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<span>Answer: </span>A swell
A wave which creates a strong undertow is called a swell. These types of waves are also known as <span>surface gravity waves.
Thank you!</span>
Answer:
(1) 14.12 m/s
Explanation:
Given:
= initial speed of the ball = 16 m/s
= angle of the initial speed with the horizontal axis = 
= initial height of the ball from where Julie throws the ball = 1.5 m
= final position of the ball where Sarah catches the ball = 1.5 m
Let us assume the following:
= horizontal component of the initial speed
= vertical component of the initial speed
= horizontal acceleration of the ball
= vertical acceleration of the ball
The given problem is projectile motion. When the ball is thrown from the air with a speed of 16 m/s at an angle 28 degree with the horizontal axis. When the ball is in the air, it experiences an only gravitational force in the downward direction if we ignore air resistance on the ball.
This means if we break the motion of the ball along two axes and study it, we have a uniform acceleration motion in the vertical direction and a zero acceleration motion along the horizontal.
Since the ball has a zero acceleration motion along the horizontal axis, the ball must have a constant speed along the horizontal at all instant of time.
Let us find out the initial velocity horizontal component of the velocity of the ball. which is given by:
As this horizontal velocity remains constant in the horizontal motion at all instants of time. So, the horizontal component of the ball's velocity when Sarah catches the ball is 14.12 m/s.
Hence, the horizontal component of the ball's velocity when the ball is caught by Sarah is 14.12 m/s.
800/20=40
So the green car would be able to drive 40m
Answer:
r = 0.0414mm
F = 757,692.3Hertz
Explanation:
If the body enters space with uniform magnetic field B, the force experienced by the object is expressed as
F = qvBsintheta... 1
Also, if the body undergoes a circular motion, the force experienced by the body in a circular path is given as
Fc = mv²/r... 2
Equating both forces
F = Fc
qvBsin theta = mv²/r
Since the body enters perpendicular to the field, theta = 90°
The equality becomes;
qvB sin90° = mv²/r
qvB = mv²/r
qB = mv/r
r = mv/qB
Given mass of the electron m = 9.11×10^-31kg
Velocity of the object v = 197m/s
Charge on the electron q = 1.6×10^-19C
Magnetic field B = 2.71×10^-5T
Substituting this value into the equation to get the radius r we have;
r = 9.11×10^-31 × 197/1.6×10^-19 × 2.71×10^-5
r = 1794.67×19^-31/4.336×10^-24
r = 413.89×10^-7
r = 0.0000414m
r = 0.0414mm
b) Frequency of the motion F = w/2π where w is the angular velocity
Since w = v/r
F = (v/r)/2π
F = v/2πr
F = 197/2π(0.0000414)
F = 757,692.3Hertz
