Answer:
Looks like you have:
a = -.324 cos 2.5 t
In this case ω^2 A = .324
ω = 2.5
f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec
Answer:
First Quarter and Third Quarter.
Explanation:
Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.
Since gravity variates with the distance:
(1)
Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.
For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.
When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).
However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.
Therefore, that happens when the Moon is at First Quarter and Third Quarter.
It's the second graph!
it's the only one with a negative gradient.
so the temperature of the ball will fall in water as it looses its heat.
activate windows,:-P
Answer:
The longest wavelength of light is 666.7 nm
Explanation:
The general form of the grating equation is
mλ = d(sinθi + sinθr)
where;
m is third-order maximum = 3
λ is the wavelength,
d is the slit spacing (m/slit)
θi is the incident angle
θr is the diffracted angle
Note: at longest wavelength, sinθi + sinθr = 1
λ = d/m
d = 1/500 slits/mm
λ = 1 mm/(500 *3) = 1mm/1500 = 666.7 X 10⁻⁶ mm = 666.7 nm
Therefore, the longest wavelength of light is 666.7 nm
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s
