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MissTica
3 years ago
12

Which has less inertia, a bicycle or a train? A. bicycle B. train C. all matter has equal inertia D. neither has any inertia on

earth
Physics
1 answer:
muminat3 years ago
5 0
Nertia is basically how difficult to stop or change an objects direction, and the more mass an object has, the more momentum it has and the greater force is needed to change it's velocity.  In other words train <span>is the answer </span>
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Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =
madam [21]

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

K = C + 273

Also,

C = K - 273

5 0
3 years ago
Read 2 more answers
In the diagram, q1 = +6.39*10^_9 C and
ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

q_1=6.39\times 10^{-9} \ C

q_2=3.22\times 10^{-9} \ C

AP=(0.150+0.250)

      =0.40 \ m

BP=0.25 \ m

Now,

At point P, the electric potential will be:

⇒ V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}

By putting values, we get

⇒     =9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]

⇒     =259.695 \ Volt

4 0
3 years ago
Identify two everyday phenomena that exhibit diffraction of sound and explain how diffraction of sound applies. Identify two eve
Naya [18.7K]

Answer:

Explanation:

Diffraction is the term used to describe the bending of a wave around an obstacle. It is one of the general properties of waves.

1. Diffraction of sound is the bending of sound waves around an obstacle which propagates from source to a listener. Two of the daily phenomena that exhibit diffraction of sound are:

i. The voices of people talking outside a building can be heard by those inside.

ii. The sound from the horn of a car can be heard by people at certain distances away.

When sound waves are produced, the surrounding air molecules are required for its transmission. This is because sound wave is a mechanical wave which requires material medium for its propagation. When a source produces a sound, the sound waves bend around obstacles on its path to reach listeners.

2. Light waves are electromagnetic waves which can undergo diffraction. Diffraction of light is the bending of the rays of light around an obstacle. Two of the daily phenomena that exhibit diffraction of light are:

i. The shadow of objects which has the umbra and penumbra regions.

ii. The apparent color of the sky.

A ray of light is the path taken by light, and the combination of two or more rays is called a beam. A ray or beam of light travels in a straight line, so any obstacle on its path would subject the light to bending around it during propagation. These are major applications in pin-hole cameras, shadows, rings of light around the sun etc.

Some significant differences between diffraction of light and that of the sound are:

i. Diffraction of light is not as common as that of sound.

ii. Sound propagates through a wider region than light waves.

iii. Sounds are longitudinal waves, while lights are transverse waves.

7 0
3 years ago
What type of system is best used to observe conservation of mass because all of the mass stays in one place?
Montano1993 [528]

Answer:

trump

Explanation:

yee hawww

8 0
3 years ago
Read 2 more answers
A star with the mass (M=2.0×1030kg) and size (R=3.5×108m) of our sun rotates once every 30.0 days. After undergoing gravitationa
Gre4nikov [31]

Answer:

r=97.22x10^{3} m

Explanation:

Using the angular formulas can determine the radius using both values neutron star and the the knowing star so

L=I*w

L_{1}=I_{1}*w_{1}=L_{2}=I_{2}*w_{2}

I_{1}*w_{1}=I_{2}*w_{2}

I=Inertia of the star

w=angular velocity

I=\frac{2*m*r^{2}}{5}

w=\frac{2\pi}{t}

Notice the angular velocity determinate by the time and the Inertia have the radius value so

\frac{2}{5}*m*r_{sn}^{2}*\frac{2\pi }{t_{1}}=\frac{2}{5}*m*r_{s}^{2}*\frac{2\pi }{t_{2}}

r_{sn}^{2}*\frac{1}{t_{1}}=r_{s}^{2}*\frac{1}{t_{2}}

r_{sn}^{2}=r_{s}^{2}*\frac{t_{1}}{t_{2}}

t_{1}=0.2s\\t_{2}=30day*\frac{24hr}{1day}*\frac{60minute}{1hr}*\frac{60seg}{1minute}=2.592x10^{6}s

r_{sn}=3.5x10^{8}m*\sqrt{\frac{0.2s}{2.592x^{6}s}}

r_{sn}=97.22x10^{3} m

8 0
3 years ago
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