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3 years ago

You can analogize the photoelectric energy to

Physics

1 answer:

Anon25 [30]3 years ago

5 0

Answer:

a useful sporting analogy ... Teachers of science routinely use analogies to help ... balls over the bar and you will probably find that.

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Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta

tester [92]

Radius of nuclei is given by formula

$R = R_oA^{1/3}$

now we can say volume of the nuclei is given as

$V = \frac{4}{3}\pi R_o^3* A$

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

$m = z*m_p + (A- z)*m_n$

$m_p = m_n = 1.008u$

$m = A*1.008u$

Now density is given as

$\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}$

here we know that

$R_0$ = 1.2 fm

$\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}$

$\rho = 2.31 * 10^{17} kg/m^3$

So from above we can say that density of all nuclei is almost same.

5 0

3 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

F = $m \times a$

= $1090 \times (\frac{21}{16})$

= 1430.625 N

Also, it is known that

$sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}$

$sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}$

$\theta$ = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

8 0

3 years ago

Need help ASAP please helppp

Setler [38]

Answer:

2.3 newtons of force

Explanation:

Divide the weight by the speed of the bike

4 0

3 years ago

If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporiz

makkiz [27]

Answer:A

Explanation:Find attached picture file for details

3 0

3 years ago

A parallel-plate air capacitor of area A = 28.1 cm2 and plate separation of d = 3.80 mm is charged by a battery to a voltage of

Bas_tet [7]

Answer:

4.52×10⁻¹⁰ C

Explanation:

The charge stored in a capacitor is given as,

Q = CV.................... Equation 1

Where Q = Charge stored in a capacitor, C = Capacitance of the capacitor, V = Voltage.

But,

C = e₀A/d.............. Equation 2

Where e₀= permitivity of free space, A = Area of the plates, d = distance of separation of the plates

Substitute equation 2 into equation 1

Q = e₀AV/d................ Equation 3

Given: A = 28.1 cm² = 0.00281 m², V = 69.0 V, d = 3.8 mm = 0.0038 m

Constant: e₀ = 8.85×10⁻¹² F/m.

Substitute into equation 3

Q =8.85×10⁻¹²×0.00281×69/0.0038

Q = 4.52×10⁻¹⁰ C.

Hence the charge on the capacitor = 4.52×10⁻¹⁰ C

5 0

3 years ago