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3 years ago

You drop a rock off the top of a building. It takes 4.5 s to hit the ground. How tall is

1 answer:

4 0

Since the rock as been dropped off the building you can assume that initial velocity equals 0m/s.

t=4.5s
acceleration due to gravity=9.8m/s^2

s=ut+1/2at^2

Since u=0, s=1/2at^2

s=1/2(9.8)(4.5)^
s=99.225m

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A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend

GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

$a=\frac{h}{5(t_1)^2}$

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = $\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2$

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

$a=\frac{h}{5(t_1)^2}$

6 0

3 years ago

Compute VO when Vin = 0.5 V, in two different ways: a) using the equation VO = G (V+-V-) with G = 106; and b) using the Golden R

Mrrafil [7]

Is b because yes thanks

7 0

3 years ago

An object located near the surface of Earth has a weight of a 245 N

marusya05 [52]

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

$m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg$

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

4 0

3 years ago

What are tectonic plates, and which of Earht's layers are they composed of?

motikmotik

The tectonic plates are made up of Earth's crust and the upper part of the mantle layer underneath. Together the crust and upper mantle are called the lithosphere. hope this helps :)

3 0

3 years ago

What is a difference between a law and a hypothesis?

AnnZ [28]

The correct answer is

C ). A hypothesis includes an explanation for why two variables affect each other, but a law only describes how they affect each other.

8 0

3 years ago