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3 years ago

What layer of the atmosphere is the most vital for supporting life on earth

1 answer:

OverLord2011 [107]3 years ago

8 0

The troposphere is most vital for supporting life because it is the onlyb layer that gives humans, plants, and animals weather and oxygen.

You might be interested in

Evaluate tan45/tan45

SIZIF [17.4K]

I'm not sure I completely understand the expression you want evaluated.

It looks like a fraction with the same exact thing in both the numerator and the denominator. A fraction like that always boils down to ' 1 '.

5 0

3 years ago

Read 2 more answers

PLEASE ANSWER ASAP!.!.!!.!!!!.!.!.

lbvjy [14]

1. Most PE, because PE is directly proportional to distance (height)
Height: 100 meters
Speed: 0 mph

2. Most KE, because KE is directly proportional to speed
Height: 10 meters
Speed: 40 mph

3. Most TE, average KE
Height: 10 meters
Speed: 40 mph

4. The skater gains thermal energy as she goes down the slope, because the speed of the skater increases, so it increases the total kinetic energy of the particles, and makes them vibrate faster, resulting in a higher temperature.

8 0

3 years ago

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is:

$\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}$

$\mu_{s} = 0.087$

The coefficient of static friction between the object and the disk is 0.087.

5 0

3 years ago

Suppose you average 75 MPH over the first halfof a drive, and your average speed for the entiredrive is 60 MPH. What was your av

alexandr402 [8]

Answer:

x = 45 MPH

Explanation:

given,

Average speed of the first half = 75 MPH

Average speed of entire ride = 60 MPH

Average speed of the second half = ?

let the average speed of the second half = x MPH

now,

average of entire ride is given as 60 mph so,

$\dfrac{75+x}{2} = 60$

$75+x= 2\times 60$

75 + x = 120

x = 120 -75

x = 45 MPH

hence, the average speed of the second half comes out to be 45 MPH.

5 0

4 years ago

A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co

exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0

3 years ago