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2 years ago

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A horizontal force of 0.9N act on a body of mass 0.1kg there is resistance of 0.15 Newton opposing the first Force what accelera

tion will be produced please help

1 answer:

natali 33 [55]2 years ago

7 0

Answer:

a = 7.5 [m/s²]

Explanation:

To solve this type of problem we must remember Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

∑F = sum of forces [N]

m = mass = 0.1 [kg]

a = acceleration [m/s²]

Now replacing:

$0.9-0.15=0.1*a\\a=7.5 [m/s^{2} ]$

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11 months ago

Convert 400 mm to m using the method of dimensional analysis

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1 year ago

At a temperature of 320K, the gas in a cylinder has a volume of 40.0 liters. If the volume of the gas is decreased to 20.0 liter

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Assuming the gas behaves ideally,
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What is the surface area to volume ratio of this cube

tamaranim1 [39]

I can't see that cube from here.

But if the length of the side of the cube is ' K ' units,
then the surface area of the cube is 6K² units², and
the volume of the cube is K³ units³.

The ratio of the surface area to the volume is

(6K² units²) / (K³ units³) = (6) / (K units) .

So for example, if the side of the cube is 2 inches, then
the ratio of surface area to volume is "3 per inch".

That's the answer. I did the whole thing in order to earn
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because I see from your username that you suck at math.
I'm sorry you decided that. Now that you've put up the
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2 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

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2 years ago