Do cherry popsicles freeze slower than orange popsicles
The calculated magnitude is 6.73 x 10³ V/m.
AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.
Potential difference = 8.20 kV= 8.20 x 10³ V
radius= 19.4/100=0.194 m
total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194
= 1.218 m
therefore Magnitude= 8.20 x 10³ / 1.218
=6.73 x 10³ V/m
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The correct answer is: Option (A) 75 J
Explanation:
First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (<em>mg</em>) of the box is given as the unit is <em>Newton</em>, not its mass (which is in kg).
As,
Potential-energy = mass * acceleration-due-to-gravity * height
PE = m*g*h --- (A)
In equation (A), mg is actually the weight of the box, which is given.
mg = 50N
h = height = 1.5m
Plug the values in equation (A):
PE = 50 * 1.5 = <em>75 J (Option A)</em>
The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin.
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
