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3 years ago

I need help ;-;

2 answers:

8 0

The First Quarter Moon is a primary Moon phase when we can see exactly half of the Moon's visible surface illuminated. If it is the left or right half, depends on where you are on Earth.

koban [17]3 years ago

5 0

Moon is a primary moon phase.

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Fill in what you can:

VMariaS [17]

Graphs because it has a lot of squares and it accelerates to 35 km/h in 20 seconds

6 0

3 years ago

6. A 73-kg woman stands on a scale in an elevator. The

olga nikolaevna [1]

Answer:

Explanation:

n = m(g +a)

n= normal force (N)

m=mass (kg)

g=acceleration of gravity

a= acceleration of elevator

rearrange:

a= n/m - g

a= (810 N/73 kg) - 9.8 m/s ^2

a= 1.3 m/s ^2 up

and the acceleration is upwards bc her weight is less than the scale reading

3 0

3 years ago

Someone help please I am suffering from these

seropon [69]

i will b i don't now but you said you was suffering so i have to help you i'm going to put a thank you on you name okay bye

3 0

4 years ago

A skier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angleof

victus00 [196]

Answer:

Explanation:

When the skier reaches the bottom of the slope , height lost by it

h = 50 sin32 m

= 26.5 m

potential energy lost

= mgh

Gain of kinetic energy

= 1/2 mv²

mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

Let μ be the coefficient of kinetic friction required.

friction force acting

= μmg

work done by friction in displacement of d (40 m ) on horizontal surface

- μmg x d

This negative work will be equal to positive kinetic energy of the skier on horizontal surface .

= μmg x d = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

8 0

4 years ago

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0

3 years ago