Answer:
<h2>8.0995×10^-21 kgms^-1</h2>
Explanation:
Mass of proton :
Speed of Proton:
Linear Momentum of a particle having mass (m) and velocity (v) :
Magnitude of momentum :
Frome equation (2), magnitude of linear momentum of the proton :
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Answer:
Explanation:
A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.
The period of a pendulum is the time it takes for the pendulum to complete one oscillation.
The period of a pendulum is given by the equation
where
L is the length of the pendulum
g is the acceleration due to gravity
From the formula, we see that the period of a pendulum does not depend on the mass.
Therefore, the only 2 factors affecting the period of a pendulum are:
- The length of the pendulum: the longer it is, the longer the period of oscillation
- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum
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<span><span>The
best and most correct answer among the choices provided by the question is </span>B.-2.71 V.</span>
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V
Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V
Since Cu is acting as the anode, the equation needs to be
reversed.
Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V
Ecell= -2.37 V+ (- 0.34 V) = -2.71 V
<span><span>
</span><span>Hope my answer would be a great help for you. </span> </span>
<span> </span>
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
<u>FOR BULB 1:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
<u>FOR BULB 2:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
<u>A₁/A₂ = 0.44</u>
