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Hoochie [10]
3 years ago
6

A measure of how far an object has moved from a starting point

Physics
1 answer:
Anvisha [2.4K]3 years ago
5 0
Volume??? velocity??????
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20. <br> All financial goals are long-term goals.<br> True<br> False
Firdavs [7]

Answer:

Hello todays answer should be False i am sorry if i am incorrect-Lucy

Explanation:

8 0
3 years ago
Read 2 more answers
A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
wolverine [178]
Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.

Let
a =  average acceleration,
t = time taken to attain final speed.

Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²

Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s

Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).
8 0
3 years ago
A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i
Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

5 0
3 years ago
A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis
romanna [79]

Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

v=\omega r

where \omega is the angular velocity and r is the radius.

\omega=\dfrac{v}{r}

The edge has a linear velocity of 10 m/s and the radius at the edge is 5 m.

\omega=\dfrac{10 \text{ m/s}}{5 \text{ m}} = 2 \text{ rad/s}

8 0
3 years ago
Identify trends in physical properties based on intermolecular forces
umka21 [38]

The Boiling point,melting point, surface tension and viscousity will increase while the Vapor pressure will decrease.

<h3 /><h3>What are intermolecular forces?</h3>

Intermolecular forces are the forces that bind two molecules together. Physical properties are affected by the strength of intermolecular forces

An increase in the strength of intermolecular forces increases will lead to an increase in force applied to break the barriers posed by the strength of the molecules.

This increased intermolecular strength will cause a rise in boiling point,melting point, viscousity and surface tension.

The Vapor pressure reduces with increasing intermolecular strength. Vapor pressure is the amount of vapor that is equilibrium with its own liquid or solid. Hence,with increasing intermolecular strength the amount of vapor that is in equilibrium with its own liquid will reduce.

To know more about intermolecular forces follow

brainly.com/question/13588164

#SPJ4

6 0
1 year ago
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