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3 years ago

PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Physics

2 answers:

____ [38]3 years ago

8 0

Nooooooooooooooooooo

horsena [70]3 years ago

5 0

It will be cloudy and there will be rain.

Hope this helps

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The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is 2.0 m from th

ASHA 777 [7]

Answer:

13 m

Explanation:

It is given that :

I got a haircut sitting at a place having two parallel mirrors at a distance = 6.5 m apart

My head is at a distance of 2 m from the nearer mirror.

Now the light from the back of my head must go to (6.5 - 2) = 4.5 m to the back mirror.

Then it must go to 6.5 m to the front mirror and 2 m from the front mirror to my eyes.

So in order to see the back of my head, it will be = 6.5 + 2 + 4.5 = 13 m away.

4 0

3 years ago

A bag is dropped from a hovering helicopter. After 42.4s, calculate the bag’s displacement.

Ann [662]

Answer:

8,809.024m

Explanation:

Displacement , y = (gt^2/2 )

g = 9.8 m/s^2

t = 42.4s

y = (9.8 (42.4*42.4)) / 2

y = 8,809.024m

3 0

3 years ago

A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.

Nataliya [291]

Answer:

Approximately $1.2 \times 10^{4}\; {\rm J}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The change in the gravitational potential energy ( ${\rm GPE}$ ) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass $m$ increased by $\Delta h$ , the ${\rm GPE}$ of that object would have increased by $m\, g\, \Delta h$ .

In this question, the height of this object increased by $\Delta h = 6.0\; {\rm m}$ . The mass of this object is $m = 200\; {\rm kg}$ . Thus, the ${\rm GPE}$ of this object would have increased by:

$\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}$ .

(Note that $1\; {\rm N \cdot m} = 1\; {\rm J}$ .)

3 0

2 years ago

What process marks the birth of a star?

Misha Larkins [42]

Answer:

Magic school bus on ya left! so basically a star is born when like....atoms are like squeezed with enough pressure and kinda squished together, once tht happens the atoms start to fuse together

Explanation:

7 0

4 years ago

Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the f

Alex17521 [72]

Answer:

g' = g/4

Explanation:

The value of the free-fall acceleration at the surface of the earth, can be obtained applying Newton's 2nd law, assuming that the only force acting on an object at the surface of the earth, is the one produced by the mass of the Earth, i.e. gravity.
This force can be expressed according the Newton's Universal Law of Gravitation , as follows:

$F_{g} = G*\frac{m_{x} *m_{E} }{r_{E}^{2} } (1)$

From Newton's 2nd Law, we have:
$F = m* a (2)$
Since the left sides in (1) and (2) are equal each other, both right sides must be equal each other also.
Simplifying the mass m, we can write the acceleration a in (2) as the acceleration due to gravity, g, as follows:

$g = G*\frac{m_{E} }{r_{E}^{2} } (3)$

Since G is an universal constant, and the mass mE remains constant, if we double the radius of the Earth, the new value for the acceleration due to gravity (let's call it g'), is as follows:

$g' = G*\frac{m_{E} }{(2r_{E})^{2} } =G*\frac{m_{E} }{4*r_{E}^{2} } = g*\frac{1}{4} =\frac{g}{4} (4)$

4 0

3 years ago