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Marizza181 [45]
3 years ago
9

A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V house

hold electricity, decreasing the resistance of the filament will _____ the current through the bulb and _____ the power dissipated by the bulb.
Physics
1 answer:
murzikaleks [220]3 years ago
5 0

Here, we're going to make use of two important equations: V = IR and P = IV.

V = IR shows us that the product of current (I) and resistance (R) must equal voltage (V). If V is held constant at 120 (which it is in the question), then, if the resistance R decreases, the current I must increase in order to make the equation work (if V = IR and R goes down, but V must stay the same, I must go up to balance the change out).

P = IV shows us that the product of current (I) and voltage (V) must equal power (P). If V is held constant at 120 (which it is in the question), then, if the resistance R decreases, the power P must decrease (because P = IV, so if I goes down and V stays the same, P must go down).

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Two moving objects collide and move apart on paths 90 degrees apart. The total momentum after the collision is _______ the total
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2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan
ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

5 0
3 years ago
Help me please, ;) I could use it
erastova [34]

Answer:

The solution(s) are in order with respect to the attachments

2.613\:\cdot10^5 Joules ; 5. Adding the same amount of heat to two different objects will produce the same increase in temperature ; 2. Same speed in both ; 2. A

Explanation:

Diagram 1 ( Liquid Nitrogen ) : So as you can see, we want our units in Joules here, and can therefore multiply the mass of gaseous nitrogen and the latent heat of liquid nitrogen, to cancel the units kg, and receive our solution - in terms of Joules. Let's do it.

q ( energy removed ) = mass of nitrogen * latent heat of liquid nitrogen,

q = 1.3 kg * 2.01 * 10⁵ J / kg = 1.3\:\cdot \:2.01\:\cdot \:\:10^5 = 10^5\cdot \:2.613 = 100000\cdot \:2.613 = 261300 Joules = 261.3 kiloJoules = 2.613 * 10⁵Joules is the energy that must be removed

Diagram 2 : The same amount of heat does not necessarily mean the same increase in temperature for two different objects. The increase in temperature depends on the specific heat capacity of the substance. Therefore your solution is 5 ) Adding the same amount of heat to two different objects will produce the same increase in temperature.

Diagram 3 : The temperatures in both glasses are the same, and hence the molecules have the same average speed. Therefore your solution is 2 ) Same speed in both.

Diagram 4 : Glass A has more water molecules, and hence has more thermal energy. Your solution is 2 ) A.

7 0
3 years ago
Read 2 more answers
A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at
docker41 [41]

Answer:

y = 4.36

Explanation:

Let the height of the ladder be L

L = 10

Also:

  • Let x = distance\ from\ the\ base\ of\ the\ ladder
  • Let y = height\ of\ the\ base\ of\ the\ ladder

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

L^2 = x^2 + y^2 --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

x = 9

Substitute 9 for x and 10 for L in L^2 = x^2 + y^2

10^2 = 9^2 + y^2

100 = 81 + y^2

Make y^2 the subject

y^2 = 100 - 81

y^2 = 19

Make y the subject

y = \sqrt{19

y = 4.36

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0
3 years ago
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