All categories

4 years ago

explain how average speed and average velocity are related to each other for an object in uniform motion

Physics

2 answers:

Tatiana [17]4 years ago

5 0

Explanation:

Average speed and average velocity both are different quantities.Speed is a scalar quantity but on the other hand velocity is a vector quantity.Speed deals with distance but on the other hand velocity deals with displacement and we know that displace is a vector quantity that is why velocity is a vector quantity.

$Average\ speed\ =\dfrac{Total\ distance}{Total\ time}$

$Average\ velocity\ =\dfrac{Total\ displacement}{Total\ time}$

If object is in uniform motion then speed and velocity will become same.

Alex787 [66]4 years ago

3 0

Both take the average speed.... however velocity also shows direction

You might be interested in

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

8 0

3 years ago

The diagram shows a compressed spring between two carts initially at rest on a horizontal frictionless surface. Cart A has a

EastWind [94]

Answer:

momentum

Explanation:

the impulse is equal for both carts, the momentum for both carts is zero both before the string is cut and after it is cut

6 0

3 years ago

A space station worker found herself floating free 100 meters from the space station because her safety line became unhooked

yuradex [85]

Answer:

Explanation: is that suppose to be the question or are you just saying?

8 0

3 years ago

Pls can anyone solve this

gayaneshka [121]

Answer:

3 pls give me brainliest

Explanation:

8 0

3 years ago

the potential energy of a 40kg cannon ball is 14000j. How high was the cannon ball to have this much potential energy?

noname [10]

We will use the formula p = mgh p is potential energy. m is mass of object in kg g is acceleration due to gravity (9.8m/s²) h is height of the objects displacement in meters. p = mgh → mgh = p → h = p / mg p is 14000j, m is 40kg and g is 9.8 m/s² h = 14000 / 40 × 9.8 → h = 1400 / 392 → h = 35.7 Therefore , the cannonball was 35.7 meters high .

7 0

4 years ago