All categories

3 years ago

Which equations can be used to solve for acceleration? Check all that apply.t =

Physics

2 answers:

alekssr [168]3 years ago

8 0

Um I would help you but I don’t see the question

lyudmila [28]3 years ago

7 0

Acc: increasing speed/ average time

You might be interested in

The value of acceleration due to gravity is less at the top of Mount Everest then that in the terai reason, why?

xxTIMURxx [149]

Answer:

The value of acceleration due to gravity is greater in terai than in mountain. In terai region the radius of earth is less as it lies close to the centre of the earth. Thus, the value of g is more in terai region.

3 0

3 years ago

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

$l_o =52 \ m$

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

3 years ago

What is potential energy called that is associated with objects that can be stretched

Umnica [9.8K]

Elastic potential energy

5 0

3 years ago

_____ is very corrosive and can cause rusting so metal tanks shouldn't be used

bagirrra123 [75]

Answer:

Iron

Explanation:

5 0

3 years ago

If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di

masya89 [10]

The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5

8 0

3 years ago

Which equations can be used to solve for acceleration? Check all that apply.t = ​

Which equations can be used to solve for acceleration? Check all that apply.t =