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3 years ago

7

Near the poles, more energy is reflected back into space than is absorbed. near the poles, more energy is reflected back into sp

ace than is absorbed.
a. True
b. False

1 answer:

VladimirAG [237]3 years ago

7 0

Correct answer: a) True.

In fact, the soil at the poles consists mainly of snow. Snow has an albedo of 90%: albedo is the fraction of light (and so, of energy) that is reflected back to the space. This means that at the poles, about 90% of the light is reflected back to the space, while only 10% is absorbed.

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Wolrd-class skiers use, rather than fight, the force of gravity to carve the most efficient path down a slope

Rudik [331]

I think it is true I think

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3 years ago

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What is the scientific saying/principle that explains why a hammer and a screwdriver, both made of steel, are used for different

deff fn [24]

Hammer and screwdriver perform different tasks because they are both simple machines.

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3 years ago

The weight of a body is 147N. what is its mass?

barxatty [35]

<u>To find the mass, with only the weight</u>:

⇒ must consider the relationship between the mass and weight

⇒ (<em>in other words</em>) we must find the equation that has both the

mass and weight

<u>Based on our physics knowledge, we know</u>:

$Weight=mass*gravitational_. acceleration$

Weight: 147N
Gravitational Acceleration: 9.8 m/s²

<u>Now let's plug in the values, and solve</u>:

$147N = mass*9.8_.m/s^2\\mass = 15_.kg$

<u>Answer: 15 kg</u>

Hope that helps!

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2 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

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3 years ago

Answer the following question about the attached diagram:

Paladinen [302]

I'm not entirely sure, but I think the first is A, and the second is inverted.

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3 years ago

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