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Sphinxa [80]
2 years ago
15

1. What is the kinetic energy of a frog that has mass of 13kg and can hop 9m/s? *

Physics
1 answer:
emmasim [6.3K]2 years ago
8 0

Answer:526.6J

Explanation:I did a test

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The law of universal gravitation states that any two objects in the universe, without exception,
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A. attract each other.

The Law of Universal Gravitation discusses the phenomenon of gravity. Remember that gravity is the force that keeps us on Earth; the Earth pulls us down, and our bodies pull back.  Gravity is the force of attraction, so the correct answer is a).
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Which of the following is an example of a Cnidarian?
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The answer to this is C jellyfish i am positive 

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A machine whose efficiency is 75% is used to lift a load of 100m.calculate the effort put into the machine if it has a velocity
aliina [53]

Explanation:

M.A = load / Effort

efficiency = M.A/V.R X 100

75 = M.A / 4 X 100

75 = 25 X M.A

M.A = 75/25 = 3

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E = 100/3 = 33.333

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3 years ago
66 POINTS! Which is true about the four atoms shown in figures A, B, C, and D?
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Answer:

A

Explanation:

A and B are isotopes of one another but the same element

C and D are isotopes of one another but the same element

However, A and B have a different proton count than C and D, indicating different elements because the proton count is equivalent to the atomic number.

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3 years ago
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A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If
Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R =  60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀  = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

I(t) = I_o(1-e^{-\frac{t}{T}})

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A

Therefore, the current in the circuit 7 ms later is 0.2499 A

6 0
3 years ago
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