Answer:
WHat do you mean by that
Explanation:
I dont get what you re asking
Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
Answer:
X-Positions: Y-Positions
x(0) = 0 y(0) = 0
x(2) = 120 m y(2) = 19.6 m
x(4) = 240 m y(4) = 78.4 m
x(6) = 360 m y(6) = 176.4 m
x(8) = 480 m y(8) = 313 m
x(10) = 600m y (10) = 490 m
Explanation:
X-Positions
- First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
- After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
- So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

- It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.
Y-Positions
- We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
- As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
- At any moment, it is subject to the acceleration of gravity, g.
- As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

- Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
- y(2) = 2* 9.8 m/s2 = 19.6 m
- y(4) = 8* 9.8 m/s2 = 78.4 m
- y(6) = 18*9.8 m/s2 = 176.4 m
- y(8) = 32*9.8 m/s2 = 313.6 m
- y(10)= 50 * 9.8 m/s2 = 490.0 m
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror