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3 years ago

9

Higher levels of co2 are more likely to be found where on the earth? question 15 options: uniformly around the globe as winds bl

ow at the south pole around the equator at the north pole

1 answer:

AlladinOne [14]3 years ago

5 0

Uniformly around the globe. it is mostly found in earths atmosphere.

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A device that uses electricity and magnetism to create motion is called a (motor magnet generator) . In a reverse process, a dev

Tems11 [23]

the answers are a. and c.

I hope I help you out!

6 0

2 years ago

Read 2 more answers

Hi so this is 15 points :D

mario62 [17]

Answer:

A. The core makes up the majority of Earth's volume.

Explanation:

Although the core and mantle are about equal in thickness, the core actually forms only 15% of earth's volume whereas the mantle takes up 84%

8 0

2 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

2 years ago

A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the

Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0

2 years ago

Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm

mixer [17]

<h2>~<u>Solution</u> :-</h2>

Here, the <u>moment arm</u> is defined as follows;

The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

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7 0

2 years ago