The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.
<u>Explanation:</u>
A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:
Here L is the inductance and I is the current.
Here, energy stored (U) is proportional to the number of turns (N) and the current (I).
mu not - permeability of core material
A -area of cross section
l - length
N - no. of turns in solenoid inductor
Now,given that the proportion always remains same:
In this way the expression
Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
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The container that contains a greater volume of water will contain a greater mass of water.
The volume of the cube is given by:
V = s³
V = volume, s = side length
Given values:
s = 3m
Plug in and solve for V:
V = 3³
<em>V = 27m³</em>
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The volume of the sphere is given by:
V = 4πr³/3
V = volume, r = radius
Given values:
r = 3m
Plug in and solve for V:
V = 4π3³/3
<em>V = 113m³</em>
<em />
The sphere contains more mass.
1) 30 degree
<span>2)3215 N </span>
<span>4)tension=8.66 N </span>
<span>force by wall=5N</span>