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natka813 [3]
1 year ago
14

Which of the following objects requires the MOST power to lift?

Physics
1 answer:
Nesterboy [21]1 year ago
5 0

Answer:

A

Explanation:

Work = F * d

Power = work/time   or    f *d / t          

  '  A '     gives the highest amount of power

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A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth
yulyashka [42]

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi*\sqrt{l/g}

hence putting values we get ,

the solution is in picture ,

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5 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
Frying an egg conduction convention radiation
Allushta [10]

Answer:

frying an egg is conduction because the pan is touching the stove and the pan is touching the eggs

Explanation:

4 0
3 years ago
I have one but can some one help with the second one?​
alisha [4.7K]

Answer:

it is a, direction

5 0
3 years ago
(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT
Stels [109]

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×10^1^3 J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
  • D + T → He + n
  • The D-T fusion reaction results in a Helium (He) and  neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59×10^6×1.6×10^-^1^9) ÷ ( 5.030×1.66×10^-^2^7)

= 3.37×10^1^4 J/Kg

= 3.0× 10^9 J/g

Rate of fuel burning in grams per hour = 3.0× 10^9 ×  3600

= 3.6×3.0×10^1^2

= 1.08 ×10^1^3 J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

#SPJ4

4 0
2 years ago
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