Answer:
Explanation:
The net force on the potatoes is given by:
F= 52 - mgSintheta
F= 52- (2×9.8× Sin70°)
F = 52 -18.4
F= 33.58N
Using Newton's 2nd law
F = ma
a=F/m = 33.58/ 2 = 16.79m/s^2
Using the equation of motion:
V^2= u^2 + 2as
V^2 = 0 + 2× 16.79 x2
V^2 = 67.16
V=sqrt(68.16)
V= 8.195m/s This is the exit velocity of the potatoes
Kinetic energy, K.E = 1/2mv^2
KE= 1/2 × 2 × 8.195^2
KE = 67.16J
Answer:
Energy stored in the capacitor is 
Explanation:
It is given that,
Charge, 
Potential difference, V = 36 V
We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :
U = 0.000027 J
So, the potential energy is stored in the capacitor is . Hence, this is the required solution.
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
If the net force is 4 N, and Frankie is pulling the rope with 7 N, Carol must be pulling the rope with 11 N (I think that Carol is going to win the tug-of-war...).
Answer:
49.85 V
Explanation:
u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s
Let the electric field is E and voltage is V.
Use second equation of motion
s = ut + 1/2 a t^2
5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2
a = 8.5 x 10^10 m/s^2
m x a = q x E
E = m x a / q
E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)
E = 887.19 V/m
V = E x s
V = 887.19 x 5.62 x 10^-2 = 49.85 V
