In a game at a carnival, a contestant rolls a ball up the slope with an initial speed vi. The object of the game is to roll the ball in such a way that it will get “stuck” in the depression at B and not return back down the slope. This will happen if the ball’s speed when it gets to point A is essentially zero. (The speed of the ball at point A really has to be greater than zero in order for the ball to make it past point A, but
the speed at point A must be greater than zero only by an arbitrarily small amount, so that we can say that the condition for the ball not to return is essentially that the speed at A must be zero.) Assuming that the ball rolls without slipping and that energy losses due to friction are negligible, find the initial speed vi required to make the
speed of the ball at point A zero. Let the mass of the ball be called M and let the radius of the ball be called R. Treat the ball as a solid sphere.

Sam and Sally are traveling aboard a spacecraft that approaches the asteroid Ceres within 14,000 kilometers. Sally will experience 1.989 × 10⁻¹¹  N of force.

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

Given data

Mass of asteroid ,m₁ = 8.7 1020 kg

Mass of sally,m₂ = 67 kg

Gravitational constant,G = 6.6 × 10⁻¹¹ kg⁻² m²

Distance of seperation,R =  14,000 km

[tex]\rm F = G\frac{m_1m_2}{R^ 2} \\\\ F = 6.6 \times 10^{-11 }\times \frac{8.71020 \times 67 }{(14,000)^2} \\\\F = 1.989 \times 10^{-11 } \ N[/tex]

Hence, the force Sally experiences will be 1.989 × 10⁻¹¹  N.

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B. Objects A and C possess charges of opposite sign.

The attraction of two objects occurs when the two objects have opposite charges.

The repulsion of two objects occurs when the two objects have same charges.

Thus, we can conclude that, objects A and C possess charges of opposite sign.

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(a) The 1,400-kg boat will accelerate at 0.425 m/sec².

(b) If the boat starts off at rest, it will travel 85 meters in 20.0 seconds.

(c)At the conclusion of the period, the speed will be 8.5 m/sec.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

[tex]\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f = 85 \ m[/tex]

c)

The velocity at the end of that time is found as;

[tex]\rm v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5[/tex] m/sec

Therefore, the boat's acceleration, the distance it travels in 20.0 seconds, and its final speed will be 0.425 m/sec², 85 m, and 8.5 m/sec.

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a)The velocity of the knife relative to the ship when it hits the deck of the ship will be 14.00 m/sec.

b)The velocity of the knife relative to a stationary observer on shore is 14.10 m/sec

c)The position at which the knife hits the deck is 82.8°.

The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

The velocity of the knife relative to the ship when it hits the deck of the ship is;

v=√2gh

v=√2×9.81 ×10.0 m

v=14.00 m/sec

b)

The velocity of the knife relative to a stationary observer on shore is found as;

v₁₂ = √[(14.00)²+(1.74)²]

v₁₂=14.10 m/sec

c)

The position at which the knife hits the deck is;

[tex]\rm \theta = tan^{-1}\frac{14.00}{1.75} \\\\ \theta = 82.8 ^ 0[/tex]

Hence, the velocity of the knife relative to the ship, the velocity of the knife relative to a stationary observer on shore, and the position at which the knife hits the deck will be 14.00 m/sec,14.10 m/sec, and 82.8° respectively.

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The upward force exerted on the board by the support is 530.8 N.

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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[tex]\Huge\boxed{\textsf{Intuitive}}[/tex]

The intuitive and observant traits are meant to describe what people do with the information they gather from the world around them.

Observant people are more interested in observable facts and straightforward outcomes, while intuitive people imagine the future potential of what they see.

For example, intuitive people are more likely to discuss views and theories of what the world could be like in the future.

Answer:

intuition

Explanation:

3200 m/s best represents the speed through solid argon.Option D is correct.

A sound wave is produced when a medium begins to vibrate. When an entity vibrates, a pressure wave is formed, which causes sound.

Through argon, sound travels at a speed of 319 m/sec.

Any chemical element in the fluid phase has one temperature-dependent value for the speed of sound.

Varying sound wave types may travel through solids at different speeds. These wave types include longitudinal transverse, and extensional.

If argon could exist as solid, then 3200 m/s is the best speed.

Hence, option D is correct.

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The angular speed of the earth, the linear speed of an object on the surface, and the acceleration of the object will be 7.288 × 10⁻⁵ m/sec,446.36 m/sec, and 31.27 m/s² respectively.

The rate of velocity change concerning time is known as acceleration.

Unit conversion;

1 hour = 3600 sec

Given data;

Velocity, v= m/s

Time elapsed, t = 23 hours 56 minutes and 4 seconds

The radius of the earth is, R= 6.37×103 km.

The total taken in the second is;

T=23  hr × 3600 sec  + 56 min  × 60 sec  + 4 sec

T= 86164 sec

The angular speed of the earth;

[tex]\rm \omega_e = \frac{2 \pi}{T} \\\\ \omega_e =\frac{ 2 \times 3.14 }{86164 \ sec} \\\\ \omega_e =7.28 8 \times 10^{-5 } \ rad /sec[/tex]

The linear speed of an object on the surface of the radius vector from the center of the earth is;

[tex]\rm v = r \times \omega \\\\ v= 6123 \ km \times 7.29 \times 10 ^{-5} \\\\ v = 446.36 \ m/sec[/tex]

The acceleration of the object on the surface of the earth is;

[tex]\rm a = \frac{v^2}{r} \\\\ a=\frac{4446.32^2}{6.37 \times 10^3} \\\\ a= 31.27 \ m/s^2[/tex]

Hence,the angular speed of the earth, the linear speed of an object on the surface, and the acceleration of the object will be 7.288 × 10⁻⁵ m/sec,446.36 m/sec, and 31.27 m/s²

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Answer:

[tex]2.5m/s^{2}[/tex]

Explanation:

We can use one of the kinematics equation to find the acceleration.

v = u + at

where v = final velocity, in m/s

u = initial velocity, in m/s

t = time in seconds

Given the information provided from the question,

v = 25m/s

u = 0m/s

t = 10s

Substitute these values into the kinematics equation.

25 = 0 + 10a

10a = 25

a = 25 / 10

= [tex]2.5m/s^{2}[/tex]

(1) The normal force exerted by the road on the car at point A is 1.06 x 10⁴ N

(2) The normal force exerted by the road on the car at point B is 7,495.8 N.

Fn(A) = mg

where;

Fn(A) = 1080 x 9.81

Fn(A) = 1.06 x 10⁴ N

The normal force exerted by the road on the car at point B is calculated as follows;

Fn = Fn(A) - mv²/r

where;

Fn = 1.06 x 10⁴ - (1080 x 19.33²)/(130)

Fn = 7,495.8 N

Thus, the normal force exerted by the road on the car at point A is 1.06 x 10⁴ N and the normal force exerted by the road on the car at point B is 7,495.8 N.

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[tex]\huge\underline{\red{A}\green{n}\blue{s}\purple{w}\pink{e}\orange{r} →}[/tex]

Sonar is an abbreviation of Sound Navigation and Ranging

Explanation:

Answer:

Equilibrium

Explanation:

Temperature is a measure of the average kinetic energy of the atoms or molecules in the system.

The zeroth law of thermodynamics states that there is no heat transferred between two objects in thermal equilibrium

This means that they are the same temperature.

The geographical region where I would expect the warmest weather is: C. in the southeast.

A weather map can be defined as a type of chart that is typically used to provide information about the average atmospheric condition of a particular geographical region over a specific period of time.

Based on the weather map shown in the image attached below, we can infer and logically deduce that the geographical region where the warmest weather is expected is in the southeast due to its very high atmospheric pressure.

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The center of mass of the boat will change to 0 meters.

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Let's think of the rowboat and man as a system. From one end of the boat to the other, the individual is walking.

To preserve momentum, the rowboat advances in the opposite direction as the person begins to move from, say, the left end to the right end.

As a result, the system's center of mass doesn't move because there isn't any external force acting on it.

Hence, the center of mass of the boat will change to 0 meters.

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a. Aster is 56.3 m at 3.16° north-east from her initial position

b. She has to head to 183.16° or 86.84° south of west to return to her initial position

Let Aster's initial position be represented by the vector r = 0i + 0j

Since she then walks walks first 70 m in the direction 37° north of east, let this displacement be represented by the vector u = (70sin37°)i + (70cos37°)j

=  (70 0.6018)i + (70 0.7986)j

= 42.13i + 55.9j

Also, she then walks  82 m in the direction 20° south of east. Let this displacement be represented by the vector v = (82sin20°)i - (82cos20°)j

= -(82 0.3420)i + (82 0.9397)j

= 28.05i - 77.05j

Finally, she walks 28 m in the direction 30° west of north. Let this displacement be represented by the vector, w = -(28sin30°)i + (28cos30°)j = -(28 0.5)i + (28 0.8660)j

= -14i + 24.25j m

So, the total displacement is R = r + u + v + w

= 42.13i + 55.9j + 28.05i + (-77.05)j +  (-14)i + 24.25j m

= 56.18i + 3.1j

So, how far she walks is the magnitude of R. The magnitude of a vector Z = xi + yj is Z = √(x² + y²)

