HEY CAN ANYONE PLS ANSWER DIS!!!!!!

If you toss a stick into the air, it appears to wobble all over the place. specifically, about what place does it wobble?

Otrada [13]

I would assume the end?

8 0

2 years ago

An object in free fall travels a distance s that is directly proportional to the square of the time t. If an object falls 1088 f

ikadub [295]

Answer:

1,700feet

Explanation:

If an object in free fall travels a distance s that is directly proportional to the square of the time t, this can be represented mathematically as;

S = kt²where;

k is the proportionality constant

K = s/t²

s1/t1²= s2/t2²= Sn/tn²= k for values of the distance and time. Using the formula

s1/t1² = s2/t2² where;

s1 is the falling distance in time t1 s2 is the falling distance in time t2

Given s1 = 1088feet, t1 = 8secs, s2 = ? t2 = 10secs

Substituting this value in the formula to get s2, we have;

1088/8²= s2/10²

64s2= 108800

s2 = 108800/64

s2 = 1,700feet

This means the object will fall a distance of 1,700feet in 10seconds

5 0

2 years ago

A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct

Taya2010 [7]

Answer:

VB − VA = g tAB & (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s$

$v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s$

$\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s$

$\frac{h}{t}=\frac{3.8}{0.02}=190\ s$

Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used

3 0

2 years ago

Observations must be _____. <br> subjective <br> objective <br> biased <br> deductive

Dmitrij [34]

Objective
because it means<span> based on measurement, or reasoning free of bias.</span>

6 0

2 years ago

Read 2 more answers

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