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3 years ago

A lightbulb has a resistance of 195 Ω and carries a current of 0.62 A.

Physics

1 answer:

ki77a [65]3 years ago

8 0

Answer:

Explanation:

Power is current times voltage:

P = IV

Voltage is current times resistance:

V = IR

Therefore:

P = I²R

Given I = 0.62 A and R = 195 Ω:

P = (0.62 A)² (195 Ω)

P ≈ 75 W

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A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

Olegator [25]

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

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3 years ago

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What would happen if you try to put two balloons with the same charge together

ivanzaharov [21]

Answer:

they would repel because they have the same charge. Charges that are different will stick together.

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3 years ago

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The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

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3 years ago

Que es la accion y reaccion

alina1380 [7]

Answer:

what is the action and reaction ?

Explanation:

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3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim

miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
We can find this distance simply applying the definition of average velocity, as follows:

$\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)$

The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

$v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)$

where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
Solving for Δx, we get:

$\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)$

So, the total distance traveled was the sum of (1) and (3):
Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

$\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)$

Δx₂, is the distance traveled while decelerating, and can be obtained using (2):

$v_{omax} ^{2} = 2* a* \Delta x_{2} (7)$

Solving for Δx₂, we get:

$\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)$

Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
Taking the positive root in the quadratic formula, we get the following value for vomax:
v₀max = 24.3 m/s.

6 0

3 years ago