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2 years ago

Which BEST describes the difference between speed and velocity?

Physics

2 answers:

kirill [66]2 years ago

6 0

Answer : The correct option is, (D) Velocity includes rate of change and direction.

Explanation :

Speed : Speed is defined as the distance traveled by an object with respect to the time taken. It is a scalar quantity that means it tell us about the magnitude of an object not direction.

Velocity : Velocity is defined as the rate of change of position of an object with respect to the time. It is a vector quantity that means it tell us about the magnitude and direction of an object.

The only difference between the speed and the velocity is that the velocity tell us about magnitude and direction but speed tell us about magnitude only.

Hence, the correct option is, (D) Velocity includes rate of change and direction.

Sliva [168]2 years ago

3 0

The magnitude of a speed completely defines it whereas you also need to state a direction to specify a velocity.

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Longitudinal waves have _____.

Yakvenalex [24]

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3 years ago

Read 2 more answers

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

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3 years ago

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3 years ago

With an increase in temperature, the viscosity of liquids will: . (a) decrease (b) increase . (c) no change

ICE Princess25 [194]

Answer:

(a) decrease

Explanation:

Viscosity is the resistance which occur to flow of the fluid.

More the inter molecular forces between particles of the liquid, more the viscosity of liquid.

<u>Effect of temperature on viscosity:-</u>

Viscosity decreases with the increase in the temperature as forces among the particles decrease on increasing temperature. The kinetic energy of the particles of the liquid increases causing to move in more random motions and thus weaker inter molecular forces and this offer less resistance to the flow.

<u>Hence, viscosity of the liquids decrease with the increasing temperature.</u>

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2 years ago

What is the metric unit for time?

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The base unit of time in the metric and SI system is the second.

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