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ANEK [815]
3 years ago
14

ANSWER ASAP

Physics
1 answer:
Hoochie [10]3 years ago
6 0
The answer to this would inFact be A
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True,when you turn the volume up on your  television , you're actually turning up the amplitude<span>!

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What is the scientific term for rocks formed from lava?
sergeinik [125]

Answer:

extrusive I'm pretty sure that's right

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Quick does the watermelon have more or less mass then the 2kg bottle?
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Answer

it will be less\

Explanation:

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Explain how to find the angle between two nonzero vectors. Choose the correct answer below. A. The angle between two nonzero vec
kozerog [31]

Answer:

θ = Cos⁻¹[A.B/|A||B|]

A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result

Explanation:

We can use the formula of the dot product, in order to find the angle between two non-zero vectors. The formula of dot product between two non-zero vectors is written a follows:

A.B = |A||B| Cosθ

where,

A = 1st Non-Zero Vector

B = 2nd Non-Zero Vector

|A| = Magnitude of Vector A

|B| = Magnitude of Vector B

θ = Angle between vector A and B

Therefore,

Cos θ = A.B/|A||B|

<u>θ = Cos⁻¹[A.B/|A||B|]</u>

Hence, the correct answer will be:

<u>A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result</u>

3 0
3 years ago
A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t
Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

T = \frac{2u Sin\theta }{g}

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}

v_{y}=u Sin 30 - 2u Sin 30

vy = - u Sin 30

The magnitude of final velocity is given by

v = \sqrt{v_{x}^{2}+v_{y}^{2}}

v = \sqrt{\left (uCos 30  \right )^{2}+\left (uSin 30  \right )^{2}}

v = u

Thus, the velocity is same as it just reaches the ground.

6 0
3 years ago
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