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inn [45]
3 years ago
9

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

the volume is constant, and it is heated to a temperature of 378 K. Find the net force on each wall of the box.
Physics
1 answer:
const2013 [10]3 years ago
5 0

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}

we know that

force = pressure * area

therefore force is

F =(\frac{T_{1}}{T_{2}}*P_{1})A

F =(\frac{378}{278}*101325)(17*10^{-2})^{2}

F = 3.98 kN

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Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p
Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation  that is

for red light

h ( c / λr ) = Ф +  kinetic energy

for green light

h ( c / λg ) = Ф +  1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф +  1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr  - 1/ λg)

put all value

Ф =( 6.63 ×10^{-34} (3 ×10^{8} )  / 0.50) × ( 1.50/ λr  - 1/ λg)

Ф =( 6.63 ×10^{-34} (3 ×10^{8} ) / 0.50 ) × ( 1.50/ 651×10^{-9}   - 1/ 521 ×10^{-9})

Ф = 1.5305  ×10^{-19} J  × ( 1ev / 1.6 ×10^{-19} J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0
3 years ago
Read 2 more answers
Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi
insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0
3 years ago
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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
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Calculate the potential energy of a 4 kg cat crouched 3 meters off the ground
Tom [10]
Gravitational potential energy = mass × gravity × height

Ep = (4)(9.81)(3)

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7 0
3 years ago
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Let the four horizontal compass directions north, east, south, and west be represented by units vectors n^, e^, s^, and w^, resp
Marrrta [24]

Answer:

c = e > b = d > a

Explanation:

Given vectors are all unit vectors, therefore they have a magnitude of 1

<h3>Let a, b be two vectors and magnitude of cross product of these two vectors is (magnitude of a) × (magnitude of b) × (sine of angle between these two vectors)</h3>

As all are unit vectors their magnitude is 1 and therefore in this case the cross product between any two vectors depends on the sine of angle between those two vectors

In option a as both the vectors are same, the angle between them will be zero and sin0° will also be 0

In option b angle between those two vectors is 135° and sin135° is 1 ÷ √2

In option c angle between those two vectors is 90° and sin90° is 1

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In option e angle between those two vectors is 90° and sin90° is 1

So by comparison of magnitudes of cross products in each option, the order will be  c = e > b = d > a

5 0
3 years ago
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