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Flauer [41]
3 years ago
7

You are less likely to see a total solar eclipse than a total lunar eclipse because a. the moon’s shadow covers all of Earth dur

ing a solar eclipse. b. new moon phases occur less often than full moon phases. c. only people on the daytime side of Earth can see a solar eclipse. d. the moon’s umbra only covers a small area on Earth’s surface
Physics
2 answers:
Bad White [126]3 years ago
8 0
D. Because the moons shadow during a total lunar eclipse is tinnier than the earth.
frozen [14]3 years ago
4 0

Answer:

You are less likely to see a total solar eclipse than a total lunar eclipse because

d. the moon’s umbra only covers a small area on earth’s surface.

Explanation:

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?
adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

\sum F = 0

F_b -W = 0

F_b = W

F_b = mg

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

\rho_w V_{displaced} g = mg

Remember the expression for which you can determine the relationship between mass, volume and density, in which

\rho = \frac{m}{V} \rightarrow m = V\rho

In this case the density would be that of the object, replacing

\rho_w V_{displaced} g = V\rho g

Since the displaced volume of water is 0.429 we will have to

\rho_w (0.429V) = V \rho

0.429\rho_w= \rho

The density of water under normal conditions is 1000kg / m ^ 3, so

0.429(1000) = \rho

\rho = 429kg/m^3

The density of the object is 429kg / m ^ 3

7 0
3 years ago
How can light energy solve our real life problem?
kifflom [539]

Answer:

It gives our light which we need for probably everything.

Explanation:

4 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP
aleksandrvk [35]

Answer:

changing from a solid to a gas without changing into a liquid. :)

6 0
2 years ago
Kieran is graduating from high school. He wants to be a urologist. What career path would be best for Kieran to follow?
faust18 [17]

Answer:

four-year degree Right arrow. graduate school Right arrow. license from the NCLEX

8 0
3 years ago
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