Answer:
The density will be 0.65 kg/m3
I think it won't float!! I am not sure though.
Answer:
8.333*10^-6 ohms
Explanation:
Resistivity of a material is expressed as;
p = RA/l
R is the resistance of the material
A is the cross sectional area
l is the length of the material
Given
R = 5 ohms
A = 0.5mm^2
A = 5 * 10^-7m^2
l = 30cm = 0.3m
Substitute into the formula;
p = (5 * 5 * 10^-7m^2)/0.3
p = 25 * 10^-7/0.3
p = 0.0000025/0.3
p = 8.333*10^-6
Hence its resistivity at 20 degrees Celsius is 8.333*10^-6 ohms
Answer:
Cause its scalar quantity
Explanation:
since speed does not take directions into consideration, it is considered to be a scalar quantity. On the other hand, the velocity of an object does not take into account direction, thus making it a vector quantity.
Answer:
0.050 m
Explanation:
The strength of the magnetic field produced by a current-carrying wire is given by
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.
In this problem, we have:
(current in the wire)
(strength of magnetic field)
Solving for r, we find 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
