In the Celsius scale each degree is one part of 100 degrees. This is because in this scale the difference between boiling and freezing temperatures of water is 100 ° - 0 ° = 100 °, so one degree Celsius is one part of 100.
In the Farenheit scale, each degree is one part of 180 degrees. This is because in this scale the difference between the boiling and freezind temperatures are 212 ° - 32 ° = 180°, so one degree Farenheti is one part of 180.
That means that 1 °C is a larger amount than 1 °C, so 20°C is a larger amount than 20°F.
Conclusion: 20 degree change represents a larger change in Celsius scale.
Answer:
the answer is a. a ball is moving towards the camera faster then slower
<span>v is perpendicular to both E and B and has a magnitude E/B</span>
Answer:
proportional to the current in the wire and inversely proportional to the distance from the wire.
Explanation:
The magnetic field produced by a long, straight current-carrying wire is given by:
where
is the vacuum permeability
I is the current intensity in the wire
r is the distance from the wire
From the formula, we notice that:
- The magnitude of the magnetic field is directly proportional to I, the current
- The magnitude of the magnetic field is inversely proportional to the distance from the wire, r
Therefore, correct option is
proportional to the current in the wire and inversely proportional to the distance from the wire.
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
