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2 years ago

You observe lighting striking and then hear to sound 8 seconds later. The speed of sound in air is 340 m/s. How

Physics

1 answer:

Leto [7]2 years ago

7 0

Answer:

2.72 Kilometers

Explanation:

8 × 340 m/s = 2720 m = 2.72 Kilometers

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What is the voltage drop across the 10.0 2 resistor?

amid [387]

Answer:

the answer is equal to 2.00v

4 0

2 years ago

Which statement is NOT true?

Nuetrik [128]

Answer:

light doesn't need a medium through which to travel because the speed of light is experimentally constant

4 0

2 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago

A train travels 200km/hr. how much distance will the train be from the station in 2.5 hours?

SCORPION-xisa [38]

Answer:

500km

Explanation:

Given parameters:

Speed = 200km/hr

Time taken = 2.5hrs

Unknown:

Distance = ?

Solution:

To solve this problem, we use the speed, time and distance equation.

Therefore;

Distance = Speed x time

So;

Distance = 200 x 2.5 = 500km

6 0

2 years ago

A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla

Yuri [45]

Answer:

$= - 5.65\times 10^{-2} J$

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18 *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

$U =- G \frac{m_1 m_2}{r}$

mass per unit length is given as

$\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m$

calculating potential energy

$dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}$

$=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}$

$=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}$

$=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}$

$=-G*m_1*\sigma ln(1.125)$

$=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)$

$= - 5.65\times 10^{-2} J$

5 0

2 years ago