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KatRina [158]
3 years ago
11

+

Physics
1 answer:
Triss [41]3 years ago
4 0

The acceleration of the boat is 2 m/s^2 (eastward)

Explanation:

We can solve this problem by using Newton's second law, which states that:

F = ma (1)

where

F is the net force on a body

m is its mass

a is its acceleration

First of all, we have to find the net force acting on the boat. We have:

- The force of the motor, 100 N eastward

- The force of the air resistance, 60 N westward

So the net force is:

F = 100 N - 60 N = 40 N (eastward)

Now we can apply eq.(1), using:

F = 40 N

m = 20 kg (mass of the boat)

and solving for a, we find the acceleration:

a=\frac{F}{m}=\frac{40}{20}=2.0 m/s^2

And the direction is the same as the net force (eastward).

Learn more about acceleration and Newton laws of motion:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

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Explanation:

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Formula:

v = fw

Solution:

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v = 30 m\ s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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Ohm's law relates the current, voltage, and resistance in a circuit. Use Ohm's law to determine what will happen to the remainin
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Answer:

   R = ½ R₀

Explanation:

This is an exercise in Ohm's law,

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           V₀ = I₀ R₀               (1)

indicates that the voltage remains constant and the current is doubled

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we substitute

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            R = ½ V₀ / I₀

we replace by equation 1

            R = ½ R₀

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All parts of the electromagnetic spectrum travel at a speed of 3 × 108 m/s when traveling through no medium. A "vacuum” means th
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Answer:

the answer is D.

Explanation:

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Listed following are several fictitious stars with their luminosities given in terms of the Sun's luminosity (LSun) and their di
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Answer:

1. Nismo: L_{a}=0.1244

2. Ferdinand: L_{a}=0.0796, Shelby: L_{a}=0.0796

3. Enzo: L_{a}=0.0398

4. Lotus: L_{a}=0.0199

Explanation:

The luminosity of a star is the amount of light it emits from its surface. The luminosity is an intrinsic property of the star. Apparent brightness is how bright the star appears to a detector on Earth. Apparent brightness is not an intrinsic property of the star; it depends on the location of the observer.

As light travels towards an observer, it spreads out and covers a larger area, reducing the intensity. Thus the apparent brightness is inversely proportionate to the square of the distance between the star and observer.

Apparent brightness can be calculated by the formula below:

L_{a}=\frac{L}{4*pi*D^{2}}

Where, L_{a} is apparent brightness, L is luminosity, D is the distance between the star and observer

Enzo: 200 LSun , 20 ly

L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{200}{4*pi*20^{2}}

L_{a}=0.0398

Ferdinand: 400 LSun , 20 ly

L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{400}{4*pi*20^{2}}

L_{a}=0.0796

Nismo: 100 LSun , 8 ly

L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{100}{4*pi*8^{2}}

L_{a}=0.1244

Lotus: 400 LSun , 40 ly

L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{400}{4*pi*40^{2}}

L_{a}=0.0199

Shelby: 100 LSun , 10 ly

L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{100}{4*pi*10^{2}}

L_{a}=0.0796

Apparent brightness of stars is summarized as below. Absolute values are considered as units are unspecified

Enzo: L_{a}=0.0398

Ferdinand: L_{a}=0.0796

Nismo: L_{a}=0.1244

Lotus: L_{a}=0.0199

Shelby: L_{a}=0.0796

Organized in order of apparent brightness

1. Nismo: L_{a}=0.1244

2. Ferdinand: L_{a}=0.0796, Shelby: L_{a}=0.0796

3. Enzo: L_{a}=0.0398

4. Lotus: L_{a}=0.0199

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