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Dennis_Churaev [7]
3 years ago
5

5. A backpack weighs 9.2 newtons and has a mass of 5 kg on the moon. What is the

Physics
1 answer:
Varvara68 [4.7K]3 years ago
3 0

Answer:

The strength of gravity on the moon is 1.84\ m/s^2.

Explanation:

We have,

Weight of the backpack, W = 9.2 N

Mass, m = 5 kg

It is required to find the strength of gravity on the moon. Weight of an object is given by :

W = mg

g is strength of gravity on the Moon

g=\dfrac{W}{m}\\\\g=\dfrac{9.2\ N}{5\ kg}\\\\g=1.84\ m/s^2

So, the strength of gravity on the moon is 1.84\ m/s^2.

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The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec
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Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

<u>λ = 1.8 x 10⁻⁷ m = 180 nm</u>

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Explanation:

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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

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Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

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\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

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10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

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