What is the intensity of the electromagnetic light waves coming from the Sun just outside of the atmosphere of Venus, Earth and Mars

Answer:

Remember, NORTH ^, EAST >, SOUTH v, WEST <

Explanation:

It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.

I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)

Answer:

The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.

Explanation:

Edmentum

The sun is not a burning fire.

It's much much much hotter than that.

The sun's energy is the result of continuous nuclear fusion in it's core. We know how to do that on Earth, but the only thing we've been able to use it for so far is hydrogen bombs and other thermonuclear weapons.

Answer:

True

Explanation:

It definetly is because, if we are biased it sets us up to fail on some things.

Answer:

40 volts

Explanation:

Use the equation [tex]V=IR[/tex]

[tex]V=(4)(10)[/tex]

[tex]V=40[/tex]

Hi there!

We can begin by finding the total time taken for the ball to reach the net using the equation:

dₓ = vₓt

12.9 = 44.7t

12.9/44.7 = t = 0.289 s

Now, we can use the following equation to solve for displacement in the Y direction:

d = y₀ + vit + 1/2at²

There is no initial vertical velocity, so:

d = y₀ + 1/2at²

Plug in known values:

d = 1.28 + 1/2(-9.8)(0.289²)

d = 0.87m

Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.

Answer:

speed of sound, speed at which sound waves propagate through different materials. In particular, for dry air at a temperature of 0 °C (32 °F), the modern value for the speed of sound is 331.29 metres (1,086.9 feet) per second

Answer:

Explanation:

Initial vertical velocity

vy₀ = 50.0sin60 = 43.3 m/s

This initial velocity is reduced to zero by gravity in a time of

t = v/a = 43.3/9.81 = 4.41 s

h(max) = ½gt² = ½(9.81)4.41² = 95.6 m

The ball will return to earth in the same amount of time

t(max) = 2(4.41) = 8.82 s

The horizontal velocity is

vx = 50.0cos60 = 25.0 m/s

d = vt = 25.0(8.82) = 221 m

That 's one heck of a kick! No air resistance of course.

Answer:

Quiet, is the adjective in this sentence

Explanation:

She had a quiet voice. This statement uses the adjective quiet. Her voice can be characterized as quiet.

What is adjectives?

A word used to characterize a person, place, or thing is called an adjective. Adjectives are used to modify nouns or pronouns.

There are many adjectives that describe quiet, including muted, peaceful, reticent, silent, soft, gentle, mild, placid, private, secluded, sedate, serene, timid, smooth, calm, tranquil, modest, restrained, sober, and subdued.

Hence, the significance of the adjectives is aforementioned.

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#SPJ2

Answer:

F12 = G M1 M2 / R12^2

F12' = G M1 M2 / R12'^2

F12' / F12 = R12'^2 / R12^2 = (1/3)^2

F12' = 1/9 F12

The new force is 1/9 the of the old force

A similar theme in both "The Stag at the Pool" and "Hans in Luck" is A. The things we think are valuable sometimes turn out not to be valuable after all.

In "The stag at the pool", a thirsty stag went to the pool to drink. As he bent over, he admired his antlers but despised his legs. He saw a lion and ran but his antlers got stuck in a tree and he was caught by the lion.

The theme in "Hans in Luck" shows that one doesn't need possessions to be happy. The character in the story had several possessions but was still unhappy. Both themes showed that the things we think are valuable sometimes turn out not to be valuable after all.

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Answer:

Hey mate.....

Explanation:

This is ur answer.....

Jelly is a colloid where colloidal particles are solids which are dispersed in the liquid. Jelly is a fluid which can not be considered as a particular one kind of state. It is in the category of 'gel' which is a colloidal form.

Hope it helps!

Brainliest pls!

Follow me! ;)

Hi there!

The formula for velocity given acceleration:

v = at

Plug in given values:

v = 6.4(7) = 44.8 m/s

[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]

Explanation:

The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as

[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]

where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is

[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]

Answer:

I think the answer is (A)...

Hope this helps!

Answer:

Explanation:

Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N

Pressure on the left

PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa

Pressure on the right

PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa

A) The force from the left tire is; FL = 342 N

B) The pressure from the left tire is; PL = 551613 N/m²

C) The force from the right tire is; FR = 342 N

D) The pressure from the right tire is; PR = 371739 N/m²

We see that;

FL and FR are upward forces

W is the downward force.

