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4 years ago

13

A train is approaching you at very high speed as you stand next to the tracks. Just as an observer on the train passes you, you

both begin to play the same recorded version of a Beethoven symphony on identical MP3 players.
According to you, whose MP3 player finishes the symphony first?
A. your player,
B. the observer's player,
C. both finish at the same time.

1 answer:

Nastasia [14]4 years ago

3 0

C.both finish at the same time.

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From what is known about spring tides and neap tides, you can conclude that:_________.

mixas84 [53]

Answer:

D

Explanation:

Because neap and spring tides happen twice a month during new moon and full moon .It occurs twice in a month which is 14 days

7 0

3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

A boy whirls a stone in a horizontal circle of radius 1.8 m and at height 1.8 m above level ground. The string breaks, and the s

umka2103 [35]

Answer:

Acceleration is $148.33\ m/s^{2}$

Solution:

As per the question:

Radius of the circle, R = 1.8 m

Height above the ground, h = 1.8 m

Horizontal distance, x = 9.9 m

Now,

The magnitude of the centripetal acceleration can be calculated as:

$a_{c} = \frac{v^{2}}{R}$

where

v = velocity

R = radius

$a_{c}$ = centripetal acceleration

Now, if we consider the vertical component of motion only, then considering the initial velocity, 'u' = 0, from kinematic eqn:

$h = ut + \frac{1}{2}gt^{2}$

$h = 0.t + \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 1.8}{9.8}}$

t = 0.606 s

Now, for the horizontal component of velocity:

x = vt

$v =\frac{x}{t}$

$v =\frac{9.9}{0.606} = 16.34\ s$

Now, we know that the centripetal acceleration is given by:

$a_{c} = \frac{16.34^{2}}{1.8} = 148.33\ m/s^{2}$

6 0

3 years ago

A satellite orbits the Earth in the same direction it rotates in a circular orbit above the equator a distance of 250 km from th

UkoKoshka [18]

Answer:

Clock on the satellite is slower than the one present on the earth = 29.376 s

Given:

Distance of satellite from the surface, d = 250 km

Explanation:

Here, the satellite orbits the earth in circular motion, thus the necessary centripetal force is provided by the gravitation force and is given by:

$\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}$

where

v = velocity of the satellite

R = radius of the earth = 6350 km = 6350000 m

G = gravitational constant = $6.674\times 10^{- 11} m^{3}/ks-s^{2}$

M = mass of earth = $5.972\times 10^{24} kg$

Therefore, the above eqn can be written as:

$v = \sqrt{\frac{GM}{R}}$

Now, for relativistic effects:

$\frac{v}{c} = \sqrt{\frac{GM}{Rc^{2}}} = 26.41\times 10^{- 6}$

Now,

r = R + 250

$\frac{v_{surface}}{c} = {\frac{1}{c}\frac{2\pi R}{24} = 1.54\times 10^{-6}$

Ratio of rate of satellite clock to surface clock:

$\frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{surface}^{2}}{c^{2}}}} = 3.43\times 10^{-10}$

Clock on the satellite is slower than the one present on the earth:

$3.43\times 10^{-10}\times 24\times 3600 = 29.376 s$

4 0

3 years ago

Which of the following will increase the resistance of a wire? Check all that apply. Check all that apply. Increasing the cross-

Lostsunrise [7]

Correct answers:

- Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

- Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

- Increasing the length of the wire will increase the resistance of the wire.

Explanation:

The resistance of a wire is given by:

$R=\rho \frac{L}{A}$

where

$\rho$ is the resistitivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

From the formula, we can notice that:

- The resistance (R) is directly proportional to both the resistivity ( $\rho$ ) and the length of the wire (L), so when one of these quantities increases, the resistance will increase as well

- The resistance (R) is inversely proportional to the cross-sectional area (A), therefore when the resistance increases when the cross-sectional area decreases.

Based on these observations, the correct statements are:

- Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

- Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

- Increasing the length of the wire will increase the resistance of the wire.

7 0

3 years ago