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2 years ago

A clock on a moving train runs _____ an identical clock at rest. A.Faster than

Physics

1 answer:

liberstina [14]2 years ago

0 0

Answer:

A.Faster than

Explanation:

Due to the Doppler effect which states that there is a change in frequency because of the relative speeds of two bodies time dilation takes place. For light the relative difference in velocity between the observer and the source needs to be considered.

Time dilation is the phenomenon where time flows differently for two bodies having different to each other. Time dilation has been verified by the readings of the satellite navigation systems.

$\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$

Where

Δt' = Time elapsed of a moving body

Δt = Time elapsed when the body is stationary

v = Velocity of the moving body

c = Speed of light

As, v is always less than c, Δt' will always be greater than Δt. This means that more time has elapsed for the body in motion.

Hence, the clock in motion will be faster than an identical clock at rest.

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A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

mote1985 [20]

Answer:

1. 6.99x 10^-6V/m

2. 18m

Explanation:

See attached file

7 0

2 years ago

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

2 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

2 years ago

Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288

Tomtit [17]

<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football.
</span>

7 0

2 years ago

Read 2 more answers

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground