11.76 cm² into two significant figures and include the appropriate units
1 answer:
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Answer:
0.9 N
Explanation:
The electric force acting on a charge is given by:
where
q is the magnitude of the charge
E is the strength of the electric field
In this problem, we have
is the charge
is the strength of the electric field
Substituting into the equation, we find
Answer:
(a)
(b)
Explanation:
(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass if it is initially at rest():
According to Newton's second law:
Replacing (2) in (1):
(b) In this case we have . So, we use the final velocity equation:
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