What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0
1 answer:
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Answer:
a) x = 40 t , y = 39 t , z = 6 + 32 t - 16 t ², b) x = 80 feet , y = 78 feet , the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ + t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t ²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
Answer:
Explanation:
Given
length of window
time Frame for which rock can be seen is
Suppose h is height above which rock is dropped
Time taken to cover
so using equation of motion
where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is
Subtract 1 and 2 we get
and from equation
so
and
so
substitute the value of in equation 2