So, the magnitude of R = √((56.18)² + (3.1)²)

= √(3156.19 + 9.61)

= √3165.8

= 56.3 m

The direction of a vector Z = xi + yj is given by Ф = tan⁻¹ (y/x)

So, the direction of R is Ф' = tan⁻¹ (3.1/56.18)

= tan⁻¹ (0.0552)

= 3.16°

So, Aster is 56.3 m at 3.16° north-east from her initial position

To return to her original position, the displacement vector is V = r - R

= 0i + 0j - (56.18i + 3.1j)

= -56.18i - 3.1j

So, the direction of V is Ф" = tan⁻¹ (-3.1/-56.18)

= tan⁻¹ (0.0552)

= 3.16°

Since this is in the third quadrant, we have that the direction she must go to return to her original position is α = 180° + 3.16°

= 183.16°

or 90° - 3.16°

= 86.84° south of west

So, she has to head to 183.16° or 86.84° south of west to return to her initial position

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The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.

The electric field is calculated as follows;

E = kq/r²

where;

midpoint of 3.08 m, x = 1.54 mm

r(1.54 mm, 2.00 mm)

|r| = √(1.54² + 2²)

|r| = 2.52 mm

E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²

E = 5.287 X 10¹³ N/C

Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.

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Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r   ⇒   v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

The workdone required to stack them over each other is 39.69 J

This is defined as the product of force and distance moved in the direction of the force.

Workdone (Wd) = Force × distance (d)

But

Force = weight = mass × acceleration due to gravity

w = mg

Distance (d) = height (h)

Thus,

Wd = weight × height

We'll begin by calculating the weight of each bricks

W = mg

W = 1.5 × 9.8

W = 14.7 N

Next, we shall determine the work done in stacking the 2nd on the 1st brick and so on..

Workdone in stacking the 2nd on the 1st

Wd = Weight × height

Wd = 14.7 × 0.06

Wd = 0.882 J

Workdone in stacking the 3rd

Wd = Weight × height

Wd = 14.7 × 0.12

Wd = 1.764 J

Workdone in stacking the 4th

Wd = Weight × height

Wd = 14.7 × 0.18

Wd = 2.646 J

Workdone in stacking the 5th

Wd = Weight × height

Wd = 14.7 × 0.24

Wd = 3.528 J

Workdone in stacking the 6th

Wd = Weight × height

Wd = 14.7 × 0.3

Wd = 4.41 J

Workdone in stacking the 7th

Wd = Weight × height

Wd = 14.7 × 0.36

Wd = 5.292 J

Workdone in stacking the 8th

Wd = Weight × height

Wd = 14.7 × 0.42

Wd = 6.174 J

Workdone in stacking the 9th

Wd = Weight × height

Wd = 14.7 × 0.48

Wd = 7.056 J

Workdone in stacking the 10th

Wd = Weight × height

Wd = 14.7 × 0.54

Wd = 7.938 J

Total workdone = 0.882 + 1.764 + 2.646 + 3.528 + 4.41 + 5.292 + 6.174 + 7.056 + 7.938

Total workdone = 39.69 J

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The WEIGHT of each brick is (m g) = 14.7 Newtons

Work to lift each brick = (force x distance) = [14.7 x (distance lifted)] Joules

Place 2nd brick on top of the 1st:  lift 6cm,  (14.7N x .06m) = 0.882 J

Place 3rd brick on top of the 2nd:  lift 12cm,  (14.7N x .12m) = 1.764 J

Place 4th brick on top of the 3rd:  lift 18cm,  (14.7N x .18m) = 2.646 J

Place 5th brick on top of the 4th:  lift 24cm,  (14.7N x .24m) = 3.528 J

Place 6th brick on top of the 5th:  lift 30cm,  (14.7N x .3m) = 4.41 J

Place 7th brick on top of the 6th:  lift 36cm,  (14.7N x .36m) = 5.292 J

Place 8th brick on top of the 7th:  lift 42cm,  (14.7N x .42m) = 6.174 J

Place 9th brick on top of the 8th:  lift 48cm,  (14.7N x .48m) = 7.056 J

Place 10th brick on top of the 9th:  lift 54cm,  (14.7N x .54m) = 7.938 J

Total work to stackum (addum up)  = 39.7 Joules

____________________________________________

Way #2  (easier):

Weight of each brick = (mg) = 14.7 Newtons

Work to stack each one = (14.7N) x (distance lifted) Joules

Number of bricks to lift  =  9

AVERAGE distance to lift them = (1/2) (54cm + 6cm) = 30cm

Average work = (weight of each brick) x (average distance lifted)

