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expeople1 [14]
3 years ago
14

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

ce of the earth
Physics
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

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Dimension of density <br>​
Rom4ik [11]

Answer:

See the explanation below.

Explanation:

We know that density is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

Therefore Ro is given in:

[kg/m^{3} ]

8 0
2 years ago
Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.
tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

s=ut+\frac{1}{2}*gt^{2}  \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m

Arrow misses the target by 5.843m ig the arrow us split horizontally

4 0
3 years ago
In which of the following situations or places would you want to reduce the
sashaice [31]

Answer:

D

Explanation:

<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>

Options A, B, and C requires that the forces of friction is increased in order to have more control.

However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.

<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>

4 0
3 years ago
A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   We have equation of motion, s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate time when displacement = 1.5 meter.

 1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 m/s^2

Substituting

     -1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0

In case of horizontal motion

    Horizontal speed of car = V cos 7 = 0.993V

    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation 4.9t^2-0.122Vt-1.5=0

    We will get 4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

7 0
3 years ago
A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which
Leviafan [203]

Answer:

The value is  a =  -3.7 \  m/s^2

Explanation:

From the question we are told that

   The  landing speed is  u =  71.9 \  m/s

   The  distance traveled is  d =  675 \  m

    The velocity it is reduced to is  v  =  11.3 \  m/s

   

Generally the average acceleration is mathematically represented as

      a =  \frac{ v^2  -  u^2 }{ 2 * d }

=>  a =  \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }

=>   a =  -3.7 \  m/s^2

5 0
3 years ago
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