All categories

3 years ago

Difference between displacement and vector quantity

Physics

1 answer:

daser333 [38]3 years ago

5 0

The only 'difference' is that they are different categories.

It's like asking "What's the difference between Susie and girl ?"

Or "What's the difference between Cadillac and car ?"

Displacement <em>IS</em> a vector quantity.

You might be interested in

Body composition refers to the amount of fat and muscle that the body has.

Varvara68 [4.7K]

Body composition refers to the amount of fat and muscle that the body has. ... Muscular strength is the amount of force a muscle or group of muscles can exert. true. Muscular endurance is fueled by anaerobic energy.

3 0

3 years ago

Read 2 more answers

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago

of 72.0 units. If the mass of Object 1 is halved AND the mass of object 2 is tripled, then the new gravitational force will be u

katen-ka-za [31]

Answer:

youtg jugre contact coming call details

6 0

2 years ago

A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an

mina [271]

Answer:

μ = 0.33

Equal to 3.2 m/s²

Explanation:

Draw a free body diagram of the block. There are three forces:

Normal force N pushing up.

Weight force mg pulling down.

Friction force Nμ pushing opposite the direction of motion.

Sum of forces in the y direction.

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction.

∑F = ma

Nμ = ma

Substitute.

mgμ = ma

μ = a/g

μ = (3.2 m/s²) / (9.8 m/s²)

μ = 0.33

As found earlier, the acceleration is a = gμ. Since g and μ are constant, a is also constant, so it does not change with velocity.

5 0

3 years ago

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

2 years ago