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3 years ago

8

How are acceleration and speed related

2 answers:

Gre4nikov [31]3 years ago

6 0

Answer:

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing. Comment on robshowsides's post “Speed is the magnitude of velocity.

Explanation:

hope it helped tee hee

Serggg [28]3 years ago

4 0

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing. Comment on robshowsides's post “Speed is the magnitude of velocity.

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What is air pressure?

dolphi86 [110]

Answer:

A. Earth's gravity pulling down on air molecules

Explanation:

Air pressure refers to the weight of the air per unit surface area. It is the amount of gravitational force which is pulling down the molecules of air.

The common unit of air pressure is: Pascal, atm

1 atm = 101325 Pa

As the column of the air above increases, the air pressure increase. This is because with the increase in amount of air, the weight increase of the air increases. This is the reason a diver feels immense pressure in the sea and cooking takes a lot of time on hilly areas because of low air pressure.

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3 years ago

(???) Is iz allergic to peanut butter jigglys, I JIGGLE IN THE WIND!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Vadim26 [7]

Uh are u okay? G tell ur parents i-

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3 years ago

Allisa [31]

<u>We are given:</u>

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

<u>Net upward force on the rocket:</u>

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

<u>Upward Acceleration of the Rocket:</u>

From newton's second law:

F = ma

<em>replacing the variables</em>

300 = 10 * a

a = 30 m/s²

5 0

3 years ago

Muscular Strength exercises focus on high______and low___

Advocard [28]

Answer:

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3 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago