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3 years ago

What do you mean by average velocity

Physics

1 answer:

mina [271]3 years ago

7 0

Answer:

Here is the answer. Hope this helps you!

Explanation:

Average velocity is the sum of initial and final velocity divided by 2. I t is the same as total Displacement divided by total time. Average velocity is calculated when the body is in non-uniform motion (also when total displacement and time is not given). The formula is as follows:

u + v/2 = $v_{av}$

Average velocity = Average speed

during motion in a straight line. therefore, the above mentioned formula can be used for calculating average speed as well, when the direction is one and only the same, that is, during motion in a straight line. The S.I unit remains the same-

m/s.

Since velocity is vector, average VELOCITY is also vector. However, Average SPEED is scalar as speed is scalar. both can be equal only when the distance = displacement and when they are following the same direction of motion.

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In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=

swat32

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

<h3>Where is the electric potential, when the particle moved?</h3>

The charge field system's electric potential energy rose. The particle experiences an electric force that is directed against the x-axis. It is pushed uphill by an outside force, which raises the potential energy.

When a charge to be moved against an applied electric field, electric potential energy is needed. A charge must be moved through a stronger electric field with more energy than it would require to carry it via a weaker electric field.

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

The electric potential energy of the charge field system:

(a) increase
(b) remain constant
(c) decrease
(d) change unpredictably

The correct option is a).

To learn more about electric potential, refer to:

brainly.com/question/21808222

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7 0

2 years ago

(ii) two locomotives approach each other on parallel tracks. each has a speed of 155 km/h with respect to the ground. if they ar

IgorLugansk [536]

You have to find an equation that would relate the two motions of the locomotives. When they meet at a certain point after being 8.5 km apart initially, then that means that their individual distances traveled is equal to 8.5 The solution is as follows:

Distance = speed*time
Total distace = 8.5 = 155t + 155t
Solving for t,
t = 0.027 hour or 98.71 seconds

4 0

4 years ago

Pam has a mass of 42.4 kg and she is at rest on smooth, level, frictionless ice. Pam straps on a rocket pack. The rocket supplie

fredd [130]

Answer:

F = 2349.6 N

Explanation:

We can solve this exercise using the relationship of momentum and momentum

I = Δp

I = F t

As the woman accelerates at a distance of 29.1 m to go from rest to 56.8 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v / 2x

a = 56.8 2/2 29.1

a = 55.43 m / s²

Let's look for the time you need to get this speed

v = v₀ + a t

t = v / a

t = 56.8 / 55.43

t = 1,025 s

Let's clear the average force momentum from the momentum

F t = m v- m v₀

F = mv / t

F = 42.4 56.8 / 1.025

F = 2349.6 N

3 0

3 years ago

A stone is thrown in a vertically

Lemur [1.5K]

Answer:

1.25 m

0.5 s

Explanation:

Given:

v₀ = 5 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (5 m/s)² + 2 (-10 m/s²) Δy

Δy = 1.25 m

Find: t

v = at + v₀

(0 m/s) = (-10 m/s²) t + (5 m/s)

t = 0.5 s

6 0

4 years ago

- How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 *10^

Sunny_sXe [5.5K]

Answer:

It will take 0.01 s or 10 ms

Solution:

As per the question:

Length of the packet, L = 1,000 bytes = $1000\times 8 = 8000 bits$

Distance, d = 2500 km = $2.5\times 10^{6}\ m$

Speed of propagation, s = $2.5\times 10^{8}\ m/s$

Transmission rate, R = 2 Mbps

Now,

Propagation time, t can be calculated as:

$t = \frac{d}{s} = \frac{2.5\times 10^{6}}{2.5\times 10^{8}} = 0.01\ s$

t = 10 ms

In general, propagation time, t is given by:

$t = \frac{link\ distance}{Propagation\ speed}$

No, this delay is independent of the length of the packet.
No, this delay is independent of the rate of transmission.

3 0

4 years ago

What do you mean by average velocity​

What do you mean by average velocity