1. The diagram below shows two pulses, each of length (I), traveling toward 1 point

What are the different ways the body senses its surroundings?

vladimir2022 [97]

Answer: Smell, touch, feel, look.

Explanation:

5 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and

vladimir1956 [14]

Answer:

a) t= 0.92 s

b) h = 0.46 m

c) v = -6.04 m/s

Explanation:

A)

In order to find the total time that the feet are in the air, we must add two times:
1) time needed to reach to the maximum height (t₁)
2) time from when starts to fall from the maximum height until her feet hit the water (t₂)
In order to get t₁, we need to take into account that at her highest point, the vertical speed will be zero.
Taking for granted the value for the acceleration due to gravity,

g = -9.8 m/s2, we can apply the definition of acceleration, and replacing by the givens, we can find t₁ as follows:

$t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)$

In order to find t₂, we need to find first the highest point above the board, which is indeed what is asked for in b).
We can use the following kinematic equation, taking into account that at the highest point, the final velocity vf will be zero.
The equation can be written as follows:

$v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h (2)$

Replacing by the givens, and solving for Δh, we get:

$\Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)$

The total height over the water will be just the sum of the takeoff point (1.40 m over the water) and the value we found in (3):
H = 1.40 m + 0.46 m = 1.86 m
Now, we can use the equation that relates the vertical displacement with the time, remembering that v₀=0, as follows:

$H = \frac{1}{2}*g*t^{2} (4)$

Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:

$t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)$

The total time will the sum of t₁ and t₂:
t = 0.3 s + 0.62 s = 0.92 s (6)

B)

As we have just found, the highest point above the board was at 0.46m above the takeoff point, so the highest point above the board is just 0.46 m.

C)

In order to find the velocity when her feet hit the water, we can use the same equation (2), taking into account that v₀=0, and Δh = H= -1.86m.
Solving (2) for vf, we get:

$v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)$

5 0

3 years ago

PLEASE HELP MEEE!! I WILL MARK YOU AS BRAINLIEST AND EVERYTHING!! THIS IS DUE TODAY AND MY GRADES ARE AT STAKE!:(

irina [24]

Answer:

They will be 140 miles apart 8 hours after the first boy started the trip or 6 hours after the second boy started the trip.

Explanation:

x = the time that the first boy travels at 14 mph

x - 2 = the time the second boy travels at 14 mph

140 the distance between them

Since one travels north and the other east (their roads are perpendicular) the distance between them can be calculated using Pythagorean Theorem

(14*x)^2 + ((x-2)*14)^2 = 140^2

the solutions of the quadratic equation are

x1 = - 6 is not the solution since x > 0

x2 = 8 h is the solution

3 0

3 years ago

Read 2 more answers

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :