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3 years ago

A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of

Physics

1 answer:

Mashcka [7]3 years ago

3 0

1.6 m/s west is the answer

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What human disease can result from exposure to ultraviolet radiation from the sun?

aev [14]

Melanoma skin cancer

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3 years ago

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A light ray strikes a flat, smooth, reflecting surface at an angle of 80° to the normal. What angle does

charle [14.2K]

Answer:

80 angle of incidence=angle of reflection

Explanation:

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3 years ago

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A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

$r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}$

$r_{cm} = {(37.26, 0)}{14.8}$

$r_{cm} = (2.52 cm, 0)$

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

$r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}$

$r_g_{cm} = {(37.26, 0)}{14.8}$

$r_g_{cm} = (2.52 cm, 0)$

So center of mass is same as center of gravity because value of gravity is constant here

3 0

3 years ago

A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car

Gekata [30.6K]

Answer:

1.63366

Explanation:

I got this answer from calculator soups physics calculators. I really recommend their website for formulas.

3 0

3 years ago

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

netineya [11]

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

6 0

2 years ago