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2 years ago

The sum of the initial and final velocity is divided by 2 to find the average.

Physics

1 answer:

4vir4ik [10]2 years ago

7 0

The average velocity is obtained by ( u + v)/2.

<h3>What is average velocity?</h3>

The term velocity refers to the rate of change of displacement with time. Velocity is a vector quantity thus it has both magnitude and direction. The velocity of a body could also be shown by the use of the vector notation.

To obtain the average velocity;

Let the initial velocity be u

Let the final velocity be v

It then follows that the average velocity = ( u + v)/2

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Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially

statuscvo [17]

Answer:

Final speed is 900.06 m/s at $0.2215^{\circ}$

Solution:

As per the question:

Mass of the first asteroid, m = $15\times 10^{3}\kg$

Mass of the second asteroid, m' = $20\times 10^{3}\kg$

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, $\theta = 20^{\circ}$

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

$\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}$

Thus applying vector addition and momentum conservation, the final velocity is given by:

$(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}$ (1)

Now,

$(m +m')v_{final} = (35\times 10^{3})v_{final}$

$(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}$

$(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}$

$2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}$

Now, substituting the suitable values in eqn (1), we get:

$v_{final} = 900.06\ m/s$

Now, the direction for the two vectors is given by:

$\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}$

$\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}$

5 0

3 years ago

A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change

lord [1]

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

$k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m$

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

$m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg$

Therefore, the mass that must be added is

$\Delta m = 1.128 - 0.500 = 0.628 kg$

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5 0

4 years ago

If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Arada [10]

The force on the fry is $0.015 N$

Explanation:

We can find the force acting on the fry by using Newton's second law:

$F=ma$

where

F is the net force on the fry

m is its mass

a is its acceleration

For the fry in this problem,

$m=3 g = 0.003 kg$

$a=5.0 m/s^2$

Therefore, the force exerted on the fry is

$F=(0.003)(5)=0.015 N$

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3 0

3 years ago

A 19.0-kg cart is moving with a velocity of 7.20 m/s down a level hallway. A constant force of -13.0 N acts on the cart and its

gulaghasi [49]

Answer:

a. -369.36J

b. -123.9J

c. 9.52m

Explanation:

From the expression for kinetic energy

K. E=1/2mv^2

Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is

K.E=1/2*19(3.6²-7.2²)

K.E= -369.36J

b. to determine the workdone by the force,we determine the distance moved.

But the acceleration is from

F=ma ,

a=f/m

a=-13/19

0.68m/s²

the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

Hence the work done is

W=force * distance

W=-13*9.52

W=-123.9J

d. the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

4 0

3 years ago

Read 2 more answers

Which type of wave forms at the boundary between air and water when you

azamat

D. Transverse wave

Hope it's correct and helps uh.

6 0

3 years ago

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