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3 years ago

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Katarina [22]3 years ago

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I don’t understand the question

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A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the

max2010maxim [7]

Answer:

V = 381.70 m³

Explanation:

ρ air = 1.28 kg / m³

ρ helium = 0.18 kg / m³

R = 4.5 m

Vb = 0.068 m³

mb = 123 kg

To determine the volume of helium in the balloon when fully inflated

V = 4 / 3 π * R ³

V = 4 * π / 3 ( 4.5 m )³

V = 381.70 m³

To determine the mass total

m = ρ helium * V

m = 0.18 kg / m³ * 381.70 m³

m = 68.70 kg

mt = ( 68.70 + 123 )kg

mt = 191.70 kg

3 0

3 years ago

If a boy (m = 50kg) at rest on skates is pushed by another boy who exerts a force of 200 N on him and if the first boy's final v

AVprozaik [17]

<span>F* t = (m x v_final) - (m x v_initial)

200 x t = 50x8 - 50x0

t= 50 x 8 /200

t= 2s </span>

8 0

3 years ago

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A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much

nekit [7.7K]

Answer:

a $W_{3.5} = 490 \ J$

b $W_{2.5} = 250 \ J$

Explanation:

Generally the force constant is mathematically represented as

$k = \frac{F}{x}$

substituting values given in the question

=> $k = \frac{16}{0.2}$

=> $k = 80 \ N /m$

Generally the workdone in stretching the spring 3.5 m is mathematically represented as

$W_{3.5} = \frac{1}{2} * k * (3.5)^2$

=> $W_{3.5} = \frac{1}{2} * 80 * (3.5)^2$

=> $W_{3.5} = 490 \ J$

Generally the workdone in compressing the spring 2.5 m is mathematically represented as

$W_{2.5} = \frac{1}{2} * k * (2.5)^2$

=> $W_{2.5} = \frac{1}{2} * 80 * (2.5)^2$

=> $W_{2.5} = 250 \ J$

5 0

3 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

3 years ago

What is a Geographic test

Genrish500 [490]

<em>It's a test on Geography!
</em>

8 0

3 years ago

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