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3 years ago

If an object has a mass of 26 g on earth would its mass be less than 26 g on the moon

Physics

1 answer:

aleksley [76]3 years ago

8 0

Answer:

Explanation:

No matter where an object is, its mass will stay the same, but its weight might change depending on gravity.

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How might spontaneous generation be possible in a reducing environment? Question options: The reducing environment was created b

melomori [17]

Please help me with the unit assessment i have no idea any of the answers

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3 years ago

If 36 columb of charge pass through a wire in 45 what current is it carrying ?

VLD [36.1K]

Current I is the rate at which charge moves through an area A, such as the cross-section of a wire

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2 years ago

A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

Georgia [21]

Answer: 10000 kg

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Now, in this case and according the conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$ (2)

Where:

$m_{1}=1000kg$ is the mass of the spacecraft

$m_{2}$ is the mass of the asteroid

$v_{1}=0$ is the initial velocity of the spacecraft

$v_{2}=0$ is the initial velocity of the asteroid (because we are told the asteroid is stationary, as the spacracft is on the sateroid it remains stationary as well)

$u_{1}=250m/s$ is the final velocity of the spacecraft

$u_{2}=-25m/s$ is the final velocity of the asteroid

Rewritting (2):

$0=m_{1}u_{1}+m_{2}u_{2}$ (3)

$0=(1000kg)(250m/s)+m_{2}(-25m/s)$ (4)

Finding $m_{2}$ :

$m_{2}=10000kg$ This is the mass of the asteroid

3 0

3 years ago

Read 2 more answers

How is friction reduced between an air hockey puck and the table.

iren2701 [21]

Answer:

An air hockey table has a very smooth and slippery surface which reduces friction by suspending the puck on cushion of air, so that this its motion is much less altered by friction, causing it to glide in a straight line at relatively constant velocity across the table.

Explanation:

6 0

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

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

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

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

4 0

3 years ago