Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.
<h3 /><h3>What is the friction force?</h3>
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically, it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.
The strength of the force of friction depends on the two factors, as;
A. The objects' sizes and weights.
D. The heat generated by the friction and the types of surfaces involved.
Hence, options A and D are correct.
To learn more about the friction force, refer to the link;
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Answer:increased
Explanation:
It is given that elevator speed is increasing while moving upward i.e.its acceleration is increasing .
This causes the apparent to be increased if measured using weighing machine.
considering upward direction to be positive
N-mg=ma
N=m(g+a)
where N=Normal reaction=Apparent weight
a=acceleration of Elevator
thus you feel as if your weight is increased.
Answer:
(a) The energy of the photon is 1.632 x
J.
(b) The wavelength of the photon is 1.2 x
m.
(c) The frequency of the photon is 2.47 x
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x
)
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x
Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x
Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x
) = (6.6 x
* 3 x
)/ λ
λ = 
= 1.213 x 
Wavelength of the photon is 1.2 x
m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x
= 6.6 x
x f
f = 
= 2.47 x
Hz
Frequency of the emitted photon is 2.47 x
Hz.
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:

The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) =
₀
V(r) =
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:

= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V