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4 years ago

The destruction of the ozone layer primarily occurs in the mesosphere. thermosphere. stratosphere. troposphere.

Physics

2 answers:

otez555 [7]4 years ago

7 0

Hello,

The correct answer is D) troposphere

Hope this helps!!!! :)

**(Vanessa)**

Fofino [41]4 years ago

4 0

I believe the answer would be Troposphere.

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What does the stirrup press against

AveGali [126]

We need a system to use those air vibrations to push against the surface of the inner ear fluid.

7 0

3 years ago

A bus travels a distance of 120 km with a speed of 40km per hour and returns with a speed of 30km per hour calculate the average

Vsevolod [243]

Answer:

35 km/hr

Explanation:

Average speed = (total of the speed)/(the sets of speeds given)

Direction does not matter in this instance since speed is only magnitude,

Average speed = (30 + 40)/2

Average speed = 70 ÷ 2

= 35 km/hr

3 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

An engineer wants to set up simple password protection with no usernames for some switches in a lab, for the purpose of keeping

Andre45 [30]

Answer:

A. A login vty mode subcommand

Explanation:

since we are protecting co-workers from connecting to the switches from their desktop PCs, we would need a Telnet line which is used to connect to devices remotely from other network devices on the same network segment as the device we want to connect to. A login local vty subcommand configures a local username for login access but since our design constraint is to configure without usernames, option A is the correct answer.

8 0

3 years ago

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

natima [27]

Answer:

The force of friction that acts on him is

$F_k=567N$

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

$F=m*a$

$F=F_w-F_k$

$m*a=F_w-F_k$

$F_w=81kg*10m/s^2=810N$

Sole to Fk

$81kg*3m/s^2=810N-F_k$

$F_k=810N-243N$

$F_k=567N$

4 0

3 years ago