational force = m g . h
The final kinetic energy = 1/2 m v²
So
Work done by all forces = Change in kinetic energy
m g h = 1/2 m v² - 0
v² = 2 g h
After one day, the rate of increase in Delta Cephei's brightness is;0.46
We are informed that the function has been used to model the brightness of the star known as Delta Cephei at time t, where t is expressed in days;
B(t)=4.0+3.5 sin(2πt/5.4)
Simply said, in order to determine the rate of increase, we must determine the derivative of the function that provides
B'(t)=(2π/5.4)×0.35 cos(2πt/5.4)
Currently, at t = 1, we have;
B'(1)=(2π/5.4)×0.35 cos(2π*1/5.4)
Now that the angle in the bracket is expressed in radians, we can use a radians calculator to determine its cosine, giving us the following results:
B'(1)=(2π/5.4)×0.3961
B'(1)≈0.46
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Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '
' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.
