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2 years ago

Is average speed a scalar or vector quantity? why?

1 answer:

3 0

The average speed is the distance per time ratio (a scalar quantity) Speed is ignorant of direction. Velocity is a vector quantity. It is aware of direction.

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Which phase comes after waxing crescent moon

12345 [234]

Answer:

new moon

Explanation:

6 0

2 years ago

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on

qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

Δx=760 m
V₀=87 m/s
t=13.6 s
a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$

3 0

2 years ago

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

Mekhanik [1.2K]

Answer:

$v\approx 16.956\,\frac{m}{s}$

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

$\Sigma F = f = m\cdot \frac{v^{2}}{R}$

The maximum speed is:

$v = \sqrt{\frac{f\cdot R}{m} }$

$v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }$

$v\approx 16.956\,\frac{m}{s}$

7 0

3 years ago

. Since the man is walking at a constant velocity, he has what acceleration

docker41 [41]

If he’s walking at a constant velocity there is no acceleration.

6 0

2 years ago

Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of

Veseljchak [2.6K]

As we know that as per Newton's II law we have

$F = \frac{dP}{dt}$

here we will have

$dP$ = change in momentum

$dt$ = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

5 0

2 years ago