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yarga [219]
3 years ago
12

At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?

Physics
2 answers:
alina1380 [7]3 years ago
5 0
1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
               = 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11  AU
               =  3.0.74 / 100  = 0.0374 AU

Radda [10]3 years ago
5 0

Explanation :

Astronomical unit is the unit to find the average distance between the earth and the sun.

1\ AU=1.496\times 10^8\ Km

So, 1\ Km=\dfrac{1}{1.496\times 10^8}\ AU

46 million kilometers means, 46\times 10^6\ Km

46\times 10^6\ Km=\dfrac{46\times 10^6}{1.496\times 10^8}

46\ million\ Km=0.3074\ AU

Hence, this is the required solution.      

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Electric charges are either positive or ____
fenix001 [56]

Answer:

Negative

Explanation:

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4 0
3 years ago
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I need help with a physics worksheet about Newtons Laws.
Viktor [21]

The diagram of the object on the inclined plane is shown below

The mass of 7kg is exerted on the plane surface. It is acted upon by gravity. The force exerted on the surface is calculated by applying the formula,

F = mass x acceleration due to gravity

Assumme acceleration due to gravity is 9.8m/s^2, then

Force = 7 x 9.8 = 68.6N

The surface exerts an opposite force of the same magnitude with the force of the object but in opposite direction. Since it is inclined at an angle of 36.9 degrees,

Normal force = mgCosθ = 68.6Cos36.9 = 54.86N

Recall, frictional force = normal reaction x coefficient of friction

Given that coefficient of kinetic friction = 0.35,

Frictional force = 54.86 x 0.35 = 19.201N

This is the force that must be overcome to keep the object moving.

The force pulling the object upwards along the inclined plane is

mgSinθ = 7 x 9.8 x Sin36.9 = 41.19N

Since the velocity is constant, it means that there is no acceleration. The net force is zero. The force required to pull the mass and make it move at constant velocity must be equal to the sum of the exerted force and the the frictional force(Since it involves coeffricient of kinetic friction, it would cause the object to keep moving)

Thus,

Required force = 19.2 + 41.19 = 60.4N

Option E is correct

8 0
1 year ago
2 strings both vibrate at exactly 220 Hz. The tension in one of them is then decreased sightly. As a result, 3 beats per second
MAVERICK [17]

Explanation:

Given that,

2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.

The tension in one of them is then decreased sightly, then f_2 will decrese.

Beat frequency, f=f_1-f_2

3=220-f_2

f_2=217\ Hz

So, the new frequency of the string is 217 Hz. Hence, this is the required solution.

3 0
3 years ago
A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
3 years ago
A small water droplet in a mist of air is approximated as being a sphere of diameter 1.25 mil. Calculate the terminal velocity a
shtirl [24]

The terminal velocity as it falls through still air is 4.65154 in/s.

The diameter of small water droplet is 1.25 mil= 1.25×0.0254×10^-3 m

                                                                            = 3.175 × 10^-5 m

Now the viscosity of still air is η = 1.83× 10⁻⁵ Pa

 So the formula for drag force is:

                Fd = 6πηrv

  where,  v is the velocity.

Now to attain terminal velocity acceleration must be zero.

            →  W = Fd

                ρVg = 6πrηv

                ρ × 4/3 πr³×g = 6πrηv

                           v = 2/9 × ρgr³/ η

                           v =  2/9 × 10³×9.81×(3.175×10^-3) / 18.6×10^-6

                           v = 0.1181  m/s

                           v = 4.65154 in/s

Learn more about terminal velocity here:

     brainly.com/question/20409472

              #SPJ4

                   

8 0
2 years ago
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