Answer:
<h2>The answer is 4 s</h2>
Explanation:
The time taken can be found by using the formula

where
d is the distance
v is the velocity
From the question we have

We have the final answer as
<h3>4 s</h3>
Hope this helps you
Answer:
13.33 K
Explanation:
Given that,
Heat absorbed, Q = 30 J
Mass of nail, m = 5 g = 0.005 kg
The specific heat capacity of the nail is 450 J/kg-°C.
We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:

So, the increase in temperature is 13.33 K.
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
Your answer should be 16.3 m 79.4º east of north
Answer:
C. amount of charge on the source charge.
Explanation:
Electric field lines can be defined as a graphical representation of the vector field or electric field.
Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.
The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.