PHYSICS COLLEGE

The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

Answers

Answer 1

Answer:

Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth = 6 × 10²⁴ kg

r = Distance between Earth and Sun = $1.5* 10^(11)\ m$

t = Time taken = 3 days

Acceleration is given by

$a=(GM)/(r^2)\n\Rightarrow a=(6.67* 10^(-11)* 6* 10^(24))/((1.5* 10^(11))^2)\n\Rightarrow a=1.77867* 10^(-8)\ m/s^2$

Velocity of the star

$v=u+at\n\Rightarrow v=0+1.77867* 10^(-8)* 3* 24* 60* 60\n\Rightarrow v=0.00461031264\ m/s$

The Sun's speed will be 0.00461031264 m/s

Related Questions

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?
If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of their masses? Express your answers using three significant figures separated by a comma.
To remove 800j of heat the compressor in the fridge does 500j of work. how much heat is released into the room?
What is the meaning of relative as a noun?
A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answers

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is $\rm 0.430 \; kg\;m^2$ .

Given :

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

$\rm s = ut + (1)/(2)at^2$

$\rm 1.8 = (1)/(2)* a* (2)^2$

$\rm a = 0.9\; m/sec^2$

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

$\rm \sum F=ma$

$\rm mg-T_b=ma$

$\rm T_b = m(g-a)$

$\rm T_b = 7* (9.8-0.9)$

$\rm T_b = 62.3\;N$

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

$\rm \sum F=ma$

$\rm T_a=ma$

$\rm T_a = 2.1* 0.9$

$\rm T_a = 1.89\;N$

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

$\rm \sum \tau = I\alpha$

$\rm T_br-T_ar = I\alpha$

$\rm I = ((T_b-T_a)r^2)/(a)$

Now, substitute the values of the known terms in the above expression.

$\rm I = ((62.3-1.89)(0.080)^2)/(0.90)$

$\rm I = 0.430 \; kg\;m^2$

For more information, refer to the link given below:

brainly.com/question/2287912

Answer:

(a) 62.3 N

(b) 1.89 N

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B. There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

= $Tensile \ strength* Area$

On substituting the values, we get

= $(3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)$

= $(3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)$

= $21195 \ N$

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

Learn more about Tensile Strength here:

brainly.com/question/14293634

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The solar system is of largely uniform composition. (T/F)

Answers

Answer:

False

Explanation:

Sun mass is dominating in Solar system as compared to other planets, asteroids and comets. Sun itself accounting for the 99.9% of the mass of the solar system. Hence the gravitational force exerted by the Sun dominates the other objects in the solar system. So we can conclude that solar system has non-uniform composition. The given statement is false

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that $v = f * d$ where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floor and a ceiling 3.0m high. What is the total force on the floor due to the air above the surface if the air pressure is 1.00 atm?

Answers

Answer:

Force, $F=2.02* 10^6\ N$

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, $P=1\ atm=101325\ Pa$

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

$P=(F)/(A)$

$F=P* l* b$

$F=101325* 4* 5$

$F=2.02* 10^6\ N$

So, the total force on the floor due to the air above the surface is $2.02* 10^6\ N$ . Hence, this is the required solution.

A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Answers

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

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The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

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