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anyanavicka [17]
3 years ago
10

Which one of the following is true?

Physics
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

(C) All spontaneous nuclear reactions are exothermic.

Explanation:

the answer is option C

Here the Spontaneous nuclear reactions occur on its own. You don't have to have any outside force. Mass defects are converted into energy during the reaction.

Mass defects are the difference between the overall product mass and the overall reactant mass.

The expression ' \Delta E= \Delta m C^2' shows both the mass defect and the energy emitted, where, C is the speed of light.

Because energy is always released in spontaneous nuclear reactions, all nuclear spontaneous reactions are always exothermic.

Hence the correct option is C

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Answer:the force that object be lower

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PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!
SVETLANKA909090 [29]

<u>Answer:</u>

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 m/s^2

<u>Explanation:</u>

1)  Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2)  Average speed = Total distance covered / Time taken for that distance to cover.

    Total distance covered = 2+0.5+2.5 = 5 km

    Time taken = 3 hours

     Average speed = 5/3 = 1.66 km /hour

3)    Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

  Change in velocity = 14 - 6 = 8 m/s

   Time = 4 seconds

   So acceleration = 8 / 4 = 2 m/s^2

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A cheetah can run at 30 m/s but only for about 12s. how far will it run in that time
Mashutka [201]
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A body is moving with velocity 30 m pr sec Wt will b the distance covered by a body in 1 sec??
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30 metres

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6 0
3 years ago
A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2
Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

4 0
3 years ago
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