Because although they cannot see it, they can see it's influence on objects that can be seen, and it's effects.
Answer:
According to the Big Bang Theory, the density and temperature of the Universe is <u>lower</u> now than in the past.
<em>Hope</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>:</em><em>)</em>
Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question ===> I = V / R .
Current = (voltage) / (resistance)
Current = (1.5 V) / (10 Ω)
<em>Current = 0.15 Ampere</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
Answer:
x = 7.14 meters
Explanation:
It is given that,
Current in wire 1, 
Current in wire 2, 
Distance between parallel wires, r = 25 cm
Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :
Let the distance is x from wire 1. So,
x = 7.14 meters
So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.
