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3 years ago

What will be the weight (acceleration due to gravity on the moon is 1/6 that of the earth

Physics

1 answer:

jeyben [28]3 years ago

7 0

Answer:

The weight will be 1/6 of whatever it is on Earth.

Explanation:

Remember F=ma

If the acceleration decreases by a factor of 6 while the mass stays constant, you can see the force (the weight) will also decrease by the same factor.

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You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find

wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0

3 years ago

Help please (20 pts) I think it's C

gregori [183]

Answer:

I'm not sure it is c I'm sure it is d

6 0

3 years ago

OK brainly if you want hard here here are sixty cups on a table. If one falls down, then how many remain

vivado [14]

Answer:

If one cup falls down then there will be 59 cups left.

5 0

3 years ago

Read 2 more answers

Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.

Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

$ma = \text{Mass}\times \text{Acceleration}$

$ma = kg \cdot \frac{m}{s^2}$

At the same time we have that

$\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2$

$\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2$

$\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}$

Therefore there is not have same units and both are not consistent and the correct answer is B.

5 0

3 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

3 years ago