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2 years ago

Help me with this question, please

Physics

1 answer:

Lunna [17]2 years ago

5 0

Answer:

First one is voltage

second is resistance

third is electric current

hope this helped!!

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Problem A spring was at its resting position where it is attached to a wall at its left side and a block at its right side as sh

puteri [66]

Answer:

F = 19.1 N

Explanation:

To find the force exerted by the string on the block you use the following formula:

$F=kx$ (1)

k: spring constant = 95.5 N/m

x: displacement of the block from its equilibrium position = 0.200 m

you replace the values of k and x in the equation (1):

$F=(95.5N/m)(0.200m)=19.1N$

Hence, the force exterted on the block is 19.1 N

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3 years ago

Pls help . Which process produces the energy that is used in photosynthesis?

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Answer: Nuclear fusion

Explanation:

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3 years ago

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antoniya [11.8K]

Answer:

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Explanation:

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3 years ago

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

Ghella [55]

Answer:

Total electric potential, $V=1.32\times 10^6\ volts$

Explanation:

It is given that,

First charge, $q_1=3\ \mu C=3\times 10^{-6}\ C$

Second charge, $q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C$

Distance of first charge from origin, $r_1=1.15\ cm=0.0115\ m$

Distance of second charge from origin, $r_2=2\ cm=0.02\ m$

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

$V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

$V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})$

$V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})$

V = 1321826.08 V

$V=1.32\times 10^6\ volts$

So, the total electric potential at the origin is $1.32\times 10^6\ volts$ . Hence, this is the required solution.

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2 years ago

A plant worker accidentally breathes some stored gaseous tritium, a beta emitter with maximum particle energy of 0.0186 MeV. The

andrey2020 [161]

Answer:

400 milli-rems

Solution:

As per the question;

Maximum energy of particle, $E_{m} = 0.0186\ MeV$

Weight, w = 1 kg

Energy absorbed, E = $4\times 10^{- 3}\ J$

Now,

Equivalent dose is given by:

$D_{eq} = \frac{E}{w} =\frac{4\times 10^{- 3}}{1} = 4\times 10^{- 3}\ J/kg$

1 Gy = 1 J/kg

Also,

1 Gy = $1\times 10^{5}\ milli-rems$

Therefore,

Dose equivalent in milli-rems is given by:

$D_{eq} = 4\times 10^{- 3}\times 1\times 10^{5} = 400\ milli-rems$

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2 years ago