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algol [13]
3 years ago
9

A hot air balloon is moving at a speed of 10 meters/second in the +x direction. The balloonist throws a brass ball with a veloci

ty of -2
meters/second with respect to himself. If the ball lands after 20 seconds, about how far does it land horizontally from the point at which it was
released?
OA 98 meters
OB. 160 meters
OC. 196 meters
OD. 420 meters
Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

160 meters

Relative of speed Vr = 10 -2 = 8 m/s    (horizontal speed)

20 sec * 8 m/s = 160 m      since ball travels 20 sec

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3 years ago
PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph
hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

x = \frac{L}{\frac{W}{F} +1}

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 = ?

L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm

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 = ?

47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8  = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W  = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g

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 = ?

50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8  = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

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