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LenKa [72]
3 years ago
7

Enample: the motion of moon around the earth Describe the motion of an object in which its speed constant but the velocity is ch

anging
​
Physics
2 answers:
Hatshy [7]3 years ago
8 0

Explanation:

Speed refers to how fast an object is moving. It can be thought of as the rate at which an object covers distance.

Velocity refers to the rate at which an object changes its position. If you picture a person moving rapidly - one step forward and one step back- always returning to the original starting position, the speed is very rapid, but the velocity is zero. Because the person always returns to the original position, this motion would bébé result in a change in position. Since velocity is defined as the rate at which the position changes, this motion results in zero velocity. To maximize velocity, every effort must be made to maximize the amount an object is displaced from its original position. Every movement should be moving the object further from where it started. Velocity is DIRECTION AWARE. When evaluating the velocity of an object, you have to keep track of direction. This is one of the essential differences between speed and velocity: speed does not keep track of direction, while velocity is directionally aware.

So, when the moon moves around the Earth, the speed remains constant, but since it's moving in an elliptical orbit, it's direction is constantly changing.

Vladimir79 [104]3 years ago
3 0

Answer:

To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed. While the speed of the object is constant, its velocity is changing. Velocity, being a vector, has a constant magnitude but a changing direction.

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The wavelength of yellow sodium light in air is 589 nm.
julsineya [31]

Explanation:

answer is attached, kindly see answer

3 0
3 years ago
John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda
NeX [460]

Answer:

They are both correct.

Explanation:

The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.

Hence considering the both statements objectively, one can say that they are both correct.

8 0
4 years ago
What is the frequency of a sound wave with a wavelength of 0.04 meter in air? What type of wave is
Viktor [21]
8500 Hz and Longitudinal


Speed = frequency x wavelength

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4 0
3 years ago
Read 2 more answers
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
3 years ago
Is there any change in the pressure of container filled with water when the volumed is increased
marshall27 [118]
Not really the volume of a container is simply length X width X depth so just how big the container unless the water is pressurized by some sort of weight or if the containers air pressure is lowered
7 0
3 years ago
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