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3 years ago

9

Which of the following is an example of kinetic energy

2 answers:

Elenna [48]3 years ago

8 0

Running because kinetic energey is the engery created by motion

musickatia [10]3 years ago

5 0

Well kinetic energy is energy<span> possessed by an object due to its motion or movement.</span>

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Can someone please help me match these up

yaroslaw [1]

Answer:

you need to give the answers choices

Explanation:

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3 years ago

an object is held 33.5 cm from a convex mirror. it creates a virtual image of magnification 0.253. what is the focal length of t

MaRussiya [10]

Answer:

Magnification= -image distance/object distance

.253=image distance/33.5

image distance= 8.48 cm

7 0

2 years ago

In which of these cases is a waiter doing work on the object

Gnom [1K]

Section 2 is right,, i think. good luck

8 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$

6 0

3 years ago