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Sever21 [200]
3 years ago
6

Brandonyaelcamposbarrios Hsjd

Physics
2 answers:
Blizzard [7]3 years ago
4 0
What does this even mean is my answer
aleksley [76]3 years ago
4 0

Answer:

What-

What are u saying?

Do u need help with hw? do u have a question?

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I think 75 m

Explanation:

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As wavelength increase, the frequency of the wave increases / decreases.
Sveta_85 [38]

Answer:

Hey heres your ans

Explanation:

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6 0
3 years ago
At what point is the velocity vector changing most rapidly?
agasfer [191]

Answer:

It is changing at the same rate at all  points.

A velocity vector represents the rate of change of the position of an object. The magnitude of a velocity vector gives the speed of an object while the vector direction gives its direction. Velocity vectors can be added or subtracted according to the principles of vector addition.

Explanation:

Velocity as a Vector Quantity

Velocity is a vector quantity that refers to "the rate at which an object changes its position." Imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity. Because the person always returns to the original position, the motion would never result in a change in position. Since velocity is defined as the rate at which the position changes, this motion results in zero velocity. If a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. Every step must go into moving that person further from where he or she started. For certain, the person should never change directions and begin to return to the starting position.

Velocity is a vector quantity. As such, velocity is direction aware. When evaluating the velocity of an object, one must keep track of direction. It would not be enough to say that an object has a velocity of 55 mi/hr. One must include direction information in order to fully describe the velocity of the object. For instance, you must describe an object's velocity as being 55 mi/hr, east. This is one of the essential differences between speed and velocity. Speed is a scalar quantity and does not keep track of direction; velocity is a vector quantity and is direction aware.

4 0
3 years ago
Why is most of an atom empty space? PLEASE ANSWER FAST!!!
Roman55 [17]
The vast majority of an atom is indeed empty space. Most of it's mass is centered in the nucleus. Flying around the nucleus are the electrons, but they're very far away. Most of the atom is the space between the nucleus and the electrons.<span />
7 0
3 years ago
A man stands on the roof of a building of height 17.0 m and throws a rock with a velocity of magnitude 33.0 m/s at an angle of 3
arlik [135]

Answer:

(A). The maximum height above the roof reached by the rock is 16.48 m.

(B). The velocity of the rock before hitting the ground is 37.6 m/s.

(C). The horizontal distance from the base of the building to the point where the rock strikes the ground is 122.7 m.

Explanation:

Given that,

Height = 17.0 m

Velocity = 33.0 m/s

Angle = 33.0°

(A). We need to calculate the maximum height above the roof reached by the rock

Using formula of height

h=\dfrac{v^2}{2g}

h=\dfrac{(v_{0}\sin\theta)^2}{2g}

Put the value into the formula

h=\dfrac{(33\times\sin33)^2}{2\times9.8}

h=16.48\ m

(B). We need to calculate the time

Using equation of motion

-h=v_{y}t-\dfrac{1}{2}gt^2

Put the value into the formula

-17.0=(v_{0}\sin\theta)t-4.9t^2

-17.0=(33.0\times\sin33)t-4.9t^2

4.9t^2-17.9t-17.0

t = 4.435\ sec

We need to calculate the  magnitude of the velocity of the rock just before it strikes the ground

Using equation of motion

v_{y}=v_{0}\sin\theta-gt

v_{y}=(33.0\times\sin33)-9.8\times4.435

v_{y}=-25.48\ m/s

The velocity of the rock before hitting the ground is

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

v=\sqrt{(v_{0}\cos\theta)^2+v_{y}^2}

v=\sqrt{(33\cos33)^2+(-25.48)^2}

v=37.6\ m/s

(C). We need to calculate the horizontal distance from the base of the building to the point where the rock strikes the ground

Using equation of motion

R=v_{x}\times t

R=v_{0}\cos\theta\times t

Put the value into the formula

R=33\times\cos33\times4.435

R=122.7\ m

Hence, (A). The maximum height above the roof reached by the rock is 16.48 m.

(B). The velocity of the rock before hitting the ground is 37.6 m/s.

(C). The horizontal distance from the base of the building to the point where the rock strikes the ground is 122.7 m.

4 0
4 years ago
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