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3 years ago

Define resistance and describe what would happen to a lightbulb if the voltage increase but the resistance stay the same?

1 answer:

7 0

The first question is how much of a voltage increase are we looking at. If it has a 110 voltage rating and you put it across a 220 source, you will see one flash and then the bulb is no more. Nothing will revive it.

If it is rated at 110 and you put 130 across it, there's no problem but the bulb will burn out sooner than it would if you just put 110 across it.

So you raise the voltage and the resistance stays the same, the current will increase. That's why it will burn out sooner.

V = I * R

The equation is a direct variation. If the voltage goes up the current goes up. If the voltage goes down, the current goes down providing that the resistance stays the same in both cases.

The second question is what is resistance? Resistance in Electricity is the ability of an electric current to go in one direction freeing up as many electrons as it can. The MORE free electrons, the lower the resistance. The FEWER free electrons the higher the resistance.

Here' the kicker. Ready? More and Less are probably the two most important words in beginning science.

The More the resistance, the Less the current flow. That's a really important consideration in battery drain in a watch (or modern day calculator). The More the Battery Drain, the Less time it will last.

Always be careful when more and Less are around.

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A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago

A 66.5kg astronaut is doing a repair in space on the orbiting space station. She throws a 2.30kg tool away from her at 3.10m/s r

MrRa [10]

Answer:

Explanation:

mass of astronaut, M = 66.5 kg

mass of tool, m = 2.3 kg

velocity of tool, v = 3.10 m/s

Let the velocity of astronaut is V.

(A) According to the conservation of moemntum

Momentum of astronaut = Momentum of tool

M x V = m x v

66.5 x V = 2.3 x 3.10

V = 0.107 m/s

(B) The direction of motion of astronaut is opposite to the direction of motion of tool.

6 0

3 years ago

Newton's law of universal gravitation states that every object in the universe attracts every other object. true or false.

Oliga [24]

True,According to Isaac Newton believes that Gravity is responsible for drawing attraction which is directly proportional to product of their masses but inversely proportional to the square of the distance between them.

5 0

3 years ago

Read 2 more answers

A projectile is fired upward at an angle θ above the horizontal with an initial speed v0. At its maximum height, what are its ve

aliya0001 [1]

Answer:

$\vec{v}_{\rm max} = v\cos(\theta)(\^x)\\|\vec{v}_{\rm max}| = v\cos(\theta)\\\vec{a} = \vec{g} = -9.8\^y$

Explanation:

The equations of kinematics will be used to solve this question:

$y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)\\v_y = v_{y_0} + a_yt$

At its maximum height, the projectile has zero velocity in the y-direction. But its velocity in the x-direction is unaffected.

First, let's apply the above equations to the x-direction.

There is no acceleration in the x-direction. So, its velocity in the x-direction is constant during the motion.

$v_x = v_{x_0} + a_xt = v_{x_0} + 0\\v_x = v_{x_0} = v\cos(\theta)$

Therefore, the velocity vector of the projectile is

$v_{max} = v_x = v\cos(\theta)$

The speed of the projectile is the same.

The acceleration vector is constant during the motion and equal to the gravitational acceleration, which is -9.8 downwards.

7 0

4 years ago

A 77.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 22.0 m/s. Suppos

Andrej [43]

Answer:

Explanation:

For elestic collision

v₁ = $\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2}$

$v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}$ [/tex]

Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg , v₁ and v₂ are velocity of goalie and puck after the collision.

v₁ = 0 + ( 2 x .15 x22 )/ 77.15

= .085 m / s

Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.

v₂ = (.15 - 77)x 22 / 77.15 +0

= - 21.91 m /s

=Velocity of puck will be - 21.91 m /s in the direction opposite to original velocity of ball before collision.

5 0

3 years ago