Introductory Astronomy1. Light is absorbed and emitted in the form ofwhich are "packets" ofelectromagnetic radiation that carry
a specific amount of energy.photonsionsexcited atomsprotons2. Light is characterized by which two properties? point)electromagnetic energy and wavelengthwavelength and frequencyfrequency and radiationOelectromagnetic radiation and gamma rays
1 answer:
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Answer:
m/s
Explanation:
Parameters given:
distance of the proton form the insulating sheet = 0.360mm
speed of the proton, = 990m/s
Surface charge density, σ = 2.34 x C/
We need to calculate the speed at time, t = 7.0 * s.
We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed on the axis.
The electric force acting on the proton moves in the y direction, so this means it is moving with velocity in the y axis.
Hence, the resultant velocity of the proton is given by:
= 990m/s from the question. We need to find
and then the resultant velocity v.
Electric field is given in terms of surface charge density, σ as:
E = σ/ε0
where ε0 = permittivity of free space
=>
E = 132 N/C
Electric Force, F is given in terms of Electric field:
F = eE
where e = electronic charge
=> F = ma = eE
∴ a = eE/m
where
a = acceleration of the proton
m = mass of proton
a = 1.3 * m/
Therefore, at time, t = 7.0 * , we can use one of the equations of linear motion to find the velocity in the y axis:
= 0 + (1.3 *
* 7.0 *
)
= 910 m/s
∴
v = 1344.69 m/s = m/s
Answer:
a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.
b) The projectile reaches maximum height 2.192 seconds after launch.
c) The maximum height of the projectile is 26.584 meters above ground.
d) The projectile will hit the ground at 4.523 seconds after launch.
e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.
Explanation:
Complete statement of problem is: <em>The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is </em><em>after t seconds. (Round your answers to two decimal places.) </em><em>(a)</em><em> Find the velocity after 2 seconds and after 4 seconds, </em><em>(b)</em><em> When does the projectile reach its maximum height? </em><em>(c)</em><em> What is the maximum height? </em><em>(d)</em><em> When does it hit the ground? </em><em>(e)</em><em> With what velocity does it hits the ground?</em>
a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:
(Eq. 1)
Where is the velocity function, measured in meters per second.
Now we evaluate this function at given times:
t = 2 s.
The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.
t = 4 s.
The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.
b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for :
The projectile reaches maximum height 2.192 seconds after launch.
c) Maximum height is calculated by evaluating height function at the time found in b). That is:
The maximum height of the projectile is 26.584 meters above ground.
d) In this case, we need to equalize the height function to zero and solve for . That is:
Roots are found by means of Quadratic Formula:
and
Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.
e) This can be found by evaluating velocity function at the time found in d):
The velocity of the projectile right before hitting the ground in -22.871 meters per second.