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FinnZ [79.3K]
3 years ago
6

How much time will it take to perform 440 joule of work at a rare of 11 w?​

Physics
1 answer:
kifflom [539]3 years ago
7 0

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

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Lower in an arrangement

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Which of the following properties of light can only be explained by wave theory and not by the Photoelectric effect? reflection
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<span>Interference can only be explained if light is a wave. Youngs Double Slit experiment proved this, proving the wave nature of light.</span>
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3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
Use ohms law to determine the battery voltage you would need to send 2.5 A of current through a light bulb with 3.6 Ω of resista
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Answer:

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Explanation:

3 0
3 years ago
A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc
BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

t=\frac{usin\alpha }{g}

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

t=40sin30/9.81\\t=2.0secs

b. the expression for the maximum height is expressed as

H=\frac{u^{2}sin^{2}\alpha  }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0
3 years ago
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