Answer:
what are they ill have a look
Explanation:
Regardless of the source's mobility, light travels at the same speed.
<h3>What makes special relativity so crucial?</h3>
In the calculating and interpretation of high-velocity phenomena, as well as on our methods of thinking, Einstein's special relativity has had a significant influence on the area of physics. Today, we have a considerably better knowledge of space and time than we did at the start of the century.
<h3>Why is special relativity thus named?</h3>
Because it exclusively uses inertial frames to apply the concept of relativity, the theory is known as "special". General relativity, which Einstein created, applies the principle broadly, that is, to any frame, and this theory takes the gravitational forces into account.
learn more about relativity here
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<h2>
Answer: x=125m, y=48.308m</h2>
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:
x-component:
(1)
Where:
is the projectile's initial speed
is the angle
is the time since the projectile is launched until it strikes the target
is the final horizontal position of the projectile (the value we want to find)
y-component:
(2)
Where:
is the initial height of the projectile (we are told it was launched at ground level)
is the final height of the projectile (the value we want to find)
is the acceleration due gravity
Having this clear, let's begin with x (1):
(3)
(4) This is the horizontal final position of the projectile
For y (2):
(5)
(6) This is the vertical final position of the projectile
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC
There is one mistake in the question.The Correct question is here
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.
Answer:
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Explanation:
Given data
time=1/2 sec to 1 sec
v(t)=-9.8t m/s
To find
Distance
Solution
As the acceleration as first derivative of velocity with respect to time
So
acceleration(-g)= dv/dt
Solve it
dv = a dt
dv = -g dt
v - v₀ = -gt
v= dy/dt
dy = v dt
dy = ( v₀ - gt ) dt
y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]
y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]
y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.