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1 year ago

14

What do echolocation and ultrasounds have in common?

1 answer:

Natali5045456 [20]1 year ago

8 0

They both are mechanical waves.

Explanation:

Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.

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What are some words that have changed meaning due to technological advances in the last 20 years?

vazorg [7]

Some words that have changed meaning due to technological advances are dial, type, tweet, drone, and spam.

8 0

1 year ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

3 years ago

A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net

Schach [20]

Explanation:

Let us first calculate long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0

2 years ago

A tuna fish body is streamlined to reduce friction. Which describes the type of friction the fish must overcome?

STatiana [176]

The answer would be Fluid Friction

7 0

2 years ago

Read 2 more answers

A 3.04 kg particle is located on the x-axis at xm = −8 m, and a 5.61 kg particle is on the x-axis at xM = 3.56 m. Find the coord

Murljashka [212]

Answer:

center of mass = −0.50 m

Explanation:

given data

mass m1 = 3.04 kg

distance xm = -8 m

mass m2 = 5.61 kg

distance xM = 3.56 m

solution

we get here center of mass for n mass of system that is express as

center of mass = $\frac{m_1x_1+m_2x_2......m_nx_n}{m_1+m_2...m_n}$ ......................1

but we have only 2 particle system so we will get

center of mass = $\frac{m1 \times xm+m2 \times xM}{m1+m2}$ .................2

put here value and we will get

center of mass = $\frac{3.04 \times (-8 )+5.61 \times 3.56}{3.04 + 5.61}$

solve it we will get

center of mass = −0.50 m

8 0

3 years ago