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nataly862011 [7]
3 years ago
11

For each situation, identify when sound would travel faster and why?

Physics
1 answer:
Rudiy273 years ago
8 0
A. Outside on a summer day, there are less particles that the sound bounces off of (snow and wind absorb the sound)

B. Water, sound travels faster through denser objects. Water has a higher density than air.

C. High air pressure, sound moves faster through matter that is closer together. This can be said because sound can't travel through space. (There is no atmosphere and consistent particles in space for sound to go through)

D. A piece of steel, It's denser than wood and water.

E. 90% nitrogen and 10% helium, nitrogen is denser than helium so it will move faster

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A 3-kg skateboard is rolling down the sidewalk at 4 m/s when it collides with a 1-kg skateboard that was initially at rest. If t
irga5000 [103]

Answer:

2 m/s

Explanation:

Parameters given:

Mass of first skateboard, m = 3 kg

Initial speed of first skateboard, u = 4 m/s

Mass of second skateboard, M = 1 kg

Initial speed of second skateboard, U = 0 m/s

Final speed of second skateboard, V = 6 m/s

Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.

Momentum is the product of mass and velocity. This implies that:

m*u + M*U = m*v + M*V

(3*4) + (1*0) = (3*v) + (1*6)

12 + 0 = 3v + 6

=> 3v = 12 - 6

3v = 6

v = 6/3 = 2 m/s

The final speed of the 3 kg skateboard is 2 m/s

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4 years ago
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A 91 kg zebra is traveling 7 m/s east. What is the zebra’s momentum? kg-m/s
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3 years ago
Does the thickness of the wire affect the strength of an electroscope
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A cardiac defibrillator stores 1275 J of energy when it is charged to 5.6kV What is the capacitance? O 11.3 pF 96.1 pF O 813 F
Angelina_Jolie [31]

Answer: 813.13(10)^{-7}F

Explanation:

The answer is not among the given options. However, this is a good example of the relation between the energy stored in a capacitor Wand its capacitance C, which is given by the following equation:

W=\frac{1}{2}CV^{2}   (1)

Where:

W=1275J

V=5.6kV=5.6(10)^{3}V is the voltage

C is the capacitance in Farads, the value we want to find

Isolating C from (1):

C=\frac{2W}{V^{2}}   (2)

C=\frac{2(1275J)}{(5.6(10)^{3}V)^{2}}   (3)

Finally:

C=0.000081313F=813.13(10)^{-7}F This is the capacitance of the cardiac defibrillator

5 0
3 years ago
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