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2 years ago

I need homework help

1 answer:

5 0

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

6) On Earth, the gravity acceleration is $g=9.81 m/s^2$ . The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
 $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$

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Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

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2 years ago

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Yuki888 [10]

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2 years ago

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Elan Coil [88]

Scientific theories are tested and proven over time; they are then considered scientific laws.

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3 years ago

Read 2 more answers

Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m

STatiana [176]

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

$F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]$