Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t
Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C
Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C
Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C
Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C
Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
i'm stuck on that question also
Answer:
The velocity of a falling object
Explanation:
The positive X axis is towards right and positive Y axis is towards up, so North direction is positive
A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.
Any vector whose magnitude is greater than 1 is never be a negative vector.
The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
Explanation:
→ Volume of cone = πr² × h/3
Here,
- Radius (r) = 13 cm
- Height (h) = 27 cm
→ Volume of cone = π(13)² × 27/3 cm³
→ Volume of cone = 169π × 9 cm³
→ Volume of cone = 1521π cm³
→ Volume of cone = 1521 × 22/7 cm³
→ Volume of cone = 33462/7 cm³
→ <u>Volume of cone = 4780.28 cm³</u>
