<u>Answer:</u>
<em>Instantaneous velocity is equal to speed of the object at that particular instant.</em>
<u>Explanation:</u>
Instantaneous velocity is the velocity of the object at that particular instant. It is also equal to speed of the object at that instant. It can be calculated by drawing a tangent to the position-time graph at that point and finding the tangent’s slope.
The first option ‘The ratio of change in position to the time interval during that change’ gives the average velocity of an object and not speed. Similarly the second option ‘the absolute value of the slope of position time graph’ gives the average speed.
Answer:
(a) 161.57 N
(b) 0.958 m/s^2
Explanation:
Force applied, F = 220 N
mass of crate, m = 61 kg
μ = 0.27
(a) The magnitude of the frictional force,
f = μ N
where, N is the normal reaction
N = m x g = 61 x 9.81 = 598.41 N
So, the frictional force, f = 0.27 x 598.41
f = 161.57 N
(b) Let a be the acceleration of the crate.
Fnet = F - f = 220 - 161.57
Fnet = 58.43 N
According to newton's second law
Fnet = mass x acceleration
58.43 = 61 x a
a = 0.958 m/s^2
Thus, the acceleration of the crate is 0.958 m/s^2.
Answer:
"How does the volume of a gas kept at constant pressure change as its temperature is increased?"
Explanation:
One possible question can be:
"How does the volume of a gas kept at constant pressure change as its temperature is increased?"
The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:
Or also written as
By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.
Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus
ΔL = 1 m
Thus, the cable stretches by 1 m.
Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
