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3 years ago

Somebody please explain how to solve this. Thanks in advance!

Physics

1 answer:

Jobisdone [24]3 years ago

5 0

work done is product of force and displacement of point of application of force

so here we have to check the product of force and displacement both

Now we will put the least to maximum work in the following order

1. -A man exerts strenuous effort in pushing a stationary wall

2. -A flea pushes a speck of dirt 1 cm

3. -A farmer pushes a 2 kg wheelbarrow 20 m

4. -A cannon launches a 3 kg cannonball a distance of 200

5. -A 2000 kg car travels 400 m down a road

6. -Space shuttle Atlantis launches from the ground into near-Earth orbit

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What is the average velocity of the particle from rest to 9 seconds?

geniusboy [140]

Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.

7 0

3 years ago

What is the surface area for a rectangular prism with the height of 6 inches, width of 8 inches, and the length of 4 inches

podryga [215]

Answer:

208 is the surface area!

Explanation:

Hope this helps! ;-)

8 0

2 years ago

Read 2 more answers

Need help ASAP

vlada-n [284]

(1) The parallel force on the Jumper is 441.42 N.

(2) The weight of the person is heavier on Earth when compared to Mars by 610 N.

<h3>Parallel force on the Jumper</h3>

The parallel force on the Jumper is determined from the parallel components of the weight of the jumper.

F = W x sinθ

F = (85 x 9.8) x sin32

F = 441.42 N

<h3>Weight of the person on Earth</h3>

W = mg

W = 100 x 9.8

W = 980 N

<h3>Weight of the person on Mars</h3>

W = mg

W = 100 x 3.7

W = 370 N

<h3>Difference between the weights</h3>

= 980 N - 370 N

= 610 N

Thus, the weight of the person is heavier on Earth when compared to Mars by 610 N.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

4 0

2 years ago

Usually the force of gravity on electrons is neglected. To see why, we can compare the force of the Earth’s gravity on an electr

miv72 [106K]

Answer:

$6.4\cdot 10^{-15} N$

Explanation:

The electric force exerted on the electron is given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the magnitude of the electron charge

E = 40000 V/m is the electric field

Substituting,

$F=(1.6\cdot 10^{-19} C)(40000 V/m)=6.4\cdot 10^{-15} N$

By comparison, the gravitational force exerted on the electron is:

$F=mg$

where

$m=9.10939\cdot 10^{-31} kg$ is the mass of the electron

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting,

$F=(9.10939\cdot 10^{-31} kg)(9.8 m/s^2)=8.93\cdot 10^{-30}N$

6 0

4 years ago

Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at

statuscvo [17]

$\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}$ $\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}$ $\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}$ $\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}$ $\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\ \end{gathered}$ $\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}$ Hence the correct answer is q= 1.73514 micro coulombs.

5 0

1 year ago