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3 years ago

All chordates are vertebrates? A. True B. False

2 answers:

8 0

All vertebrates are chordates but all chordates are not vertebrates.....So its false...an example is a lancelet...Please mark me brainliest

makvit [3.9K]3 years ago

3 0

This is true chordates are vertebrae

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Reasons why inductors opposes charges passing through it

saul85 [17]

Answer:

é isso daí msm

Explanation:

oq eu falei

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2 years ago

Which is the best location for storing radioactive wastes?

Lubov Fominskaja [6]

The answer is in chambers carved into the rock of a mountain. Radioactive waste should be well disposed where they are little to no anthropogenic activity that would uncover them. This is especially critical for high-level radioactive waste should be buried deep up to 2000 meters. While low-level radioactive waste can be buried 100 meters into the ground such as in sand, the risk is that it could contaminate the water table.

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3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Scorpion4ik [409]

Answer:

protect equipment by stopping the flow of electric current; protect your house from fire.

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3 years ago

A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,

masha68 [24]

Answer:

a) The car was moving at a speed of $29.167\ m.s^{-1}$

b) The negative sign of $v_o$ denotes that the observer is coming towards the police car which is the source of the sound.

c) $f_o=1283.33\ Hz$

Explanation:

Given:

original frequency of the source, $f=1200\ Hz$

speed of the source, $v_s=0\ m.s^{-1}$

velocity of the obstacle car be, $v_o$

speed of sound, $s=350\ m.s^{-1}$

observed frequency, $f_o=1100\ Hz$

Using the equation from the Doppler's effect:

$\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}$

$\frac{1100}{1200} =\frac{(350+v_o)}{350-0}$

$v_o=-29.167\ m.s^{-1}$

The car was moving at a speed of $29.167\ m.s^{-1}$

The negative sign of $v_o$ denotes that the observer is coming towards the police car which is the source of the sound.

Now when, $v_s=50\ m.s^{-1}$

Then, $f_o=?$

Using the Doppler's eq.:

$\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}$

$f_o=1283.33\ Hz$

3 0

3 years ago

Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a

mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where, taking upward as positive direction:

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the pebble is launched horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

$v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s$ (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

$v_x = 20.0 m/s$

So, the final speed of the pebble as it strikes the ground is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s$

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0

3 years ago