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natita [175]
3 years ago
8

Colton the Insulator of electricty.

Physics
1 answer:
maria [59]3 years ago
3 0

Answer:

lemon juice , vinegar

Explanation:

Chemical Effects of Electric Current

Two substances other than water which conducts electricity are: (i) Lemon juice. (ii) Vinegar.

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2. How do the phytochemicals present in various foods help us?<br>​
vredina [299]
<h2>Answer:</h2>

Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.

8 0
3 years ago
A 10 kg ball is held above a building with a height of 30 m. What is the
Darina [25.2K]

Answer: 2940 J

Explanation: solution attached:

PE= mgh

Substitute the values:

PE= 10kg x 9.8 m/s² x 30 m

= 2940 J

6 0
4 years ago
What is normal for a spring that obeys hook's law ?
Bezzdna [24]
A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm
7 0
3 years ago
Read 2 more answers
A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.
blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N

According to Hooks Law

F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m

The part b and c of this question is done in the attachment

7 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
4 years ago
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