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gavmur [86]
2 years ago
12

PLzzz helpppp

Physics
2 answers:
ioda2 years ago
5 0

Answer: Charging.

Explanation: ...

finlep [7]2 years ago
5 0

Answer:

elbowing, boarding

Explanation:

Any player who skates, jumps, or charges into an opponent will receive minor penalty.

Elbowing and boarding are one among minor penalty

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How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?
Zarrin [17]
<span>The kinetic energy is the work done by the object due to its motion. It is represented by the formula of the half the velocity squared multiply by the mass of the object. In this problem, you have two  vehicles, the other one is large and the other one is small. Let us assume that they travel with the same velocity. Note that the kinetic energy is proportional to the mass of the object. So when you increase the mass of the other, it also increases the kinetic energy of that object. The same holds true for the two vehicles. The larger the vehicle, its kinetic energy is also large and therefore its stopping distance will be longer than that of the smaller vehicle.</span>
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3 years ago
What is the average rate of change for this exponential function for the interval from x = 2 to x= 4 ?
irina [24]

The answer is D. 6 . This is because it is a positive slope so you can cancel out B and C and then, you count your boxes

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3 years ago
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What is common between the quantities of area, density and speed?​
Rainbow [258]

Answer:

They are all s

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3 years ago
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Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What
IgorLugansk [536]

Answer:

Frequency is 213.04\ s^{-1}.

Explanation:

Distance between source 1 from the receiver , S_1 =\sqrt{10^2+22^2}=24.17\ m.

Distance between source 2 from the receiver , S_2=\sqrt{5^2+22^2}=22.56\ m.

Now ,

Path difference , r = S_1-S_2=24.17-22.56=1.61\ m.

We know, for constructive interference path difference should be integral multiple of wavelength .  

Therefore, r=n\times \lambda

It is given that n = 1,

Therefore, \lambda=1.61\ m.

Frequency can be found by , \nu=\dfrac{v}{\lambda}= \dfrac{343}{1.61}=   213.04\ s^{-1} .

Hence, this is the required solution.

5 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
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