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elena-14-01-66 [18.8K]

3 years ago

15

The temperature in the upper atmosphere is -2°C, and the temperature on Earth’s surface is 5°C. Which type of weather is most li

kely in these conditions?

2 answers:

Troyanec [42]3 years ago

8 0

Snowing

I hope this helps and have a wonderful day filled with joy!!

Morgarella [4.7K]3 years ago

4 0

Well, I assume it would be snowing.

Snow forms when the atmospheric temperature is at or below freezing (0 degrees Celsius<span> or </span>32 degrees Fahrenheit<span>) and there is a minimum amount of moisture in the air. If the ground temperature is at or below freezing, the snow will reach the ground.
</span>
So, to be exact. It will be snowing since the temperature in the atmosphere is below freezing and the snow will reach and stay on the group since the ground is also below freezing.

Hope this helped :)

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Two wheels have the same mass and radius of 4.4 kg and 0.48 m, respectively. One has (a) the shape of a hoop and the other (b) t

abruzzese [7]

Answer:

1) $\sum T_{externalhoop}=0.4418Nm$

2) $\sum T_{externaldisc}=0.2209Nm$

Explanation:

Using the second equation of angular motion we have

$\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}$

Since the wheels start from rest we ahve $omega _{o}=0$

Applying the given values in the equation we have

$14=\frac{1}{2}\alpha \times 8^{2}\\\\\therefore \alpha =\frac{28}{64}=0.4375rad/s^{2}$

Now by newton's second law of motion in angular motion we have

$\sum T_{external}=I\alpha$

1) For Hoop We have

$I_{hoop}=Mr^{2}\\\\\therefore I_{hoop}=4.4\times(0.48)^{2}=1.01376kgm^{2}$

Thus

$\sum T_{external}=1.01376\times 0.4375$

$\sum T_{external}=0.4418Nm$

2)For disc We have

$I_{disc}=\frac{Mr^{2}}{2}\\\\\therefore I_{hoop}=2.2\times(0.48)^{2}=0.506kgm^{2}$

Thus

$\sum T_{external}=0.506\times 0.4375$

$\sum T_{external}=0.2209Nm$

5 0

4 years ago

_______________ involves removing the top layer of earth, mining the coal, and then replacing the earth back on the surface

aniked [119]

Surface mining is the method of removing the overburden parts of the earth which are the earth and rocks. It also involves removal and loading out of the coal, and then placing back the collected earth back into a surface once again. It's also a method that uses equipment or explosives.

5 0

3 years ago

I’LL MARK YOU BRAINIEST IF U ANSWER IT

Kryger [21]

4 is the answer to this question

6 0

3 years ago

Read 2 more answers

(PLEASE HELP! Will give Brainliest Answer)

soldi70 [24.7K]

Answer:

B

Explanation:

872 / 20 = 43

4 0

4 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago