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melamori03 [73]

2 years ago

6

A 100 W lightbulb produces 5.0 W of visible light. (The other 95 W are dissipated as heat and infrared radiation.) What is the l

ight intensity on a wall 2.5 m away from the lightbulb?

2 answers:

Arada [10]2 years ago

6 0

17.5 would be your answer

lesantik [10]2 years ago

3 0

Answer:

17.5

Explanation:

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When the car was stopped by the tree, its change in velocity during the collision was-6 meters/second. This change in

HACTEHA [7]

Answer: -3 meters/second^2.

Explanation:

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2 years ago

Ball A is allowed to fall and strike ball B. Assume that all of ball A's energy is transferred to ball B at point I, and that th

Alika [10]

Answer:

Explanation:

If no Energy is lost ( either as Sound energy or any other energy) the total energy of ball A equals the total energy of ball B

Therefore the Kinetic Energy of Ball B equals the Total Energy of Ball A minus the Potential Energy of Ball A

K. E = T. E - P.E

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3 years ago

Read 2 more answers

Bowen’s reaction series illustrates relations between:

Finger [1]

C. Temperature, chemical composition and mineral structure

Explanation:

The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.

The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.

The left part is the discontinuous end while the right side is the continuous series.
From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
At a relatively low temperature, minerals with frame work structures begins to form . The magma is more enriched with felsic minerals and late stage crystallization occurs here.

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5 0

2 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0

2 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

2 years ago