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Svetllana [295]
2 years ago
9

The speed of sound is 344 m/sec when the air is 20 degrees Celsius how far is the source of the sound if it takes 8 seconds for

the sound to reach you?
Physics
1 answer:
pishuonlain [190]2 years ago
5 0

Answer: 2752 m

Explanation:

344*8=

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A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
The diagram shows a spectrometer being used to analyze a sample of a solution.
fredd [130]

Answer

B. X

A plate of glass or metal ruled with very close parallel lines, producing a spectrum by diffraction and interference of light.

diffraction grating is an optical component that splits and diffracts light into several beams travelling in different directions. The light emerges with its colours separated.

4 0
3 years ago
Read 2 more answers
Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take
KiRa [710]

Answer:

\frac{V_{e}}{V_{h}}=0.428*10^{2}

Explanation:

From conservation of energy states that

K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}

5 0
3 years ago
A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
3 years ago
You fill a car with gasoline. The car now has...
ValentinkaMS [17]

Answer:

  • Fuel or a provided source of energy for combustion
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