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2 years ago

15

Compared to the amount of radiation received from the Sun, about how much radiation does the surface of the earth receive from t

he atmosphere

1 answer:

babymother [125]2 years ago

5 0

Answer:

https://earthobservatory.nasa.gov/features/EnergyBalance#:~:text=The%20atmosphere%20absorbs%2023%20percent,surface%20radiates%20only%2012%20percent.

Explanation:

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Which resource would be the best choice to learn more information about studying martial arts?

zhuklara [117]

Answer:

Dance studio

Explanation:

Martial art use to defend ourself from any dangerous. Dance is a way to learn it

3 0

3 years ago

A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?

givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

a = -25 \ 30 = -0.833 m\s^2

the object is slowing down 0.83 meter every second

8 0

3 years ago

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0

3 years ago

DNA is coiled into chromosomes in a cell's

tia_tia [17]

Answer: Cell's nucleus

Explanation :

The full form of DNA is Deoxyribonucleic acid. The genetic information of a cell is organized in the DNA. It is inherited from parents by their children.

DNA is coiled into chromosomes in a cell's nucleus. It has a shape like a double helix. It is twisted in the form of spirals.

Hence, DNA is coiled into chromosomes in a cell's Nucleus.

6 0

3 years ago

Read 2 more answers

What is the value of the equivalent resistance for the three resistors connected in series?

Harman [31]

The value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

To find the answer, we have to know more about the equivalent resistance.

<h3>What is meant by equivalent resistance?</h3>

equivalent resistance is the total value of the resistance connected in a circuit.
If n resistors are connected in series, then the equivalent resistance will be,

$R_E=R_1+R_2+..........+R_n$

In our question we have three resistors. Thus, the equivalent resistance will be,

$R_E=R_1+R_2+R_3$

Thus, we can conclude that, the value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

Learn more about the equivalent resistance here:

brainly.com/question/11603204

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7 0

2 years ago