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Karolina [17]
3 years ago
15

Compare the catching of two different water balloons. Case A:​ A 600-mL water balloon moving at 8 m/s is caught and brought to a

stop. Case B:​ A 150-mL water balloon moving at 8 m/s is caught with the same technique and brought to a stop. The collision time is the same for each case. Which variable is different for these two cases?
Physics
2 answers:
Vadim26 [7]3 years ago
8 0

Answer:

jduruhrjrhryyrurjjrjrh4h5ghrrhrhhrhrhrhrrhrhrh sus

Explanation:

jdhshhehehehje

muminat3 years ago
5 0
It is case A, which you need to give an example about of it to get the answer .
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You used the right-hand rule to determine the z component of the angular momentum, but as a check, calculate in terms of positio
KengaRu [80]

Answer:

case x py   L is in the positive z direction

case y px   L the negative z direction

Explanation:

The angular amount is defined by the relation

         L = r x p

the bold are vectors, where r is the position vector and p is the linear amount vector.

The module of this vector can be concentrated by the relation

         L = r p sin θ

the direction of the vector L can be found by the right-hand rule where the thumb points in the direction of the displacement vector, the fingers extended in the direction of the moment p which is the same direction of speed and the palm points in the direction of the angular momentum L

in the case x py

the thumb is in the x direction, the fingers are extended in the direction and the palm is in the positive z direction

In the case y px

the thumb is in the y direction, the fingers are in the x direction, the palm is in the negative z direction

7 0
3 years ago
Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate
Helen [10]

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

<u>Explanation:</u>

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.  

                            P=e \sigma A\left(T^{4}-T_{c}^{4}\right)

So this law is application for calculating power absorbed by any surface.

4 0
3 years ago
The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using
Crazy boy [7]

Answer:

6.04788 in

115.82654\ in^3

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in

Diameter is given by

d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in

The diameter is 6.04788 in

6.04788\times 2.54=15.3616152\ cm

Volume of sphere is given by

v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3

The volume is 115.82654\ in^3

115.82654\times \dfrac{1}{1728}=0.06702\ ft^3

Fall velocity is given by

V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s

The velocity of the fall will be 78.38779 m/s

Mass is given by

m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb

1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg

The mass is 0.88159 kg

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J

The kinetic energy is 34.55294 J

4 0
3 years ago
Hydrogen line spectrum lies entirely within visible range
mart [117]

No, that's silly.

You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level.  All of those transitions ploop out
photons at Infrared wavelengths.

THEN next you get your Balmer series, where the electrons fall in
to the 2nd level.  Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.

And then there's the Lyman series, where electrons fall all the way
down to the #1 level.  Those are ALL in the ultraviolet. 
6 0
3 years ago
Can somebody answer this for me asap!
andreev551 [17]
It shows all except the types of precipitation and created for newspapers
7 0
3 years ago
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