The strength of the electric field on the point charge at this distance will be 4000 V/m.
<h3>What is the strength of the electric field?</h3>
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;
Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
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Answer:
900.48925 Hz
979.9785 Hz
Explanation:
= Relative velocity of ambulance =
= Velocity of wall = 0
v = Velocity of sound in air = 343 m/s
From doppler effect we have
The frequency of sound is 900.48925 Hz
When the wall acts like a source
The frequency of sound is 979.9785 Hz
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
<u>c. The incident light must have at least as much energy as the electron work function</u>
Answer:
A. Final pressure P2
P2/P1 = (T2/T1)^n/n-1
P1 = 4bar
T1 = 438K
T2 = 300K
Polytropic index, n, = 1.3
P2 = 4 (300/438)^1.3/1.3-1
P2 = 4 (300/438)^4.333
P2 = 4 * 0.19400
P2 = 0.776bar.
B. The work done is;
W = mR/ n-1 (T1 -T2)
Where, R = 0.1889kJ/kg.K, m = 1
W = 1 * 0.1889/ 1.3-1 * (438-300)
W = 86.89kJ/kg.
C. The heat transfer, Q
Q = W + ΔU
Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk
Q = 86.89 + 1 * 0.743 (300-438)
Q = 86.89 + (-102.534)
Q = -15.644kJ/K
Q = 15.64kJ/K
Answer:
The total elongation for the tension member is of 0.25mm
Explanation:
Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:
(1)
Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:
(2)
Here L is the member extension and δL the change total longitudinal elongation.
Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:
Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:
Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):
