A = 59.35cm
B = 196.56g
C = 74.65g
<u>Explanation:</u>
We know,

and L = x+y
1.
Total length, L = 100cm
Weight of Beam, W = 71.8g
Center of mass, x = 49.2cm
Added weight, F = 240g
Position weight placed from fulcrum, y = ?

Therefore, position weight placed from fulcrum is 59.35cm
2.
Total length, L = 100cm
Center of mass, x = 47.8 cm
Added weight, F = 180g
Position weight placed from fulcrum, y = 12.4cm
Weight of Beam, W = ?

Therefore, weight of the beam is 196.56g
3.
Total length, L = 100cm
Center of mass, x = 50.8 cm
Position weight placed from fulcrum, y = 9.8cm
Weight of Beam, W = 72.3g
Added weight, F = ?

Therefore, Added weight F is 74.65g
A = 59.35cm
B = 196.56g
C = 74.65g
186282 miles. Hope it helps
The answer is:
Transverse dune
The explanation:
Transverse dune : is abundant barchan dunes It may merge into barchanoid ridges, which then grade into linear .
The transverse dunes is called that because they lie transverse, or across, the wind direction, with the wind blowing perpendicular to the ridge crest.
It is large, very asymmetrical, elongated dune lying at right angles 90° to the prevailing wind direction.
Transverse dunes have a gently sloping windward side and a steeply sloping leeward side.
They general form in areas of sparse vegetation and abundant sand are transverse dunes.
Answer:
The current decreases.
Explanation:
Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is
, where V is voltage, I is current and R is resistance.
Rearranging this equation, you get:
and

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation
, the value of V remains constant the value of I decreases, then the value of R, resistance will increase.
Thus, it can be seen that the current will decrease as resistance increases and vice versa.
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s