This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster
The electric field between plates is 4000V/m.
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.
The voltage between points A and B is
V=E.d
E =V/d (uniform E- field only)
where d is the distance from A to B, or the distance between the plates.
Given:
distance d = 3 cm
voltage V = 120 V
Electric field E = V/d
E = 120 V / 3cm
E = 40 V / 1 cm [ 1 cm = 1/100 m ]
E = 4000 V/m.
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I think this is correct, but I am not entirely certain.
Find the force constant of the spring:
F = - KX
(0 - 62.4) = -K(0.172m)
-362.791 = -K
362.791 N/m = K
Find the work done in stretching the spring:
W = (1/2)KX
W = (1/2)(362.791)(0.172m)
W = 31.2 J
Answer:
a)
b)
c) 0 J/K
d)S= 61.53 J/K
Explanation:
Given that
T₁ = 745 K
T₂ = 101 K
Q= 7190 J
a)
The entropy change of reservoir 745 K

Negative sign because heat is leaving.

b)
The entropy change of reservoir 101 K


c)
The entropy change of the rod will be zero.
d)
The entropy change of the system
S= S₁ + S₂
S = 71.18 - 9.65 J/K
S= 61.53 J/K
Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance


We need to calculate the focal length
Using formula of lens

put the value into the formula



The focal length of the lens is 35.52.
(B). Given that,
Object distance = 95 cm
Focal length = 29 cm
We need to calculate the distance of the image
Using formula of lens

Put the value in to the formula




We need to calculate the magnification
Using formula of magnification



The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.