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2 years ago

Acceleration is a change in?

Physics

2 answers:

vodomira [7]2 years ago

5 0

Acceleration is the change in velocity.

konstantin123 [22]2 years ago

3 0

Acceleration is any change in speed or direction of motion.

Speed and direction together comprise 'velocity'.

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4. A woman releases one egg every month for 37 years. Calculate how many

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so, 444 eggs would have been released in 37yrs

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2 years ago

. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T

Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

I = F t = Δp

in this case the final velocity is zero

F t = 0 -m v₀

F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

F₁ = -p₀ / 0.1

F₁ = - 10 p₀

barrier collapses

F₂ = -p₀ / 1

let's look for the relationship of the forces

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

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2 years ago

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Fiesta28 [93]

Explanation:

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2 years ago

The graph represents the gravitational attraction between two equal masses vs. their distance apart, which statement BEST descri

swat32

C would be the answer based on reass

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2 years ago

Read 2 more answers

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago