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liraira [26]
2 years ago
15

Amanda goes shopping for light bulbs. In the light bulb aisle, she finds the following incandescent light bulbs. Can you help he

r determine which light bulb is the least efficient? Wattage Lumen output 25 W 210 40 W 445 75 W 1180
Physics
1 answer:
Luden [163]2 years ago
4 0

Answer:

The least efficient light bulb is the first one (25 W - 210 Lumen)

Explanation:

Efficiency can be defined as what you want to obtain over what you need to produce it. In this case Eff= Wattage / Lumen. For each light bulb, their efficiency is: 8.4 / 11.12 / 15.7 Lum/W

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Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air
expeople1 [14]
U = 0, initial vertical velocity

Neglect air resistance, and g = 9.8 m/s².

The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s

Answer:  2.0 s
8 0
2 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
2 years ago
A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
2 years ago
A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to
White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation E= \frac{V}{d}

where E= electric field (force per coulomb)

            V= voltage

            d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

3 0
2 years ago
A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0
2 years ago
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