How is energy transformed to chemical energy?

Light of frequency 5*10^14hz liberates electron with energy 2.31*10^-19 joule from a certain surface .what is wave length of ult

stealth61 [152]

Answer: Light of frequency 5 x 1014 HZ liberates electrons with energy 2.3 x 10-19from a certain metallic surface.

Explanation:

5 0

2 years ago

A person who normally weighs 580 N is riding in an elevator that is moving upward, but slowing down at a steady rate. If this pe

Zanzabum

Answer:

M au = Fs - M g au = upwards acceleration; Fs = scale reading

Fs = M (au + g) scalar quantities where g is positive downwards and au is positive upwards - Fs is the net force acting on the person

If the acceleration is zero Fs = M g and the scale reads the persons weight

If the elevator is decelerating then au is negative and the scale reading Fs = (g - au) M and the scale reading is less than the weight of the person

5 0

2 years ago

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$ rev/sec

$=\frac{2 \pi \times 7.5}{8}$ rad/sec

= 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

$$$=1640 \times (5.89)^2 \times 7.5$

= 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

= 427126.9 x 7.5

= 3,203,451.75 J

6 0

3 years ago

9. A body is a particular amount of matter. It can be a solid, liquid or gas. It can be

soldier1979 [14.2K]

Answer:

B. Time and Space

Explanation:

Matter is a pshyical or corporeal susbtance in general, whether solid, liquid, or gaseous especially as distinguised from incorpeal substance, as spirit or mind, or form qualities, actions, and the

Within this time-and-space boundary, all matter exists. All matter is made up of atoms. Because these atoms take up space in the universe, solids, liquids, and gases are recognized as existing and may be found in all three dimensions.

Time is a figurative term. Space has three dimensions but time only has one.

7 0

2 years ago

Read 2 more answers

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago