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lord [1]
3 years ago
8

If the mass of the products measured 120 g, what would be the mass of the reactants?

Physics
2 answers:
aksik [14]3 years ago
7 0
According to law of conservation of mass, mass of reactant = mass of products

So answer is C

Hope it helps!
Basile [38]3 years ago
3 0

Answer: The correct answer option(C).

Explanation:

In a balanced chemical reaction mass of the reactant are always equal to mass of the products. Also known as Law of Conservation of Mass  which states that " mass can nor be created nor be destroyed in a chemical reaction."

So, the mass of the reactant will be equal to the mass of products.That is 120 grams.

Hence, the correct answer option(C).

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A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d
DedPeter [7]

Answer:

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

x-component = -1.50 m/s

y-component = 3.90 m/s

Explanation:

Relative velocity of a body A relative to another body B, Vab, is given as

Vab = Va - Vb

where

Va = Relative velocity of body A with respect to another third body or frame of reference C

Vb = Relative velocity of body B with respect to that same third body or frame of reference C.

So, relative velocity can be given further as

Vab = Vac - Vbc

Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.

Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form

What is Pauli's velocity relative to Daniel?

Vpd = Vp - Vd

(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)

Vpd = Vpn - Vdn

Vpn = -Vnp = -(1.50î) m/s

Vdn = -Vnd = -(3.90ĵ) m/s

Vpd = -1.50î - (-3.90ĵ)

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

Hope this Helps!!!!

5 0
3 years ago
A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron
alex41 [277]

Answer:

KE=3.529\times10^{−27}\ J

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

E=\dfrac{hC}{\lambda }

Now by putting the values

h=6.6\times 10^{-34}\ m^2.kg/s

C=3\times 10^{8}\ m/s

E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }

E=1.03\times 10^{-18} J

We know that

Kinetic energy given as

KE=\dfrac{P^2}{2m}

KE=\dfrac{E^2}{2mC^2}

KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}

KE=3.529\times10^{−27}\ J

5 0
3 years ago
What is the field outside the capacitor plates in a parallel capacitor?​
FrozenT [24]

Answer is zero

Plz mark me brainlist

7 0
3 years ago
Read 2 more answers
Please Help !!!!!!!!!!!!!!!!!!!!!!!!!!
larisa [96]

As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.

Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.

As per Planck's quantum theory -

                                                      hf =E'-E

          Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.

The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.

We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

       v=\lambda f

[here \lambda is the wavelength of the wave ]

It is given that wavelength =21 cm=0.21 m

 The frequency is calculated as -

                                                         f=\frac{v}{\lambda}

                                                                 =\frac{c}{\lambda}

                                                                 =\frac{3*10^{8} m/s }{0.21 m}

                                                                  =14.2857*10^{8} s^{-1}

 Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz

Here Hertz[ Hz] is the unit of frequency.

7 0
3 years ago
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