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Naily [24]
3 years ago
12

What are day and night are produced by?

Physics
2 answers:
yuradex [85]3 years ago
7 0

Answer:

Day and night are due to the Earth rotating on its axis, not its orbiting around the sun. The term 'one day' is determined by the time the Earth takes to rotate once on its axis and includes both day time and night time.

Explanation:

Lelechka [254]3 years ago
4 0

Answer:

Day and night are due to the Earth rotating on its axis, not its orbiting around the sun. The term 'one day' is determined by the time the Earth takes to rotate once on its axis and includes both day time and night time.

note:

your welcome and please answer my question its called part 1 math work if you could do that i will give you 40 points

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What is gravitonal force
zhuklara [117]

Answer:

its something that hold the air for forceing liy by the exgen

Explanation:

8 0
3 years ago
A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)
stepladder [879]

a) F=(3675i-4543k)N

b) 5843 N

Explanation:

a)

The position of the UFO at time t is given by the vector:

r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k

Therefore it has 3 components:

r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2

We start by finding the velocity of the UFO, which is given by the derivative of the position:

v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t

And then, by differentiating again, we find the acceleration:

a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6

The weight of the UFO is W = 12,500 N, so its mass is:

m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg

Therefore, the components of the force on the UFO are given by Newton's second law:

F=ma

So, Substituting t = 2 s, we find:

F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N

So the net force on the UFO at t = 2 s is

F=(3675i-4543k)N

b)

The magnitude of a 3-dimensional vector is given by

|v|=\sqrt{v_x^2+v_y^2+v_z^2}

where

v_x,v_y,v_z are the three components of the vector

In this problem, the three components of the net force are:

F_x=3675 N\\F_y=0\\F_z=-4543 N

Therefore, substituting into the equation, we find the magnitude of the net force:

|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N

7 0
3 years ago
How do you double the speed in physics
Mariulka [41]

You measure the open distance between the floor and the bottom surface of the gas pedal.  Then you press the gas pedal down 1/2 of that distance.

3 0
3 years ago
What is peer review.why is it important​
madam [21]

Acc to

Publisso

Peer review is subjecting the author's scholarly work and research to the scrutiny of other experts in the same field to check its validity and evaluate its suitability for publication.

A peer review helps the publisher decide whether a work should be accepted.

6 0
3 years ago
Read 2 more answers
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
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