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ra1l [238]
2 years ago
12

Astronomers have divided the planets into two groups. The inner, “terrestrial,” planets are Mercury, Venus, Earth, and Mars. The

outer, “Jovian,” planets are Jupiter, Saturn, Uranus, and Neptune. Which statement best compares the differences between the inner and outer planets?
A. The outer planets are warmer and larger than the inner planets.
B. The inner planets have rings and are cooler than the outer planets.
C. The inner planets are warmer and relatively small compared to the outer planets.
D. The outer planets have rings and are larger and warmer than all of the inner planets.
Explain
Physics
1 answer:
algol [13]2 years ago
3 0

Answer:

it's letter C because inner planets are warmer because they are close to the sun and they are smaller than the Jovian or Giant Gas Planets which have rings, and cooler.

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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
1500 kg Peugeot car is traveling at 16.67 m s ⁻¹ and accelerates to 30.56 m s ⁻¹ for 2 minutes. Calculate the impulse of the car
Elis [28]

Answer:

2084 kg*m/s

Explanation:

Impulse is change in momentum

Mathematically;

Impulse, J= F*t=mΔv   where

F= ma = 1500 * { 30.56 - 16.67}/2*60

F= 1500 *0.11575

F=  174 N

J=F*t

J= 174*120*0.1

J= 2084 kg*m/s

5 0
2 years ago
The magnitude of the charge of the electron is:
brilliants [131]

Answer:

a. Exactly the same as the magnitude of the charge of the proton.

Explanation:

The elementary charge (e) is the smallest electric charge that can exist in the universe. Any positive or negative electric charge can be expressed as a multiple of the elementary charge, since is the electric charge carried by a single proton or, equivalently, the magnitude of the electric charge carried by a single electron (-1e).

3 0
3 years ago
An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste
sesenic [268]

Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

According to work energy theorem work done by all the forces is equal to change in kinetic energy of object

W=\frac{1}{2}mv^2---1

where m=mass of object

v=velocity of object

When the object is already have velocity v then the final speed is given by work energy theorem

W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2

From 1 and 2 we get

\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2

2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2

v_f^2=2v^2

v_f=\sqrt{2}v                

8 0
3 years ago
You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an
zheka24 [161]

The ball's horizontal position in the air is

x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t

It hits the wall when x=17.5\,\mathrm m, which happens at

17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s

Meanwhile, the ball's vertical position is

y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2

where g is the acceleration due to gravity, 9.80 m/s^2.

At the time the ball hits the wall, its vertical position (relative to its initial position) is

y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}

5 0
3 years ago
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