Answer:
Explanation:
Use the one-dimensional equation:
which says that the final velocity of a falling object is equal to its initial velocity times the acceleration of gravity times the time it takes to fall. We have the final velocity, -14.5 (negative because its direction is down and down is negative), initial velocity is 0 (because it was held still by someone before it was dropped), and acceleration is -9.8 (negative again, because direction is down while acceleration increases). Filling in:
-14.5 = 0 - 9.8t and
-14.5 = -9.8t so
t = 1.5 seconds
Answer: ohdela student i will be calling home.
Explanation:
Answer:
A) Magma emplacement
Explanation:
Sedimentary structures forms during deposition of sediments. It can also form after sediments have been deposited. Sedimentary structures can only be found in sedimentary rocks. Some examples include mud cracks, ripple marks, cross stratification, potholes, etc
Magma emplacement is an igneous process which describes the different mechanisms by which magma can be emplaced. It is only typical of igneous rocks.
Answer:
a. Point A
b. 20 V
c. 100 J
Explanation:
a. Point A is at a higher potential because there is a positive sign in front of its magnitude. Since it is a positive integral value, and has a higher magnitude than point B which is at -4, point A is thus at a higher potential than point B.
b. The potential difference between the two points ΔV = A - B
= +16 V - (-4 V)
= +16 V + 4 V
= + 20 V
c. The work done, W in moving a charge Q across a potential difference ΔV is W = QΔV
So, since Q = 5 C and ΔV = + 20 V
W = QΔV
= 5 C × (+ 20 V)
= 100 J

Let's start by finding the time it takes for the dog to reach a velocity of
m/s.
We can use the following equation, where
is initial velocity,
is final velocity,
is time, and
is acceleration.

We're trying to solve for
first, so divide both sides by
.

Substitute in the known values.



Now, we can use the following formula to find the distance.

Substitute in the known values.

Anything multiplied by
is

Just simplify from there.



