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Alina [70]
3 years ago
6

(b) An example of an inverse square field is the gravitational field F = −(mMG)r/|r|3. Use part (a) to find the work done by the

gravitational field when the earth moves from aphelion (at a maximum distance of 1.52 108 km from the sun) to perihelion (at a minimum distance of 1.47 108 km). Use the values m = 5.97 1024 kg, M = 1.99 1030 kg, and G = 6.67 ✕ 10−11 N·m2/kg2. (Round your answer to two decimal places.)
Physics
1 answer:
enyata [817]3 years ago
4 0

Answer:

To solve this, we need to recall that Work done W,

W = F x dₐ                         Eqn 1

where F is the gravitational field force and d is the distance travelled by the earth.

But F = −(mMG)r/|r|³                    Eqn 2

then, W = −∫ (mMG)r/|r|³ x dₐ

Integrating this expression gives us

W = -4mMG ((1/d1) -(1/d2))

m = 5.971024 kg, M = 1.991030 kg G = 6.67 ✕ 10⁻¹¹ N·m²/kg²

d2 = 1.52x10¹¹m, d1 = 1.47x10¹¹m  make the units in S.I units and similar

Substituting these value in to the eqn above gives

Now , W = 1.77 x 10³²J  (2 Decimal Places)

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How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h
tresset_1 [31]

Answer:

The  time taken is  t =  32.5 \  s

Explanation:

From the question we are told that

   The  speed  of  first car is  v_1  =  66.7 \ km/h  =  18.3 \  m/s

    The  speed  of  second car is v_2  =  52.7 \ km/h  =  14.64 \  m/s

   The  initial distance of separation is  d =  119 \ m

The distance covered by first car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  0 m/s

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_1 * t

So

     d_t =  0 \  m/s  +  (v_1 * t )

     d_t =  0 \  m/s  +  (18.3 * t )

The distance covered by second  car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  119 m

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_2* t

       d_t =  119  + 14.64 *  t

Given that the two car are now in the same position we have that

    119  + 14.64 *  t  =   0   +  (18.3 * t )

   t =  32.5 \  s

6 0
3 years ago
A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of
Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

Workdone = force * distance

Making "distance" the subject of formula, we have;

Distance = \frac {workdone}{force}

Substituting into the equation, we have;

Distance = \frac {195}{56.7}

Distance = 3.4 meters.

8 0
3 years ago
A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . A
yaroslaw [1]

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

v_{f}^{2}=v_{0}^{2}+2gy\\v_{f}=\sqrt{0+2(9.8\frac{m}{s^{2}})(145m)}=53.31\frac{m}{s}

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

F=(3.8*10^{8})(1.602*10^{-19}C)(53.31\frac{m}{s})(0.205T)=6.65*10^{-10}N

B)

b.  south - north (by the rigth hand rule)

I hope this is usefull for you

regards

8 0
3 years ago
Give a example of scalar quantity
Serhud [2]

speed, volume, mass, temperature and power

7 0
3 years ago
Read 2 more answers
A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre
AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz

3 0
3 years ago
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