Speed in any given direction is called A. Trust B.velocity C.mass D.acceleration

The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward

ch4aika [34]

The object's initial velocity is equal to $-10.59\frac{m}{s}$

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

$y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}$

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

So, isolating the initial velocity, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}$

Finally, substituting and calculating, we have:

$v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}$

Hence, we have that the initial velocity of the object is $-10.59\frac{m}{s}$

Have a nice day!

7 0

4 years ago

Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocit

ser-zykov [4K]

The reaction time of Boris is t(r), so before that, Boris will not have jumped. Thus, H(b)(t) = 0
The vertical displacement will simply be
D(t) = H(a)(t)

7 0

3 years ago

The earth exerts the necessary centripetal force on an orbiting satellite to keep it moving in a circle at constant speed. Which

jenyasd209 [6]

Answer:

E) The centripetal force is always perpendicular to the velocity.

Explanation:

Due to gravity and inertia, the satellite follows a uniform circular motion. In this movement, the velocity is always tangent to the orbit and the centripetal force is directed towards the center. Therefore, there is no net acceleration in the same direction of velocity, which implies that it remains constant.

8 0

4 years ago

A mass moves back and forth in simple harmonic motion with amplitude A and period T.

Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

$1 T : 4 A = t : 2 A$

and solving for t we find

$t=\frac{(1T)(2 A)}{4A}=0.5 T$

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

$1 T : 4 A = t : 5 A$

And by solving for t, we find

$t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T$

6 0

3 years ago

When a beam of light passes at an oblique angle into a material of lower optical density, the angle of incidence is

Anastasy [175]

The correct answer is letter B. less than the angle of refraction. When a beam of light passes at an oblique angle into a material of lower optical density, the angle of incidence is less than the angle of refraction. The reason for that because the beam of light passes through an oblique angle.

5 0

4 years ago

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