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madreJ [45]
3 years ago
13

How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?

Physics
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

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(a) We can find the current flowing between the walls by using Ohm's law:
I= \frac{\Delta V}{R}
where \Delta V=69 mV=0.069 V is the potential difference and R=6.8\cdot 10^9 \Omega is the resistance. Substituting these values, we get
I=1.01 \cdot 10^{-11} A

(b) The total charge flowing between the walls is the product between the current and the time interval:
Q=I \Delta t
The problem says \Delta t=0.86 s, so the total charge is
Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C

The current consists of Na+ ions, each of them having a charge of e=1.6 \cdot 10^{-19} C. To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions
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3 years ago
Wolfgang pauli hypothesized an exclusion principle. This principle says two electrons in an atom cannot have the same what?.
tatiyna

No two electrons in an atom or molecule may have the same four electronic quantum numbers, according to the Pauli Exclusion Principle. Only two electrons can fit into an orbital at a time, hence they must have opposing spins.

<h3>What is Pauli's exclusion principle ?</h3>

According to Pauli's Exclusion Principle, no two electrons in the same atom can have values for all four of their quantum numbers that are exactly the same. In other words, two electrons in the same orbital must have opposing spins and no more than two electrons can occupy the same orbital.

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Learn more about Pauli's exclusion principle here:

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A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air
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Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

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Answer:

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