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Scorpion4ik [409]
2 years ago
12

As a 5.0 × 10^2-newton basketball player jumps

Physics
1 answer:
AysviL [449]2 years ago
4 0
     This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
     Using the Newton's Secound Law, we have:

F_{R}=ma \\ W+E-N=ma \\ N =W+E \\ N=5*10^2+10*10^2 \\ \boxed {N=1.5*10^3N}

Number 3

If you notice any mistake in my english, please let me know, because i am not native.
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A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?
iren2701 [21]

Answer:

K.E=29.403125J

Explanation:

From the question we are told that

Mass M=100

Height 50-20=30m

Generally the equation for velocity before impact is is is mathematically given by

v=\sqrt{2gh}

v=\sqrt{2*9.8*30}

v=24.25

Generally the equation for Kinetic Energy is is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*100*(24.25)^2\\

K.E=29403.125J

K.E=29.403125J

8 0
2 years ago
What if is a forecasting game, below are actions on your document and all you have to do is predict what will happen after the a
Talja [164]

1. Triple click on the paragraph.

By extension, the triple click can allow you to select a line and the quadruple click a section (paragraph).

2. Pressing Ctrl + A.

Ctrl+A | Ctrl+E: Select all. Depending on the programs and the translations, it will be one or the other.

3. Double click within the word.

In most text editors, a double-click on a word selects it in its entirety.

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If you press one of the arrow keys with the Shift key pressed, you can select a section of text.

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Please mark me as brainliest.

<h3>➥ I hope I have helped you, greetings! </h3><h3>Atte: ღTheGirlSadღ </h3>
7 0
2 years ago
Read 2 more answers
If a body p with a positive charge is placed in contact with a body q (initially uncharged), what will be the nature of the char
uysha [10]

If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.

There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object.

Let's know, how a element gain positive charge?

A positive charge occurs when the number of protons exceeds the number of electrons. A positive charge may be created by adding protons to an atom or object with a neutral charge. A positive charge also can be created by removing electrons from a neutrally charged object.

To learn more about Positive charge here

brainly.com/question/2903220

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5 0
2 years ago
The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by
Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}

we have

a. for c = 20 cm

m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.

Evaluated at c = 20 cm, we have

\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0
3 years ago
In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, an
Tju [1.3M]
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
8 0
3 years ago
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