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Sergeeva-Olga [200]
2 years ago
15

Help ASAP). Draw additional masses to show how the block can be made to move toward point A. Be sure to draw the string and labe

l the mass of the block. For full credit, draw the minimum number of blocks required to make the block move toward point A.​ Will Mark Brainliest​

Physics
1 answer:
chubhunter [2.5K]2 years ago
4 0

Answer:

Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.

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Answer:

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2 years ago
Use the periodic table and your knowledge of isotopes to complete these statements.
klasskru [66]
<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.

When polonium-210 emits an alpha particle, the child isotope has an atomic mass of </span><span> ⇒ 206</span>.</span>

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2 years ago
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A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu
Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 m/s^2

Explanation:

Mass (m) of the sailboat =  2000 kg

Force acting on the sailboat due to ocean  tide is F_1 = 3000N

Eastwards means takes place along the positive x direction

ThenF_{1x} = 3000N and F_{1y}= 0

Wind Force acting on the Sailboat isF_2  = 6000N directed towards the northwest that means at an angle  45 degree above the negative x axis

Then  

F_{2x} = -(6000N) cos 45 degree = -4242.6 N

F_{2y}  = (6000N) cos 45 degree = 4242.6 N

Hence  , the net force acting on the sailboat in x direction is  

F_x = F_{1x}+ F_{2x}

=  - 3000 N + 4242.6 N

=  - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is  

F_y = F_{1y}+ F_{2y}

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

\sqrt{(1242.6)^2 + (4242.6)^2

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

4420.8 N

F = ma

a = \frac{F}{m}

a =\frac{4420.8}{ 2000}

=2.2 m/s^2

4 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

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The correct answer is:
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