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atroni [7]
4 years ago
10

Which of the following particles has mass? a. protons only b. electrons only c. both protons and electrons

Physics
1 answer:
nikdorinn [45]4 years ago
6 0
<span>c. both protons and electrons</span>
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An engine absorbs 2150 J as heat from a hot reservoir and gives off 750 J as heat to a cold reservoir during each cycle. What is
tia_tia [17]

Answer:

65.2 %

Explanation:

Let Q1 = Heat absorbed by the system

Q2 = Heat released by the system

e= (1 - (Q2/Q1)) x 100

e= (1 - (750/2150)) x 100

e= (1 - 0.348) x 100

e= 0.652 x 100

e= 65.2 %

6 0
4 years ago
Light of wavelength 550 nm falls on a slit that is 3.50 x 10^-3 mm wide. How far from the central . maximum will the first diffr
myrzilka [38]
Λ= 550 * 10^{-9}\] m.&#10;&#10;D = 3.50*10^-6 m.
d = 10 m. &#10;&#10;m = 1&#10;&#10;2x = ?

Then you need to find x when n=1
or 
sinθ=λ/w=524x10^-9/0.0033=1.59x10^-4
≈θ for θ<<1 , so θ=1.59x10^-4 radians

Then you can find rhe distance of<span> bright fringe from the center:</span>d= θD=1.59x10^-4x10m ≈ 1.6 mm
6 0
4 years ago
The c-cl bond dissociation energy in cf3cl is 339 kj/mol. What is the maximum wavelength of photons that can rupture this bond?
ruslelena [56]

Answer:

3.53*10^{-7} m

Explanation:

Photon that can rupture the bonds are those with the energy of the bond dissociation energy. If we want to know the energy for each molecule we have to take into account that:

1mol=6.022*10^{23}molecule

Hence, we have

E_d=339\frac{10^{3}J}{mol}*\frac{1mol}{6-022*10^{23}molecules}=5.62*10^{-19}J/molecule

but the energy is also:

E_d=h\nu =\frac{hc}{\lambda}\\\\\lambda=\frac{hc}{E_d}

where h is the Planck's constant and c is the speed of ligth. By replacing we obtain:

\lambda=\frac{(6.62*10^{-34}Js)(3*10^{8}m/s)}{5.62*10^{-19}J}=3.53*10^{-7}m

hope this helps!

7 0
3 years ago
Read 2 more answers
PLEASE ANSWER You are in a vehicle waiting at a railroad crossing and notice a train going by at a constant speed of 5 kph. You
Dominik [7]

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5 0
3 years ago
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Two children (m = 40 kg each) stand opposite each other on the edge of a merry-go-round. The merry-go-round, which has a mass M
s344n2d4d5 [400]

Answer:

Explanation:

Initially the two boys were sitting on the periphery , total moment of inertia

= 1/2 M  r² + 2mr²     ; M is mass of the merry go round , m is mass of each boy and r is the radius

1/2 x 200 x 2² + 2 x 40 x 2²

= 400 + 320

I₁ =  720 kg m²

Finally  the two boys were sitting at the middle  , total moment of inertia

= 1/2 M  r² + 2m( r/2)²     ; M is mass of the merry go round , m is mass of each boy and r is the radius

1/2 x 200 x 2² + 2 x 40 x 1²

= 400 + 80

I₂  = 480

Now the system will obey law of coservation of angular momentum because no torque is acting on the system.

I₁ω₁ =  I₂ω₂ ,         I₁  and ω₁  are moment of inertia and angular velocity of first case and  I₂  and ω₂ are of second case.

720 X 12 = 480 ω₂

ω₂ = 18 rad / s

3 0
3 years ago
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