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3 years ago

The temperature on the moon is

2 answers:

7 0

When the sun shines, it illuminates only one part of the moon. The article states, "<span>Daytime on one side of the moon lasts about 13 and a half days, followed by 13 and a half nights of darkness. When sunlight hits the moon's surface, the temperature can reach 253 degrees F (123 C). The "dark side of the moon" can have temperatures dipping to minus 243 F (minus 153 C).</span><span>Oct 22, 2012". Therefore, b is correct</span>

Nuetrik [128]3 years ago

4 0

B.

It can go from very hot to very cold, it depends on the area of the moon and where the sunlight hits.

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Which of these is an example of temperature changing over time? A. a pan cools off on the stove B. a frozen dinner sits in the f

lesya [120]

Is this middle school

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3 years ago

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Which mineral test was used if a geologist refers to a mineral as metallic?

natita [175]

They would use a magnet

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3 years ago

The current in the wires of a circuit is 60 milliamps. If the resistance of the circuit were doubled (with no change in voltage)

anygoal [31]

Answer:30

Explanation:

Current=60 milliamps

Current=(voltage)/(resistance)

60=(voltage)/(resistance)

Doubling the resistance means multiplying both sides by 1/2

60x1/2=(voltage)/(resistance) x 1/2

30=(voltage)/2(resistance)

Therefore the resistance would be 30 milliamp if we double the resistance

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2 years ago

The cannon on a battleship can fire a shell a maximum distance of 26.0 km.

Alla [95]

It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is $\theta = 45^{\circ}$ .
In fact, the laws of motions on both x- and y- directions are
$S_x(t)= v_0 cos \theta t$
$S_y(t)= v_0 \sin \theta t - \frac{1}{2} gt^2$
From the second equation, we get the time t at which the projectile hits the ground, by requiring $S_y(t)=0$ , and we get:
$t= \frac{2 v_0 \sin \theta}{g}$
And inserting this value into Sx(t), we find
$S_x(t) = 2 \frac{v_0^2}{g} \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)$
And this value is maximum when $\theta=45^{\circ}$ , so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
$S_x(t)= \frac{v_0^2}{g} \sin (2\theta)$
And by using $S_x = 26 km=26000 m$ , we find the value of the initial velocity v0:
$v_0 = \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s$

8 0

3 years ago

A race finishes at the same place as it started. The contestants start from rest and cross the finish line running. Which one of

valentina_108 [34]

Answer:

b).average speed ≠ 0 m/s, average velocity = 0 m/s, average acceleration ≠ 0 m/s2

Explanation:

When a costentant starts a race, he/she acquire speed, velocity and acceleration.

Speed is just a quantity that describes position changing over time. Velocity is similar to speed, but here we have to take on account direction because velocity is a vectorial unit. To describe velocity of a body you must have the speed plus the direction the body is moving at a certain direction.

On the other hand, acceleration is the rate at which velocity changes over time. Like speed, is only a quantity and does not need a direction to be fully described.

So, in this problem, the constestants start from rest and finish running in the same point, It means that at the end the contestants' speed is diferent from 0. Then, the average velocity of a contestant, because of its vectorial nature, and because the contestant starts and finish in the same point, the sum of all velocity vectors throughout the race is equall to 0.

Finally, assuming the contestant accelerates at the beginning (starts with speed = 0 m/s and runs up to speed = x m/s )and keeps a stable speed throughout the race, and because it finishes running, the average acceleration is different from 0.

6 0

3 years ago