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Ierofanga [76]
4 years ago
14

A marble, a bowling ball, a basketball, and a baseball are rolling across the floor at 10 m/s. Which one has the greatest

Physics
1 answer:
goldenfox [79]4 years ago
4 0

Answer:

The bowling ball

Explanation:

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

In the problem, all the balls have same speed v: so their difference in kinetic energy will depend on their mass only.

We don't know the exact masses of the balls, however we can assume that the bowling ball has the largest mass among them: therefore, the bowling ball will also have the greatest kinetic energy.

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A 26-cm-long wire with a linear density of 20 g/m passes across the open end of an 86-cm-long open-closed tube of air. If the wi
damaskus [11]

Answer: T = 472.71 N

Explanation: The wire vibrates thus making sound waves in the tube.

The frequency of sound wave on the string equals frequency of sound wave in the tube.

L= Length of wire = 26cm = 0.26m

u=linear density of wire = 20g/m = 0.02kg/m

Length of open close tube = 86cm = 0.86m

Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as

L = 3λ/4

0.86 = 3λ/4

3λ = 4 * 0.86

3λ = 3.44

λ = 3.44/3 = 1.15m.

Speed of sound in the tube = 340 m/s

Hence to get frequency of sound, we use the formulae below.

v = fλ

340 = f * 1.15

f = 340/ 1.15

f = 295.65Hz.

f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.

The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as

L = λ/2

0.26 = λ/2

λ = 0.52m

The speed of sound in string is given as v = fλ

Where λ = 0.52m f = 295.65 Hz

v = 295.65 * 0.52

v = 153.738 m/s.

The velocity of sound in the string is related to tension, linear density and tension is given below as

v = √(T/u)

153.738 = √T/ 0.02

By squaring both sides

153.738² = T / 0.02

T = 153.738² * 0.02

T = 23,635.372 * 0.02

T= 472.71 N

3 0
3 years ago
a 10 kg solid disk of radius 0.5 m is rotated about an axis through its center.the disk accelerates from rest to angular speed o
mezya [45]

Answer:

τ ≈ 0.90 N•m

F =

Explanation:

I = ½mR² = ½(10)0.5² = 1.25 kg•m²

α = ω²/2θ = 3.0² / 4π = 0.716... rad/s²

τ = Iα = 1.25(0.716) = 0.8952... ≈ 0.90 N•m

τ = FR

Now we have the unanswered question of reference frame.

80° from what?

If it's 80° from the radial

F = τ/Rsinθ = 0.90/0.5sin80 = 1.818...  ≈ 1.8 N

If it's If it's 80° from the tangential

F = τ/Rcosθ = 0.90/0.5cos80 = 10.311...  ≈ 10 N

There are an infinite number of other potential solutions

7 0
3 years ago
Which of the following describes the time over which a periodic wave repeats?
sineoko [7]

Answer:

b

Explanation:

4 0
4 years ago
Consider two stars that are identical (same size, temperature and luminosity). Star A is 10 pc away from us and Star B is 20 pc
Mkey [24]

Answer:

Star A would be brighter than Star B

Explanation:

The apparent brightness of a star as perceived on Earth is dependent on its temperature, size, luminosity and distance from the Earth. Apparent brightness is the visible brightness to the eye at the surface of the Earth, while luminosity is the true brightness at the surface of the star.

A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.

Brightness and distance from earth have an inverse square relationship.

brightness∝\frac{1}{distance^{2} }

Assuming the star is a point source of radiation, as distance from the source is increased, the radiation is distributed over a surface proportional to the distance form the source. As distance is further increased, the radiation is distributed over a larger surface area reducing the effective luminosity.

If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be \frac{1}{2^{2} } of Star A.

Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.

3 0
4 years ago
(Write the answer in fair test) You have to investigate whether surface area affects how fast some water evaporates
Damm [24]

Answer:

Evaporation increases with an increase in the surface area

If the surface area is increased, then the amount is of liquid that is exposed to air is larger. More molecules can escape with a wider surface area.

Explanation:

7 0
3 years ago
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