If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be
Average acceleration: [final velocity - initial velocity ] /Δ time
Instantaneous acceleration = d V / dt =slope of the velocity vs t graph
If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.
If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.
If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.
Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.
That is why the only true option and the answer is the option D. only at constant accelerations.
Answer:
False
Explanation:
Please see the attached file
Answer:
if u are caught by the jews u die
Explanation:
The heat lost by the water will be equivalent to the energy gained by the alcohol. Thus:
maCaΔT = -mwCwΔT
400 x 2.64 x (T - 10) = 400 x 4.186 x (88 - T)
T = 57.8 °C
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
