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vlabodo [156]
3 years ago
9

45 grams in kilo grams

Physics
2 answers:
Hunter-Best [27]3 years ago
7 0

Answer:

0.045 kg

Explanation:

1000 grams = 1kg

Agata [3.3K]3 years ago
4 0

Answer:

0.045 kilograms ...........

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Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

5 0
3 years ago
Write four (4) strategies on how can we manage land use around South Florida. For each strategy, explain it's benefits to the en
a_sh-v [17]

Answer:

What are some ways we can improve on our land use?

- Avoid deforestation and close the forest frontier.

The first-order priority is to end deforestation by closing the "forest frontier" or intact forests, to development.

- Increase agricultural productivity.

- Restore forests and landscapes.

- Reduce food loss and waste.

- Improve diets

3 0
3 years ago
What is the answer to this question?
icang [17]

Answer:

Explanation:

In a velocity/time (aka acceleration) graph, the slope of a line indicates the value of the acceleration in m/s/s. Acceleration is the change in velocity over the change in time. From 0 - 2 seconds, there is no change in velocity, so the acceleration during this interval is 0 (which is the same as the slope of the line). From 2 - 4 seconds, the slope of the line is -2, so the acceleration during the time interval from 2 to 4 seconds is -2 (negative because David is slowing down but is still going the same direction: to the right).

8 0
3 years ago
A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8
mylen [45]

Answer:

f = 2.6 \times 10^{-13}

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

N = \frac{m}{29} \times 6.02 \times 10^{23}

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

N_e = m(6.02 \times 10^{29})

now diameter of the ball is 7.0 mm

density of the ball = 8900 kg/m^3

now we have

m = (\frac{4}{3}\pi r^3)(8900)

m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)

m = 1.6 gram

now we have

N_e = 9.63 \times 10^{23}

now the charge on the copper ball is 40 nC

so the number of electrons removed

Q = ne

40 \times 10^{-9} = n(1.6 \times 10^{-19}

n = 2.5 \times 10^{11}

so the fraction of number of electrons removed is given as

f = \frac{n}{N_e}

f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}

f = 2.6 \times 10^{-13}

7 0
3 years ago
Sam heats an 8kg sample of sand, with a specific heat of 664 J/kg·C°, from 20° to 40°. What is the change in thermal energy?
zysi [14]

Answer:

106.24 kJ.

Explanation:

Given that,

Mass of sample of sand, m = 8 kg

Specific heat of sand, c = 664 J/kg-°C

The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :

Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ

So, the change in thermal energy is 106.24 kJ.

7 0
3 years ago
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