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3 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

2 answers:

7 0

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

Mars2501 [29]3 years ago

3 0

Answer: Hello there!

in this case, you have a graph of velocity vs time, then in this graph is represented the function v(t).

The acceleration is defined as the integral with respect to the time of the velocity v(t), then we could calculate the mean acceleration between two times in a next way:

$a = \frac{v(t2) - v(t1))}{t2 -t1}$

this is the slope between the times t2 and t1, where t2>t1 and V(t2) is the velocity at the time t2 (the final velocity) and v(t1) is the velocity at the time t1 (the initial velocity). In this case, a is the mean acceleration between these times.

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If a 16 N net force makes an object accelerate at 4 m/s2, what is the mass of the object?

12345 [234]

Answer:

4 kg

Explanation:

Acceleration = 4 m/s^2

Force = 16 N

Force = mass * acceleration

mass = force / acceleration

mass = 16 / 4

mass = 4 kg

4 0

2 years ago

Which of the following are true for acceleration?

trapecia [35]

D. Acceleration describes how the velocity changes in time.

3 0

3 years ago

Read 2 more answers

Help me solve this please

Leto [7]

Since it's a projectile being launched the only force acting upon it is gravity, since the object is in free fall once it's launched

so to calculate time you'd utilize the general formula of
$xf = xi + vxi(t) + \frac{1}{2} (a) {t}^{2}$
and then solve using time and make it into the y axis, so change the x's to y's, which will change a to g.
since Vyi is always usually 0, you can drop that out of this equation so the formula to find time would be
$t = \sqrt{ \frac{2(y)}{g} }$
So you'll plug in and it'll be
$t = \sqrt{ \frac{2( - 49m)}{ - 9.81 \frac{m}{s {}^{2} } } }$
to find the maximum height you'll have to do some trigonometry to solve it.
To make it easier draw a triangle
put the 60° mark as shown in the picture.
Then you'll need to find the hypotenuse or horizontal to find the vertical
So the hypotenuse would be the 113m/s
so then you'll use
$\ \sin( 60) = ( \frac{o}{h})$

plug in the numbers
$113( \ \sin (60) ) = o$
now that you have the vertical

use the formula
${vf}^{2} = {vi}^{2} + 2gd$
solve for d which will give you the hypotenuse
$d = { (-vertical \frac{m}{s} )}^{2} \div 2( - 9.81 \frac{m}{ {s}^{2} } )$
The "vertical" is what you found in the previous step.
Vf^2 is equal to 0 so you can just drop that number out since it's 0
then once you have that then youre not done yet
since you're on a cliff of 49 m you'll have to add 49m to the previous answer that you found d to find the maximum height.

I hope this helps!

6 0

2 years ago

Which of the following is not true? A) The energy of the sun comes from nuclear fusion. B) The fuel for nuclear fusion comes fr

denpristay [2]

Answer:

Option B) The fuel for nuclear fusion comes from sea water

Explanation:

Nuclear fusion can not be carried out by using any type of water as water is composed of stable water molecules ( $H_{2}O$ which is quite difficult to be decayed by any simple method.

Moreover, fusion reactions demands the combination of nuclei of lighter elements for which it makes use of the heavy isotopes of Hydrogen that are Deuterium and Tritium as fuel to carry out the nuclear fusion.

4 0

3 years ago

A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

$\xi = Acos(\omega t +\phi)$

PART A) Our equation corresponds to

$y = -5cos(4\pi t)$

Therefore the value of omega is equivalent to that of

$\omega = 4\pi$

From the definition we know that the period as a function of angular velocity is equivalent to

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{4\pi}$

$T = \frac{1}{2}$

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the $cos\theta$ Is equivalent to . So the maximum speed that the body can reach is,