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Svetllana [295]
2 years ago
13

12. a) What is the gravitational attraction of the sun

Physics
1 answer:
emmasim [6.3K]2 years ago
3 0

Answer:

About 3.54x1022 N.

Explanation: The gravitational force between the Sun and the Earth is about 3.54x1022 N. This force keeps the Earth orbiting around the Sun. The gravitational force from the other planets does slightly affect the Earth's orbit, but the gravitational pull from the other planets and the Moon is still very small. Sorry if I get this wrong. I am in 5th grade! ♥

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Which of these terms means an object's resistance to changing its motion?
Mila [183]

Answer:

Inertia

F=ma

Action, reaction

All of the above

A heavy object requires more force to push than a lighter object.

8 0
3 years ago
Help awnser physics​
kirill115 [55]

Answer:

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3 0
3 years ago
KE 1/2mv^2 is a for kinetic energy (KE) in terms of mass, m, and the velocity, v, of a moving object. Which choice shows a corre
aivan3 [116]

Answer:Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

7 0
3 years ago
Read 2 more answers
28. Ken and Musa shared a cake such that Ken got twice the size
ratelena [41]

Answer:

Musa = \frac{1}{3}

Ken = \frac{2}{3}

Explanation:

Given

Ken = 2 * Musa --- Ken's share

Required

The fraction each got

Since they both shared a cake, we have:

Ken + Musa = 1

Substitute: Ken = 2 * Musa

2 * Musa+ Musa = 1

Factorize

Musa(2+ 1)= 1

Musa(3)= 1

Divide both sides by 3

Musa = \frac{1}{3}

Recall that: Ken = 2 * Musa

Ken = 2 * \frac{1}{3}

Ken = \frac{2}{3}

3 0
3 years ago
For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
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