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Svetradugi [14.3K]
3 years ago
13

Pls help me quickly ......​

Physics
1 answer:
marishachu [46]3 years ago
3 0

Answer:

the last one: weight force

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A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0
3 years ago
Define Momentum in detail.
Harrizon [31]

Explanation:

Momentum Is defined as the product of of mass and its velocity

Momentum (M) =mass *velocity

SI unit of momentum is kgm/s

The rate of change in momentum

=change in momentum / time

=(mv-mu)/t

8 0
3 years ago
Read 2 more answers
The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o
Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

E_n=\frac{-13.6eV}{n^2}

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

E_{k}=11.1eV-10.2eV=0.9eV

6 0
3 years ago
The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and
Blizzard [7]

Answer:  Frequency factor  A = 8 × 10⁹

activation energy Ea = 15.5 KJ/Mol

Explanation: to begin, let us first define the parameters given;

K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹

K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹

K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹

also T₁ = 300.3 K

T₂ = 341.2 K

T₃ = 392.2 K

we know that;

㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]

where R is given as 8.314 J/mol-k

Ea = activation energy

K₁, K₂ = rate constant

T₁, T₂ = Temperature

therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]

this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol

∴ Ea = 15.5 KJ/ Mol

also given that K = A e⁻∧Ea/RT

here A = frequency factor

∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)

A = 7.99 × 10⁹ = 8 × 10⁹

3 0
2 years ago
PLZZ HELP ME ASAP!! If you get it correct I will give brainliest and 50 POINTS!!
pav-90 [236]

Answer:

cccccccccccccccccccccc

6 0
3 years ago
Read 2 more answers
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