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2 years ago

Does Anybody Know The Answers?

Physics

1 answer:

ch4aika [34]2 years ago

5 0

Answer:

I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry

Explanation:

$9^{2} + 12^2 = x^2\\81 + 144= x^2\\\sqrt{225} = \sqrt{x} \\ 15=x\\\\ 2.\\x^2+12^2+=13^2\\x^2+144 =169\\x^2 = 25\\\sqrt{x^2 =\sqrt{25\\\\$

x=5

$3.\\12^2+32^2 = x^2\\34.176= x$

5,12,13

$\frac{x}{4} ,\frac{12}{4} ,\frac{20}{4}\\\\\frac{x}{4},3,5 \\\\x=16\\\\12. \\x^2 + 48^2=50^2\\\\x^2=196\\x=14$

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An astronaut in a space craft looks out the window and sees an asteroid move pas a backward direction at 68 mph relative to the

Annette [7]

Answer: -194 mph

Explanation:

Taking into account the Sun as the center (origin, point zero) of the reference system, the velocity of the spacecraft relative to the Sun $V_{R-S}$ is:

$V_{R-S}=126mph$ Note it is positive because the spacecraft is moving away from the Sun

Taking into account the spacecraft as the center of another reference system, the velocity of the asteroid relative to the spacecraft $V_{A-R}$ is:

$V_{A-R}=-68mph$ Note it is negative because the asteroid is moving towards the spacracft.

Now, the velocity of the asteroid relative to the Sun $V_{A-S}$ is:

$V_{A-S}=V_{A-R}-V_{R-S}$

$V_{A-S}=-68mph-126mph$

Finally:

$V_{A-S}=-194mph$ This is the velocity of the asteroid relative to the Sun and its negative sign indicates it is moving towards the Sun.

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3 years ago

For general projectile motion, which of the following best describes the horizontal component of a projectile's acceleration? As

malfutka [58]

Answer:

E ) The horizontal component of a projectile acceleration is zero.

Explanation:

In case of a projectile , force of gravity acts in vertically downward direction so acceleration will act in vertically downward direction . Its direction never changes during course of its journey. So horizontal component of acceleration will always be zero at all points of its journey.

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3 years ago

Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance bet

Elena-2011 [213]

Answer:

243 N

Explanation:

The formula for electromagnetic force is F= Kq1q2/r^2

where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N

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3 years ago

Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o

jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

= -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28 × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

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3 years ago

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Are there any answer choices?
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Hope this helps! :)

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3 years ago

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