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bezimeni [28]
2 years ago
3

Neptune has a mass that is about 17 times the mass of Earth. The distance between the Sun and Neptune is about 30.1 times the di

stance between the Sun and Earth. If the gravitational force between the Sun and Earth is 3.5 x 1028 N, which is closest to the force between Neptune and the Sun?
Physics
1 answer:
Verdich [7]2 years ago
8 0

Given:

Mass of Neptune (M₁) = 17 M₂

Here, mass of the Earth is M₂ and the distance between the Earth and the Sun is R₂. The gravitational force between the Sun and Earth (F₂) = 3.5 x 10²⁸ N

The distance between the Sun and Neptune(R₁) = 30.1 R₂

Let M be the mass of the Sun

Apply, Newton's Law of attraction

The force of attraction between the Earth and the Sun

                 (F₂) = G× M×M₂ / R₂² ---------------(i)

The force of attraction between the Neptune and the Sun

                 (F₁) = G× M×M₁ / R₁² ---------------(ii)

Now, substituting the value of M₁ and R₁ in equation (ii)

                    (F₁) = G× M× 17 M₂  / (30.1 R₂ )²

or,                (F₁) = [17 / (30.1)²] × [G× M×M₁ / R₁²]

or,                (F₁) = [17 / (30.1 )²] ×F₂

or,                (F₁) = [17 / (30.1 )²] ×3.5 x 10²⁸ N

or,                (F₁) = 6.6 x 10²⁶ N

Hence, the required force of attraction between the Neptune and the Sun will be 6.6 x 10²⁶ N

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A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
1 year ago
luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that
Andreas93 [3]

Answer:

a)  W = 643.5 J, b) W = -427.4 J  

Explanation:

a) Work is defined by

       W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

       W = F x

let's calculate

       W = 165  3.9

        W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

       sin θ = Wₓ / W

       cos θ = Wy / W

        Wₓ = W sinθ = mg sin θ

        Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

      W = Wₓ x cos 180

      W = - mg sin 34  x

     

let's calculate

       W = -20 9.8 sin 34 3.9

        W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0
2 years ago
The chart shows data for four moving objects. A 4 column table with 4 rows. The first column is labeled Object with entries, W,
KatRina [158]

Answer:

y

Explanation:

I took the test

3 0
2 years ago
The same physics student jumps off the back of her Laser again, but this time the Laser is
soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

p_i = p_f\\0=mv+MV

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s

So, the final speed of the student is 1.07 m/s.

b)

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

(m+M)u=mv+MV

where

m = 42 kg is the mass of the Laser

M = 59 kg is the student's mass

u = 3.1 m/s is the initial velocity of the student and the Laser

V = -2.1 m/s is the velocity of the student after the jump (she jumps backward)

v is the final velocity of the Laser

And solving for v, we find

v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s

Learn more about momentum:

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3 0
2 years ago
Read 2 more answers
You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro
ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N

The spring obeys Hook's law:

F=k\Delta x

where k is the spring constant and \Delta x is the stretching of the spring. Since we know \Delta x=2.92 cm=0.0292 m, we can re-arrange the equation to find the spring constant:

k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance \Delta x is equal to the elastic potential energy stored by the spring, given by:

W=U=\frac{1}{2}k\Delta x^2

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

5 0
2 years ago
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