We know that in equilibrium;

Sum of upward forces = sum of downward forces

Thus;

FL + FR = W

We are given W = 684 N

Since W is at the center, it means that FL = FR. Thus;

FL = FR = 684/2

FL = FR = 342 N

We are given;

Contact area of left tire; AL = 6.2 × 10⁻⁴ m²

Contact area of right tire; AR = 9.2 × 10⁻⁴ m²

Formula for pressure is;

Pressure = Force/Area

Pressure on the left tire;

PL = FL/AL

PL = 342/(6.2 × 10⁻⁴)

PL = 551613 N/m²

Pressure on the t right tire;

PR = FR/AR

PR = 342/(9.2 × 10⁻⁴)

PR = 371739 N/m²

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Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.

Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.

Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar  

The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;

= 1000 MB - 976 MB

= 24 MB

Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.

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Answer:

Albert Einstein WAS a very well known Theoretical physicist

The USA claims he did not ever get his hands directly on an atomic bomb but in fact, other country textbooks like in Germany say he did.

Explanation:

A fun fact is that he was hired by the United States to make the Atomic bomb. Albert Einstein was a german yet many believed him to be a Smart American, they were wrong.

Hi there!

We know that:

I = Δp = m(vf - vi)

Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.

I = 1.1(5 - (-23)) = 30.8 Ns

Hi there!

We can use the equation:

d = v₀t + 1/2at²

Where:

v₀ = initial velocity downward

a = acceleration due to gravity

t = time

Plug in given values:

d = 4(12) + 1/2(9.8)(12²)

d = 48 + 705.6 = 753.6 m

Answer:

Explanation:

F = ma

F = 2.60(14.0)

F = 36.4 N

Hi there!

Part A:

The only quantities explicitly given to us are:

3. mass (m)

4. Maximum velocity (vmax)

5. Amplitude (A)

8. Position x from equilibrium

Part B:

To solve, we must begin by calculating the force constant, 'k'.

We can use the following relationship:

[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]

We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:

[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]

[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]

Now, we can find the velocity at 4.7cm (0.047m) using the equation:

[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]

Plug this value into the equation for kinetic energy:

[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]

Part C:

The potential energy of a spring is given as:

[tex]U = \frac{1}{2}kx^2[/tex]

Find 'k' using the derived quantity above:

[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]

Now, calculate potential energy:

[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]

Answer:

M1 + M2 = 0   total momentum before explosion = momentum after explosion where M1 is the lighter fragment

M1 V1 + 7 M1 V2 = 0

V2 = -V1 / 7      

The lighter fragment will slide 7 times as far - 39.2 mm because it must have 7 times as much velocity - assuming the distance slid is proportional to the original velocity

Explanation:

1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2

Answer:

200J

Explanation:

work done = force x distance

                  = 20 x 10

                  = 200J

Answer:

Overcurrents

Circuit breakers ("fuses" back in the day) are supposed to trigger when you are requesting past it more power (thus more current, since the tension is fixed and power is the product of voltsge times current). So whoever made that part of the grid did not foresee that many appliances.

Another option could be a short circuit somehwere but I think it is unlikely since the breaker would not trigger only when all three loads are powered on.

Answer:

12345678910i don't know

The applied torque increases the angular speed by the application of an

angular acceleration.

Reasons:

The speed of the flywheel at the axis = 4 rev/s

The torque applied, T = 25 N·m

The time the torque is applied, t = 10 sec

Moment of inertia of the flywheel, I = 1.2 kg·m²

Required:

The speed at the end of the 10 seconds

Solution:

T = I·α

Where;

α = Angular acceleration

[tex]\displaystyle \alpha = \frac{T}{I}[/tex]

Therefore;

[tex]\displaystyle \alpha = \frac{24 \ N\cdot m}{1.2 \ kg \cdot m^2} = \mathbf{20\frac{5}{6} \ s^{-2}}[/tex]

The rotational speed, ω = ω₀ + α·t

Which gives;

[tex]\displaystyle \mathrm{The \ angular \ speed, } \ \omega = \frac{2 \cdot \pi \times 4 \ rad }{s} = \frac{8 \cdot \pi \ rad }{s}[/tex]

ω₀ = 8·π rad/s

Which gives;

[tex]\displaystyle \omega = \mathbf{8 \cdot \pi +2 \frac{5}{6} \times 10} = 233.47[/tex]

The speed of the wheel in revolution per second is therefore;

[tex]\displaystyle Speed \ in \ rev/s = \frac{8 \cdot \pi +2 \frac{5}{6} \times 10}{2\cdot \pi} \approx 37.16[/tex]

The speed after 10 seconds is approximately 37.16 rev/s.

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