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hammer [34]
3 years ago
10

Energy Calculations

Physics
1 answer:
MaRussiya [10]3 years ago
6 0

Answer: 20,734.69 N/m

Explanation:

The elastic potential energy (ELPE) of the rubber band is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x = 0.035 m is the stretching of the rubber band

E = 12.7 J is the ELPE of the rubber band

Substituting the numbers and re-arranging the equation, we find

k=\frac{2E}{x^2}=\frac{2\cdot 12.7 J}{(0.035 m)^2}=20,734.69 N/m

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Three heavy rods are all made of the same uniform material. These rods have lengths 3 m, 4 m, and 5 m, and compose the sides of
Sonbull [250]

Answer:

[ 2.67 , 1 ] m

Explanation:

Given:-

- The side lengths of the rods are as follows:

                             a = 4 m , b = 4 m , c = 5 m

                             a = Base , b = Perpendicular , c = Hypotenuse

- All rods are made of same material with uniform density. With  

Find:-

Find the coordinates of the center of mass of the triangle.

Solution:-

- The center of mass of any triangle is at the intersection of its medians.

- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).

  • Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:

                                       bx−ay=0

  • Median from (a,0) to midpoint (0,b/2) of opposite side has equation:

                                      bx+2ay=ab

  • Median from (0,b) to midpoint (a/2,0) of opposite side has equation:

                                     2bx+ay=ab

  • Solve all three equations simultaneously:

                                     bx−ay=0  , bx = ay

                                     ay + 2ay = ab , 3ay = ab , y = b/3

                                     bx = b/3

                                     x = a / 3

  • So the distance from the median to each leg of the triangle is 1/3 length of other leg.

- So the coordinates of the centroid for right angle triangle would be:

                                   [ 2a/3 , b/3 ]

                                   [ 2.67 , 1 ] m

                                                         

                                 

3 0
3 years ago
Please don’t troll. I need someone to actually answer these questions.
mash [69]

Answer:

-3+3 i think this is the answer

Explanation:

i think you can ask someone else sorry

4 0
2 years ago
Giving brainiest to correct answer.
mixas84 [53]

Answer:

5.33\ m/s

Explanation:

We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s

Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s

5 0
3 years ago
A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

Hence, it is the required solution.

7 0
3 years ago
A man pushes a 10kg box with a constant acceleration of 5m/s2. What force is applied to the box?
drek231 [11]
FORMULA

\boxed {F = m \times a}

F = force
m = mass
a = acceleration

Using the formula

F = 10 \times 5

Multiply

\boxed {\textsf {F = 10 N}}
8 0
2 years ago
Read 2 more answers
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