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3 years ago

The main difference between distance and displacement is tat

Physics

2 answers:

vladimir2022 [97]3 years ago

8 0

Answer:

Distance is the length of the path taken by an object whereas displacement is the simply the distance between where the object started and where it ended up.

Explanation:

Hope this helps! Good Luck on the rest of your assignments! :)

melomori [17]3 years ago

6 0

Answer:

Distance is quite literally the total amount of distance you have covered. Displacement is the total displacement from your origin. For example, if you were to move 5m north then 2m south, your total displacement from your origin would be 3m north.

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Calculate the mass of an 86kg object on the moon

Vikentia [17]

If an object that is 86kg on the moon then that is the answer, 86kg.

3 0

3 years ago

A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr

7nadin3 [17]

Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88

Hope this helps :)

7 0

3 years ago

How to solve number 10

Vinil7 [7]

The 'formulas' to use are just the definitions of 'power' and 'work':

Power = (work done) / (time to do the work)

and

Work = (force) x (distance) .

Combine these into one. Take the definition of 'Work', and write it in place of 'work' in the definition of power.

Power = (force x distance) / (time)

From the sheet, we know the power, the distance, and the time. So we can use this one formula to find the force.

Power = (force x distance) / (time)

Multiply each side by (time): (Power) x (time) = (force) x (distance)

Divide each side by (distance): Force = (power x time) / (distance).

Look how neat, clean, and simple that is !

Force = (13.3 watts) x (3 seconds) / (4 meters)

Force = (13.3 x 3 / 4) (watt-seconds / meter)

Force = 39.9/4 (joules/meter)

<em>Force = 9.975 Newtons</em>

Is that awesome or what !

6 0

3 years ago

A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. On the same track, a 75-gram stainless steel ball is

elena-s [515]

Answer:

4.2 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v

315 g m/s − 525 g m/s = -525 g m/s + (75 g) v

315 g m/s = (75 g) v

v = 4.2 m/s

5 0

4 years ago

Read 2 more answers

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago