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3 years ago

11

Can y'all help me please?

2 answers:

topjm [15]3 years ago

5 0

North!!

hope this helps you!

fenix001 [56]3 years ago

3 0

Answer:

North

Explanation:

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Orion is visible on winter evenings in the northern hemisphere but not summer evenings because of_______________.

Bess [88]

Answer:

c. the location of Earth in its orbit.

Explanation:

It is known that the earth orbits the sun once every 365 days, during this time the<u> constellations that are visible in the sky are changing</u>.

This is due to the fact that depending on the position, we have access to different parts of the celestial tomb, that is, the stars and constellations visible from earth.

This is why Orion is visible in winter but not in summer, since in summer it is in a <u>different part of the orbit</u> and with visibility towards different constellations.

the answer is c. the location of Earth in its orbit.

3 0

3 years ago

16. Kinematic equations can only be used if...

Effectus [21]

the acceleration or the next force acting on the body is constant

7 0

3 years ago

a fire truck drives 3 miles to the east, then 1.2 miles to the west.what is the displacement of the fire truck

blagie [28]

Then he is 1.8 miles west

6 0

3 years ago

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

3 years ago

Two very long uniform lines of charge are parallel and are separated by 0.300 m. each line of charge has charge per unit length

Alenkasestr [34]

linear charge density of system of two line charges is given as

$\lambda = 5.20 \muC/m$

now as we know that electric field due to a line charge at some distance from it is given by

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

so here we will first find the electric field of first line charge at the position of other line charge

$E = \frac{5.20 * 10^{-6}}{2 \pi * 8.85 * 10^{-12}* 0.300}$

$E = 312000 N/C$

now as we know that

$F = qE$

here q = charge on the line charge system at which force is required

E = electric field on that system of charge where force is required

now we can find the charge by

$q = \lambda * L$

$q = 5.20 * 10^{-6}* 0.05 = 0.26 * 10^{-6} C$

Now using the above formula

$F = qE$

$F = 0.26 * 10^{-6} * 312000$

$F = 0.0811 N$

so force on the part of wire is F = 0.0811 N

8 0

3 years ago