Answer:A block rests on a horizontal, frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an…
Explanation:
Answer:
V = f λ speed of wave in terms of frequency and wavelength
t = S / V time for wave to travel a distance S
t = 91.4 m / 344.5 m/s = .265 sec time to travel 91.4 m
Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.
When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>
</span>
Answer: 0.01 m
Explanation: The formulae for capillarity rise or fall is given below as
h = (2T×cosθ)/rpg
Where θ = angle mercury made with glass = 50°
T = surface tension = 0.51 N/m
g = acceleration due gravity = 9.8 m/s²
r = radius of tube = 0.5mm = 0.0005m
p = density of mercury.
h = height of rise or fall
From the question, specific gravity of density = 13.3
Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³
Hence density of mercury = 13.3×1000 = 13,300 kg/m³.
By substituting parameters, we have that
h = 2×0.51×cos 50/0.0005×9.8×13,300
h = 0.6556/65.17
h = 0.01 m
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:

Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,






The hall voltage is given by

Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,

