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3 years ago

6

a train travles at a speed of 30m/s. the train starts at an initial position of 1000 meters and travels for 30 seconds. what is

its final position?

1 answer:

ycow [4]3 years ago

5 0

Answer:

1900 meters

Explanation:

30m/s x 30 second = 900 meters

+ 1000 meters starting position

= 1900meters

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Two bullets have masses of 3.0 g and 6.0 g respectively. Both are fired with a speed of 40.0 m/s.(a) Which bullet has more kinet

Stolb23 [73]

Answer:

(a) The bullet with a mass of 6.0 g

(b) $\frac{K_2}{K_1}=2$

Explanation:

(a) Kinetic energy is defined as:

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

So, we have:

$K_1=\frac{3*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_1=2.4J$

$K_2=\frac{6*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_2=4.8J$

The bullet with a mass of 6.0 g have more kinetic energy

(b) We have $m_2=2m_1(1)$ and $v_1=v_2(2)$ . So:

$K_1=\frac{m_1v_1^2}{2}\\K_2=\frac{m_2v_2^2}{2}\\\frac{K_2}{K_1}=\frac{\frac{m_2v_2^2}{2}}{\frac{m_1v_1^2}{2}}(3)$

Replacing (1) and (2) in (3):

$\frac{K_2}{K_1}=\frac{\frac{2m_1v_1^2}{2}}{\frac{m_1v_1^2}{2}}\\\frac{K_2}{K_1}=2$

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3 years ago

How does convection affect our atmosphere? our layers of earth? our oceans?

Xelga [282]

Answer: In convection, there is a transfer of heat due to the bulk movement of molecules within the fluid. It affects our atmosphere, layers of earth and our ocean due to the temperature difference.

Explanation:

Convection is a transfer of heat due to bulk movement of molecules within the fluid.

Convection affects our atmosphere. There is a temperature difference in the atmosphere. Warmer air rises up in compare to the cooler air.

Convection affects our layers of earth. It is limited to troposphere. Due to the high transmission of clear sky to bulk incident energy from sun, troposphere is subjected to convection.

Convection affects our oceans. There will be temperature difference in the ocean water. When the water of ocean heats up then it becomes less dense.

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3 years ago

In which regions can the gravitational field strength due to the two planets be zero? Explain.

yan [13]

Answer:

I believe its a and c but my notes are all kinds of messed up so im sorry if its wrong

Explanation:

7 0

3 years ago

You estimate the radius of the big wheel to be 19 m

posledela

Explanation:

first, find the circumference of the wheel by using the formula 2(pi)(r):

2(pi)(19) = 119.380521

divide by 25 secs

119.380521/25 = 4.77522083

round to the nearest tenth is 4.8, so the speed is 4.8mm/sec

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4 years ago

A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric

gayaneshka [121]

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

$KE = \fract{1}{2}mv^2$

$PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))$

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

$m = 910 Kg$

$r_1 = 1200 + 6371 km = 7571km$

$r_2 = 6371 km,$

Replacing we have,

$\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})$

$v = 4456 m/s$

Therefore the speed of the object when striking the surface of earth is 4456 m/s

3 0

4 years ago