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Pani-rosa [81]
3 years ago
15

If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha

lf what it is now, what would be the acceleration due to gravity at the surface of the Earth?
Physics
2 answers:
vovikov84 [41]3 years ago
6 0
If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
Neporo4naja [7]3 years ago
4 0
G=GM/r^2
Here are 2 ways to look at this: 
1. Since the r is both inverted and squared on the right while acceleration due to gravity (g) is on the top on the left, any factor applied to r will be inverted and squared as applied to g. In this case, r is multiplied by 1/2, so g will be multiplied by (2/1)^2, which is 4. Assuming your teacher uses 9.8 m/s^2 as the acceleration due to gravity, it would be 39.2 m/s^2. <span>
2. </span>Let r(1) be the radius of the earth, r(2) the smaller radius, g(1) is the acceleration due to gravity on earth and g(2) is the new acceleration due to gravity. In this case, r(2) = (1/2R(1))^2
<span>You can re-arrange the equation for things that will not change on one side and things that are changing on the other. 
This would be gr^2=GM. If GM is not changing, we can say GM = g(1)r(1)^2 = g(2)r(2)^2. Plug in what you know, 9.8*r(1)^2=g(2)*(1/2r(1)^2)
square the 1/2: </span>9.8*r(1)^2=g(2)*1/4r(1)^2
the r(1)^2 cancels out: 9.8=g(2)*1/4
then multiply by 4 on each side: 4*9.8=g(2) = 39.2 m/s^2
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Illusion [34]

In order to lift the fat (306 lb) physics professor 89 meters up to
the rim, he'll need more potential energy, equal to

      (mass) x (gravity) x (height) = (139 x 9.8 x 89) = 121,236 joules .

If the faithful horse delivers 1 constant horsepower = 746 watts,
AND if the cute-as-a-button student has instantly figured out a
way to keep the rope sliding around the edge without any friction,
then the soonest Prof. Tubby can arrive at the rim is

    (121,236 joules) / (746 joules/sec) = 162.5 seconds . 

Nowhere in this tense drama has the student needed her linguistics
skill yet, but I'll bet it comes in handy as she attempts gamely to
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4 0
3 years ago
In a little league baseball game, the 145 g ball enters the strike zone with a speed of 14.0m/s . the batter hits the ball, and
Novosadov [1.4K]

Answer:

5365 N

Explanation:

v = Final velocity = 23 m/s

u = Initial velocity = -14 m/s (opposite direction)

m = Mass of ball = 145 g

t = Time taken = 1 ms

Impulse is given by

J=m(v-u)

Impulse is also given by

J=Ft

Ft=m(v-u)\\\Rightarrow F=\dfrac{m(v-u)}{t}\\\Rightarrow F=\dfrac{0.145\times (23-(-14))}{1\times 10^{-3}}\\\Rightarrow F=5365\ N

The magnitude of the average force exerted by the bat on the ball is 5365 N

8 0
3 years ago
An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef
Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field B=5.0\times10^{-5}\ T

(a). We need to calculate the induced emf

Using formula of emf

E = Blv\sin\theta

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}

E=1.398\times10^{-3}\ V

We need to calculate the induced current

E =IR

I=\dfrac{E}{R}

Put the value into the formula

I=\dfrac{1.398\times10^{-3}}{3.0}

I=0.5\ mA

(b). We need to calculate the power dissipated through the resistor

Using formula of power

P=I^2 R

Put the value into the formula

P=(0.5\times10^{-3})^2\times3.0

P=7.5\times10^{-7}\ Watt

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

6 0
3 years ago
Read 2 more answers
A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t
Zina [86]

Answer:

The value of spring constant is 266.01 \frac{N}{m}

Explanation:

Given:

Mass of pellet m = 2.01 \times 10^{-2} kg

Height difference of pellet rise h_{f} - h_{o} = 6.03 m

Spring compression x = 9.45 \times 10^{-2} m

From energy conservation law,

Spring potential energy is stored into potential energy,

  mg(h_{f} -h_{o})  = \frac{1}{2} kx^{2}

Where k = spring constant, g = 9.8 \frac{m}{s^{2} }

  k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }

  k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }

  k = 266.01 \frac{N}{m}

Therefore, the value of spring constant is 266.01 \frac{N}{m}

6 0
3 years ago
blank refers to the method of spreading fertilizer evenly over the entire field by hand it is done at the blank stage
Andrej [43]

Answer:

Broadcasting is the method, not sure about the stage it is done in

Explanation:

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