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Leno4ka [110]
3 years ago
7

Which is heavier: 30 kilogram or 300 milligrams?

Physics
1 answer:
Ainat [17]3 years ago
5 0

Answer:

30 killograms is 85.9803

and 300 milligrams is 0.000661387 pounds

Explanation:

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algol13

Answer:

Generator

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3 years ago
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What causes the mechanical advantage of the inclined plane to be less than that of the lever? 30 points !!!
gayaneshka [121]
Moving an object up an inclined plane<span> requires </span>less<span>force </span>than<span> lifting it straight up, at a cost of an increase in the distance moved. The </span>mechanical advantage<span>of an </span>inclined plane<span>, the factor by which the force is reduced, is equal to the ratio of the length of the sloped surface to the height it spans.</span>
8 0
3 years ago
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
(A) If an atom of uranium-238( U ) undergoes alpha decay, what daughter isotope is produced? (B) If this radioisotope from (A) t
salantis [7]

Answer:

Th-234, Pa-234

Explanation:

In alpha decay, 2 protons and 2 neutrons are lost.  So U-238 would become Th-234.

In beta decay, one neutron is turned into a proton.  So Th-234 would become Pa-234.

7 0
3 years ago
A 5.00X10^5 kg rocket is accelerating straight up. Its engines produce 1.250X10^7 N of thrust, and air resistance is 4.50X10^6 N
katrin [286]
<span>6.20 m/s^2 The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be 9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N Add in the atmospheric drag and you get 4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N Now subtract that total drag from the thrust available. 1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So 3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2 Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
8 0
3 years ago
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