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allsm [11]
3 years ago
14

An 8 kg toddler is running at a speed of 10 m/s. how much energy does he have?

Physics
1 answer:
wlad13 [49]3 years ago
8 0
To find the Kinetic energy this is the equation


KE=1/2mv^2

so we know the mass (m) which is 8kg

we also know the velocity (v) which is 10 m/s

so we plug that into the equation and it should look like this - 1/2(8kg)(10m/s)^2 now we solve

so 1/2 x 8 = 4 so then we just do (4)(10)^2 which gets us 400.

Since kinetic=Joules

The answer is 400 J
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To achieve a speed of 2 m/s, the bottle must be dropped at m. To achieve a speed of 3 m/s, the bottle must be dropped at m. To a
klio [65]

Answer:

\begin{array}{l|l}\text{Speed}\; \mathrm{(m\cdot s^{-1})} & \text{Minimum Height\;(m)}\\\cline{1-2}\\[-1em] 2 & 0.204\\3&0.459\\4 & 0.815\\5 & 1.27 \\6 & 1.83\end{array}.

Assumptions:

  • The object is dropped in a free fall.
  • There's no air resistance.
  • The downward acceleration due to gravity is \rm 9.81\;m\cdot s^{-2}

Explanation:

Consider the "SUVAT" equation

\displaystyle \frac{v^{2} - u^{2}}{2a} = x,

where

  • v is the final velocity,
  • u is the initial velocity,
  • a is the acceleration of the object, and
  • x is the change in the object's position.

For example, if the bottle needs to achieve a speed of v = \rm 2\; m\cdot s^{-1} by the time it reaches the ground,

  • u = 0 since the statement that the bottle is "dropped" implies a free fall.
  • a = g = \rm 9.81\;m\cdot s^{-2}.

Apply the previous equation to find the minimum height, x:

\displaystyle x = \frac{v^{2} - u^{2}}{2a} = \rm \frac{\left(2\; m\cdot s^{-1}\right)^{2}}{2\times 9.81\; m\cdot s^{-2}} \approx 0.204\; m.

Replace the v value and apply the formula to find the minimum height required to reach different final speeds.

8 0
3 years ago
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Nami conducts an investigation on plants. She places a grow light on a timer to give the plants different amounts of light to se
elena-s [515]

Answer:

It is independent variable

Explanation:

7 0
3 years ago
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Article 5 of the Fundamental Orders of Connecticut is MOST LIKELY related to which idea?
Sauron [17]

Answer:

Option A, Rule of law

Explanation:

The complete question is

Article 5 of the Fundamental Orders of Connecticut is MOST LIKELY related to which idea? A) rule of law B) personal freedom C) individual rights D) limited government

Solution

It is the first colonial constitution and also the first written constitution in America. It was first applicable to the inhabitants and Residents of Windsor, Harteford and Wethersfield

The article V of the Fundamental Orders of Connecticut is subjected to the Courte of Election of the towns. The towns were responsible for maintaining their own form of government through election with the involvement of general public. The 5th article is about the Judicial Department and the process of selection of judges who further governs the city. This abide by law of equality by vesting equal right of choosing the judges.

3 0
3 years ago
For elevator questions, does the weight of a person increase when the elevator is going up or if it's going down-
TiliK225 [7]

The weight of a person increase when the elevator is going up.

<h3>Weight of the person in the elevator</h3>

The weight of the person in the elevator is calculated as follows;

<h3>When the person is going up</h3>

F = ma + mg

F = m(a + g)

where;

  • a is acceleration of the person
  • g is acceleration due to gravity

<h3>When the person is going down</h3>

F = mg - ma

F = m(g - a)

Thus, the weight of a person increase when the elevator is going up.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

6 0
2 years ago
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

8 0
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