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3 years ago

Does a light bulb with a greater wattage have a greater brightness

2 answers:

8 0

Only within the same technology. / / / If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.

Bond [772]3 years ago

7 0

Yes. If a lamp develops a higher power certainly produce more luminosity, more brightness.

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Hello, I am so confused about this problem, could you help ?

iVinArrow [24]

Given:

The masses of the balls, m₁=1 kg

m₂=2 kg

The height of 1 kg ball, h₁=6 m

The height of 2 kg ball, h₂=3 m

To find:

Which of the given statements are true?

Explanation:

From the law of conservation of energy, the energy can neither be created nor be destroyed. As the air resistance is negligible, the initial potential energy of the balls will be equal to their kinetic energy when they reach the ground.

Thus, the kinetic energy of the balls when they reach the ground is given by,

$KE=PE=mgh$

Where m is the mass of the balls, h is their respective height, and g is the acceleration due to gravity.

Thus the kinetic energy of mass m₁ is,

$\begin{gathered} KE_1=1\times9.8\times6 \\ =58.8\text{ J} \end{gathered}$

The kinetic energy of mass m₂ is,

$\begin{gathered} KE_1=2\times9.8\times3 \\ =58.8\text{ J} \end{gathered}$

The velocities of the balls will be given by the equation of kinetic energy.

Thus,

$KE=\frac{1}{2}mv^2$

Where v is the respective velocities of the balls when they reach the ground.

On rearranging the above equation, the velocities will be given by,

$v=\sqrt{\frac{2\times KE}{m}}$

On substituting the known values, the velocity of the mass m₁ is

$\begin{gathered} v_1=\sqrt{\frac{2\times58.8}{1}} \\ =10.84\text{ m/s} \end{gathered}$

The velocity of the mass m₂ is,

$\begin{gathered} v_2=\sqrt{\frac{2\times58.8}{2}} \\ =7.7\text{ m/s} \end{gathered}$

Thus the balls will have the same kinetic energies when they reach the ground. But the 1 kg ball will have a greater velocity than the 2-kg ball.

The time interval the ball takes is dependent on the height only and not on the mass. Thus the balls will not reach the ground at the same time.

Final answer:

Thus the correct answer is options are option B and option D.

6 0

1 year ago

1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of

laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

constant velocity of the elevator, u₁ = 10 m/s

initial velocity of the ball, u₂ = 20 m/s

height of the boy above the elevator floor, h₁ = 2 m

height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

$v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m$

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m + 45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

$h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s$

To learn more about projectile calculations please visit: brainly.com/question/14083704

6 0

3 years ago

I need help FAST!!! It’s on my grade

Stells [14]

Solar energy, because of solar panels, hydroelectricity from a dam, or fossil fuels. fossil fuels would be the most ideal because it is more commonly used

3 0

3 years ago

Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional

Alexxx [7]

Answer:

165.529454

Explanation:

According to the Pythagorean Theorem for calculating the lengths of a right angle triangle's sides, a^2 + b+2 = c^2, where c is the longest side (and the side opposing the right angle). So in your case it would be 150*150 + 70*70 = 27400. And √ 27400 is your answer.

7 0

2 years ago

All ions are atoms with a

Cloud [144]

All ions are atoms with a charge

3 0

3 years ago