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Gennadij [26K]
3 years ago
9

______ energy is defined as the energy of motion.

Physics
2 answers:
MatroZZZ [7]3 years ago
7 0
Kinetic energy is defined as the energy of motion. On the other hand, potential energy is the energy of non-motion.
Hope that helped =)
jekas [21]3 years ago
5 0

Answer:

kinetic

Explanation:

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It takes a slug 20 minutes to travel from the grass to the trash can , a trip of 15 meters. How far could the slug travel in 60
inn [45]

Answer:

45 meters

Explanation:

20 min = 15 meters

So if 20 x 3 = 60

you have to do 3 x 15 !

- which equals to 45 <3

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5 0
4 years ago
A gymnast with mass m1 = 44 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =
SVEN [57.7K]

Answer:

Explanation:

e. At the right edge of the beam

Check attachment for other solution

4 0
3 years ago
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 7.2 m from this surface, the potenti
Yakvenalex [24]

Answer:

The radius of the sphere is 3.6 m.

Explanation:

Given that,

Potential of first sphere = 450 V

Radial distance = 7.2 m

If the potential of sphere =150 V

We need to calculate the radius

Using formula for potential

For 450 V

V=\dfrac{kQ}{r}

450=\dfrac{kQ}{r}....(I)

For 150 V

150=\dfrac{kQ}{r+7.2}....(II)

Divided equation (I) by equation (II)

\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}

3=\dfrac{(r+7.2)}{r}

3r=r+7.2

r=\dfrac{7.2}{2}

r=3.6\ m

Hence, The radius of the sphere is 3.6 m.

3 0
3 years ago
What is the technology used behind scanning probe microscopes?
Wittaler [7]

hola ya sabes  la respuesta si si ,me la dices porfa

5 0
3 years ago
Read 2 more answers
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
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