A ) The displacement:
d = v o t - (gt²) / 2 =
= 19.6 m/s × 1 s - ( 9.8 m/s² x 1 s² ) / 2 =
= 19.6 m - 4.9 m = 14.7 m
b ) v = v o - g t
0 = 19.6 m/s - 9.8 t ( when the ball is at the highest point )
9.8 t = 19.6
t = 19.6 : 9.8
t = 2 s
h = v o t - (gt²)/2 = 19.6 x 2 - ( 9.8 x 4 ) / 2 = 39.2 - 19.6
h = 19.6 m
c ) h = gt² / 2
19.6 = 9.8 t²/2
9.8 t² = 39.2
t² = 39.2 : 9.8
t² = 4
t 2 = 2 s ( and we know that t 1 = 2 s )
t = t 1 + t 2 = 2 s + 2 s = 4 s
Answer:
The object distance is 20cm
Explanation:
Given
Magnification = -1.5
Image distance = 30 cm.
Required
Object Distance
We can calculate the object's distance using magnification formula;
M = -V/U
Where M = Magnification = -1.50
V = Image Distance = 30cm
U = Object Distance
Substitute the above parameters in the formula above.
-1.50 = -30/U
Multiply both sides by -1
-1.50 * -1 = -30/U * -1
1.50 = 30/U
Multiply both sides by U
1.50 * U = 30/U * U
1.50U = 30
Divide through by 1.50
1.50U/1.50 = 30/1.50
U = 30/1.50
U = 20cm
Recall that U represented the object distance.
Hence, the object distance is 20cm
Answer:
so height is 0.1283 m
Explanation:
given data
height = 28 cm
diameter = 11 cm
cross-sectional area = 1.55 cm2
water flow rate = 2.46×10^−4 m3/s
to find out
How high will the water in the bucket rise
solution
we know that here
potential energy = kinetic energy
mgh = 1/2 mv²
multiply both sides by the 2 and we get
2mgh=mv²
solve it we get
√(2gh) = v ....................1
h = v²/2g ...............2
and
flow rate = A V
2.46×10^−4 = V 1.55×10^−4
V = 1.5870 m/s
so from 2
h = v²/2g
h = 1.5870²/ 2(9.81)
h = 0.1283 m
so height is 0.1283 m
Answer:
A car traveling on a highway takes to stop without skidding. What happens to its kinetic energy? ... The energy goes to parts of the brakes. No, the energy can only be reused if there is a heat pump present