Total work = (number of bricks) x (weight of each) x (average lift)

Total work = (9 bricks) x (14.7N) x (0.3m) = 39.69 Joules

The angular acceleration of the rod, just before it touches the horizontal surface is mathematically given as

w = 3.6 rad/s

Generally, the equation for Torque is  mathematically given as

[tex]T = F r[/tex]

Therefore

[tex]T = (mg Cos\theta) (L/2)\\\\T = I \alpha[/tex]

[tex]\alpha = (1.5) (9.8 Cos65.2) (2.10)[/tex]

[tex]\alpha = 12.95 rad/s2[/tex]

Generally, the equation for conservation of energy is  mathematically given as

P.E=K.E

Therefore

[tex]mg (L/2) sin65.2 = (0.5) I w2[/tex]

[tex](0.5) (9.8) Sin65.2 = (0.5) (2.1/3) w2[/tex]

w = 3.6 rad/s

In conclusion, the angular acceleration

w = 3.6 rad/s

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Let's check

[tex]\\ \rm\dashrightarrow F=\dfrac{k}{q_1q_2}{r^2}[/tex]

[tex]\\ \rm\dashrightarrow F\propto Q[/tex]

[tex]\\ \rm\dashrightarrow F\propto \dfrac{1}{r^2}[/tex]

So

Option A and C can be used

Answer:

Decreasing the distance between the particles by a factor of 2

Explanation:

To double the force between the two charge particles, the distance between the particles should be reduced by a factor of two.

According to coulombs law "force is directly proportional to the potential between the two charges and inversely proportional to the square of their distances".

       [tex]\sf{F  = \dfrac{kq_{1} q_{2} }{r^{2} }}[/tex]

F is the electric force

k is the coulomb constant

q is the charge

r is the distance or separation

As the separation between the charges is reduced the force increases and vice versa.

The prompt above requires an explanatory essay. The focus of an explanatory essay is to give clarity to a concept. See the sample below.

From history, it is clear that Science builds on itself. For avoidance of doubt, this means that improvements in science are incremental with each improvement feeding off the last.

From the lesson we have examples of this phenomena such as that of Galileo. Recall that the text indicates that He improved the telescope; and while carrying out the same research that his predecessor Copernicus had done in relation to the stars and planets, Galileo used math to justify his position.

It is also on record that Johannes Kepler who lived in the same period as Galileo took interest in Astronomy. In this case, science built on itself because, Kepler's work served as the foundation for Sir Isaac Newton's work. This is also another example of how science builds on itself.

From the above textual evidence, it is given beyond reasonable double that science indeed builds on itself with the discovery of each generation serving as the foundation for the advancement of the next.

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1. The speed (in light-years per year) of the debris is 4.67×10⁻² light-years / year

2. The speed (in Km/h) of the debris is 50436000 Km/h

3. The speed of light is 21.4 times the speed of the debris

Speed is the distance travelled per unit. Mathematically, it can be expressed as:

Speed = distance / time

Speed = distance / time

Speed = 0.7 / 15

Speed = 4.67×10⁻² light-years / year

1 light-years / year = 1.08×10⁹ Km/h

Therefore,

4.67×10⁻² light-years / year = 4.67×10⁻² × 1.08×10⁹

4.67×10⁻² light-years / year = 50436000 Km/h

Comparison = Speed of light / speed of debris

Speed of light / speed of debris = 3×10⁸ / 14010000

Speed of light / speed of debris = 21.4

Cross multiply

Speed of light = 21.4 × Speed of debris

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Answer:

ORGANISM

Explanation:

The types of fusion that occurs in stars are as follows:

Nuclear fusion is the process by which nucleus of small atoms combine together to produces atoms of larger nucleus.

The types of nuclear fusion that can occur in a star are;

The proton-proton fusion occur in stars which have core temperatures less than 15 million Kelvin having minimum masses of 0.08 solar mass.

The carbon cycle fusion occurs in stars with core temperatures greater than 15 million Kelvin having minimum masses of 4.0 solar mass.

The helium fusion occurs in stars with core temperatures greater than 100 million Kelvin having minimum masses of 0.5 solar mass.

In conclusion, fusion reactions occurring in the stars produce the energy released by stars.

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A firefighter is using a hose and the flow rate of the water leaving the hose. the speed of the water leaving the hose is mathematically given as

u = 9.947 m/s

Generally, the equation for Volume flow rate is  mathematically given as

(V)= (A) * (u)

Where

A= area

v=velocity

Therefore

V = A*u

V=0.032 m3/s

A=πR2

A= π*0.013

u=V/A

u=0.032/π*0.0322

u = 60.27m/s

In conclusion, the speed of the water leaving the hose

u = 60.27m/s

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a)The velocity of the second runner relative to the first will be 0.8 m/s.

b)The second runner will win.

c) The distance ahead will the winner be when she crosses the finish line will be 55.6 m.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

The relative distance between the runner, S =  45.0 m.

The velocity of the front runner,v₁ = 3.45 m/s

The velocity of the second runner,v₂ = 4.25 m/s,

The distance of front runners from the finish line  =250 m

The velocity of the second runner relative to the first is,v₂₁=?

The first runner's time to cross the finish line is,t₁

The second runner's time to cross the finish line is recorded,t₂

The distance ahead will the winner be when she crosses the finish line is, S₂₁

a)

The velocity of the second runner relative to the first is found as;

v₂₁ = v₂-v₁

v₂₁ =4.25-3.45

v₂₁ = 0.8 m/s in the runners' direction of travel.

b)

The first runner's time to cross the finish line is;

t₁ = 250/3.45

t₁=72.5 secs.

The second runner's time to cross the finish line is recorded;

t₂=(250+45)/4.25

t₂=69.4 secs.

So, the winner is the second runner.

c)

The winner will be in front when she crosses the finish line by a distance, The second runner travels a distance of within this time.

S₂ = 250+45

S₂=295 m.

The distance ahead will the winner be when she crosses the finish line is;

S₂₁ = 295-239.4

S₂₁=55.6

Hence, the velocity of the second runner relative to the first will be 0.8 m/s, the second runner will win. and the distance ahead will the winner be when she crosses the finish line will be 55.6 m.

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The running speed of Jenny willl be 2.5 times the velocity she walks.

The change of distance concerning to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Condition 1;

She takes 25 minutes to run 1,000 m and walk 1,600 m The total time to run as well as the walk is 25 minutes.

[tex]\rm t = t_r +t_w \\\\ 25 = t_r +t_w[/tex]

As we know that;

Distance = speed × time

[tex]\rm x_r = v_r \times t_r \\\\ t_r = \frac{x_r}{v_r}[/tex]

For walking;

[tex]\rm t_w = \frac{x_w}{v_w}[/tex]

[tex]\rm t = t_r +t_w \\\\ 25 = \frac{x_r}{v_r} +\frac{x_w}{v_w} \\\\ 25 = \frac{1000 \ m}{v_r} +\frac{1600 \ }{v_w} \\\\ 40 v_w +64 v_r = v_rv_w[/tex]

Same as for condition 2;

[tex]\rm 100 v_w + 40 v_r = v_rv_w[/tex]

[tex]\rm 40 v_w +64 v_r =\rm 100 v_w + 40 v_r \\\\ v_r = 2.5 v_w[/tex]

Condition 3;

[tex]\rm d_w = 800 \ m[/tex]

[tex]\rm d_w = v_w \times t_w[/tex]

[tex]\rm t_w = \frac{d_w}{v_w } \\\\ \rm t_w = \frac{800}{v_w } \\\\[/tex]

Hence their running speed of Jenny willl be 2.5 times the velocity at she walks.

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(a) The initial gravitational potential energy of the system is 0.5 (m₁ + m₂)gL.

(b) The final kinetic energy of the system is 0.5 (m₁ + m₂)gL .

P.E = mgh

where;

Let the length of the incline = L

h = L sinθ

P.E = mg(L sinθ)

P.E = (m₁ + m₂)g (L sinθ)

P.E = 0.5 (m₁ + m₂)gL

The final kinetic energy of the system is equal to the initial potential energy since the inclined plane is frictionless.

K.E(f) =  (m₁ + m₂)g (L sin30)

K.E(f) = 0.5 (m₁ + m₂)gL

Thus, the initial gravitational potential energy of the system is 0.5 (m₁ + m₂)gL and the final kinetic energy of the system is0.5 (m₁ + m₂)gL .

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In a fusion nuclear reactor, a drop in temperature would most likely result in a decline in the rate of energy output.

The utilization of nuclear reactions to generate energy is known as a nuclear power. Nuclear fission, nuclear decay, and nuclear fusion processes are all sources of nuclear energy.

Nuclear power facilities currently produce the great bulk of the electricity generated by nuclear fission of uranium and plutonium.

The environment must be hot enough for the deuterium and tritium ions' kinetic energies to be sufficient to break through the Coulomb barrier and fuse together.

Hence a decrease in temperature would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor.

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Answer:

A: a drop in the temperature in the reactor

Explanation:

right on edg