Answer:
E = 3.54 x 10⁻¹⁹ J
Explanation:
The energy of the photon can be given in terms of its wavelength by the use of the following formula:
where,
E = energy = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ Js
c = speed of light = 2.998 x 10⁸ m/s
λ = wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m
Therefore,
<u>E = 3.54 x 10⁻¹⁹ J</u>
Answer:
2 seconds
Explanation:
The frequency of a wave is related to its wavelength and speed by the equation
where
f is the frequency
v is the speed of the wave
is the wavelength
For the wave in this problem,
v = 2 m/s
So the frequency is
The period of a wave is equal to the reciprocal of the frequency, so for this wave:
This means that the wave takes 4 seconds to complete one full cycle.
Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:
Answer:
A concave mirror is used as a torch reflector. ... When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector. These rays are then reflected to produce a strong, parallel-sided beam of light.
Explanation:
Resistance = (voltage) / (current) =
(120 V) / (0.5 A) = <em>240 ohms</em>
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
