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3 years ago

7

There are many types of factors that have an effect on the force of gravity on an elephant. Which factor will LEAST affect the f

orce of gravity between the earth and an elephant?

2 answers:

Pavlova-9 [17]3 years ago

7 0

The gravitational potential energy
I hope this helps!

Fudgin [204]3 years ago

3 0

The elephant's name or color will hardly affect it at all.

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I need help witha worksheet over circuitsin physics could someone help??

garik1379 [7]

Yes u can help I need to see th worksheet to help tho

5 0

3 years ago

Which of the following is an example of a properly written testable hypothesis? *

sveta [45]

Answer:

D. if it is dark, then an owl will find a mouse by the sound the mouse makes

3 0

3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li

Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

$L=L_0\sqrt{1-\frac{v^2}{c^2} }$

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

$L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }$

= 1.136×10^17 m

L= $= \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}$ m

so L= 12 light years

8 0

3 years ago

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal

kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

3 0

3 years ago