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inessss [21]
3 years ago
12

The Sun is only an average sized star. Why does it appear so much brighter than any other star?

Physics
1 answer:
aivan3 [116]3 years ago
8 0

Because it's so much closer to us than any other star.

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You have an electrically neutral toy that you divide into two pieces. You notice that at least one of those pieces has an electr
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Answer:

The pieces will attract one another

Explanation:

From the law of conservation of energy, we know that energy can neither be created nor destroyed, but transformed. If one piece of the toy that was neutral ends up having an electric charge (positive or negative), from the conservation of energy, the other piece must have a charge opposite to that on the other charged piece but equal in magnitude. These two pieces which are oppositely charged attracts each other, this shows that electric charge is conserved.

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Which term refers to a group or community that forms the focal point of study and consists of all the people considered for eval
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It is the variable that is the answer
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A ball is dropped from the roof of a 25-m-tall building. What is the velocity of the object when it touches the ground? Suppose
Wewaii [24]

Answer:

a)  h=25m

b)  v=19.8m/sec

Explanation:

From the question we are told that:

Height h=25m

Bounce Height h'=20m

Generally the Kinematic equation is mathematically given by

V=\sqrt{2gh}\\\\V=\sqrt{2*9.81*25}

V=22.1m/sec

Therefore Height

h=\frac{V^2}{2g}\\\\h=\frac{22.1^2}{2*9.81}

h=25m

b)

Generally the Kinematic equation is mathematically given by

v^2=2ah

v^2=2*9.8*20

v=\sqrt{2*9.8*20}

v=19.8m/sec

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3 years ago
I need help please i really do appreciate it
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Answer:

Inter molecular forces

Explanation:

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An aging coyote cannot run fast enough to catch a roadrunner. He purchases on eBay a set of jet-powered roller skates, which pro
Xelga [282]

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + v_{oy} t - ½ g t²

             

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = v_{oy} - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

   

         v = (45.83 i ^ -109.56 j ^) m / s

4 0
3 years ago
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