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GenaCL600 [577]
4 years ago
14

When an object like a tree is illuminated by the sun, and you are looking toward the tree, light rays leave the object _________

___.a.from every point on the surface of the tree, and in every directionb. only from points at the top and base of the tree, but only toward your eyesc. only from points at the top and base of the tree, but in every directiond. from every point on the surface of the tree, but only toward your eyes
Physics
1 answer:
Dima020 [189]4 years ago
6 0

Objects absorb and reflect light differently depending on their physical characteristics, such as their shape or composition. Thanks to the reflection we can see the objects. Reflection can be defined as the change of direction of a wave, which, when in contact with the separation surface between two changing means, returns to the point where it originated. When the light illuminates the object, such as the tree, the rays of light will disperse in all directions allowing observation.

The correct answer is A. From every point on the surface of the tree, and in every direction

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A supply bag is dropped from a rescue plane. After the bag falls for 3.2 seconds , what is the velocity of the bag?
loris [4]

Answer: -31.36 m/s

Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

V_{f}=V_{o}+a.t  (1)

Where:

V_{f} is the final velocity of the supply bag

V_{o}=0 is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

a=g=-9.8m/s^{2} is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

t=3.2s is the time

Knowing this, let's solve (1):

V_{f}=0+(-9.8m/s^{2})(3.2s)  (2)

Finally:

V_{f}=-31.36m/s  Note the negative sign is because the direction of the bag is downwards as well.

8 0
3 years ago
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

6 0
3 years ago
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