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3 years ago

7

What would be the effect on the annual march of the seasons if earth's axis were not inclined relative to the plane of the eclip

tic?

1 answer:

victus00 [196]3 years ago

6 0

The poles wouldn't get light all year round.

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9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?

kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

U 2 can help me by marking as brainliest........

5 0

2 years ago

Read 2 more answers

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

A black hole in the universe is a An empty region in space b- A massive collapsed star c- A moon that is always turned to its da

marusya05 [52]

Answer:

b. a massive collapsed star

Explanation:

A black hole in the universe is nothing but a massive collapsed star. When the size of the star crosses a particular limit it cannot holds its mass and it collapses under it own self. This is called supernova. A black hole is actually a region in space where gravity is so strong that even light cannot escape through it. Gravity so strong because the matter has been pressed into a tiny space. hence option b is correct

4 0

2 years ago

I beggggg youu to solve this for me, I'm giving you 60 POINTS !!!

djyliett [7]

Answer:

but where is the question ?

Explanation:

<em>hope</em><em> it</em><em> </em><em>works</em><em> out</em>

6 0

2 years ago

Read 2 more answers