What’s the difference between distance and displacment

REY [17]

1: Boiling Water
2: Ice Melting
3: Hot Air Balloon
4: Frozen Material Thawing
5: Radiator
6: Steaming cup of hot coffee
7: Campfire
8: Putting wet shoes on floor vent

5 0

3 years ago

A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine o

noname [10]

Answer:

The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

Explanation:

Using the equations of motion,

When the rocket is fired from the ground,

u = initial velocity = 0 m/s (since it was initially at rest)

a = 10 m/s²

The engine cuts off at y = 0.5 km = 500 m

The velocity at that point = v

v² = u² + 2ay

v² = 0² + 2(10)(500) = 10000

v = 100 m/s

The velocity at this point is the initial velocity for the next phase of the motion

u = 100 m/s

v = final velocity = 0 m/s (at maximum height, velocity = 0)

y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?

g = acceleration due to gravity = - 9.8 m/s²

v² = u² + 2gy

0 = 100² + 2(-9.8)(y)

- 19.6 y = - 10000

y = 510.2 m = 0.510 km

So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km

Hope this helps!!!

5 0

3 years ago

Pretend you are explain net force to someone much younger than you. Explain how to calculate net force and how the net force det

jeka57 [31]

Answer:

you can tell them it is the amount of "push" on an object then you can explain using a soccer ball being kicked or a cart being pushed

Explanation:

4 0

3 years ago

Using this formula Vj = V; + at, If a vehicle starts from rest

Furkat [3]

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

$v_f=0+4.5*8$

$v_f=36\ m/s$

The final speed of the vehicle is 36 m/s

5 0

3 years ago

5. A 5.5 x10-6 C charge is located 0.28 m from a -3.5 x 10-6 C charge.

ra1l [238]

(a) The magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

Explanation:

The force acting between two charged particles separated by a distance is termed as Coloumb's force or electrostatic force. It can be termed as electrostatic force of attraction if the the force acting between the charges are oppositely charged. And it can be termed as electrostatic force of repulsion if the charges are similar or like charges.

In the present case, there is a positive and negative charge, so electrostatic force of attraction will be acting between them. As per Coloumb's law, the electrostatic force of attraction is directly proportional to the product of charges and inversely proportional to the square of distance of separation.

$F = \frac{kQq}{d^{2} }$

Here, k is the constant of proportionality which is equal to 9 × $10^{9}$ and Q, q are the two charges, d is the distance of separation.

So here Q = 5.5 × $10^{-6} C$ and q = - 3.5 × $10^{-6} C$ and d = 0.28 m

Then, $F=-\frac{9*10^{9}*5.5*10^{-6} * 3.5 * 10^{-6} }{(0.28)^{2} } = 2209.82*10^{-3}$

So the magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

3 0

4 years ago