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Tju [1.3M]
3 years ago
7

A mass of gas under constant pressure occupies a volume of 0.5 m3 at a temperature of 20°C. Using the formula for cubic expansio

n, what will be the volume at a temperature of 45°C without a change in pressure?
Physics
1 answer:
Vsevolod [243]3 years ago
4 0
P1v1/t1 = p2v2/t2
p(0.5 m^3)/(20+273 K) = pv2/(45+273 K)
find v2
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What is the energy of a single photon of visible light with a wavelength of 500.0 nm? 1 nm=10^−9 m
Dmitrij [34]

Answer:

3.97×10^−19 J

From the planks equation

E=hv

V= c/ wave length

V= 3×10^8/500×10^-9

=6×10^14

E= hv

6.63×10^-34×6×10^14

= 3.97×10^−19 J

Explanation:

8 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
3 years ago
The proper IUPAC naming convention for the compound symbol 'KI' is
Darya [45]
Proablyly a or c idk 
6 0
3 years ago
The driving force behind an efficient supply chain is __________.
Doss [256]
Customer satisfaction is considered to be the "driving force" in order to achieve an efficient supply chain. An efficient supply chain takes place when the organization, or the company itself, meets with the demands of the consumers to improve and provides services that satisfies the people.
5 0
3 years ago
What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?
Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s.

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

p=mv

p=3400\ kg\times 201.16 \ m/s

p=683944\ kg-m/s

or

p=6.83\times 10^5\ kg-m/s

So, the magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s. Hence, this is the required solution.

7 0
3 years ago
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