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Ivenika [448]
2 years ago
6

The gravitational force between the Sun (mass = 1.99 × 1030 kg) and Mercury (mass = 3.30 × 1023 kg) is 8.99 × 1021 N. How far is

Mercury from the Sun?
Physics
2 answers:
Svet_ta [14]2 years ago
8 0
The answer is B - 6.98 x 10^7 km
STALIN [3.7K]2 years ago
4 0

Answer : Distance, d=6.98 × 10⁷ Km

Explanation :

Given that,

Mass of the sun, m₁ = 1.99 × 10³⁰ kg

Mass of Mercury, m₂ = 3.30 × 10²³ kg

Gravitational force between the sun and mercury, F = 8.99 × 10²¹ N

According to Universal law of gravitation,

F=G\dfrac{m_1m_2}{d^2}

d is the distance of mercury from the sun

d=\sqrt{\dfrac{Gm_1m_2}{F}}

d=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.99\times {30}\times 3.30\times 10^{23}}{8.99\times 10^{21}}}

d=\sqrt{4.87\times 10^{21}} m

d=6.98\times 10^{10}\ m

or

d=6.98\times 10^{7}\ Km

So, mercury is 6.98\times 10^{7}\ Km far from the sun.

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In the lab setup above, block 1 is being pulled across a smooth surface by a light string connected across a pulley to a falling
Nuetrik [128]

Answer:

As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.

Explanation:

As block 2 goes down , work is done by gravity on block 2 . This is converted

into kinetic energy of block 1 and block 2 . Work done by gravity is mgh which can be measured easily . kinetic energy of both the blocks can also be measured.  

8 0
2 years ago
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

4 0
2 years ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
2 years ago
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