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1 year ago

Why would north america will not be able to view the eclipse

1 answer:

4 0

The North America will not be able to view the eclipse because of its location on the earth caused by the tilting of the earth.

<h3>Effect of tilting of the Earth</h3>

The tilt of the Earth is what causes seasons to occur. These are the seasons in relation to the Northern Hemisphere.

The tilting of the Earth causes the difference in the amount of sun reaching each region of the earth like in the North America continent.

Thus, the North America will not be able to view the eclipse because of its location on the earth caused by the tilting of the earth.

Learn more about eclipse here: brainly.com/question/8643

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A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about

tankabanditka [31]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

6 0

2 years ago

Well-designed weight-training programs only target two or three body areas.<br> a. True<br> b. False

Sloan [31]

False<span>, well designed weight training programs actually target most of the muscles in the body. A good weight training program includes many compound exercises to activate multiple muscle groups and promote muscle hypertrophy. Some isolation exercises may also be included to target a specific muscle and help it grow.</span>

4 0

3 years ago

Read 2 more answers

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:

lutik1710 [3]

Answer:

At $t = 1\; \rm s$ , the particle should have a velocity of $-8\; \rm m \cdot s^{-1}$ .
At $t = 2\; \rm s$ , the particle should have a velocity of $-1\; \rm m \cdot s^{-1}$ .
At $t = 3\; \rm s$ , the particle should have a velocity of $\displaystyle -\frac{8}{27}\; \rm m \cdot s^{-1}$ .

For $a > 0$ , at $t = a \; \text{second}$ , the particle should have a velocity of $\displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}$ .

Explanation:

Differentiate the displacement of an object (with respect to time) to find the object's velocity.

Note that the in this question, the expression for displacement is undefined (and not differentiable) when $t$ is equal to zero. For $t > 0$ :

$\begin{aligned}v &= \frac{\rm d}{{\rm d}t}\, [s] = \frac{\rm d}{{\rm d}t}\, \left[\frac{4}{t^2}\right] \\ &= \frac{\rm d}{{\rm d}t}\, \left[4\, t^{-2}\right] = 4\, \left((-2)\, t^{-3}\right) = -8\, t^{-3} =-\frac{8}{t^3}\end{aligned}$ .