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Svetach [21]
1 year ago
5

Which would require the greater energy; slowing down of the orbital speed of the Earth so it crashes into the sun, or speeding u

p the orbital speed of the Earth so it escapes the sun
Physics
1 answer:
tresset_1 [31]1 year ago
3 0

Answer: Speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Explanation: To find the answer, we need to know more about the Orbital and escape velocities.

<h3>What is Orbital and Escape velocity?</h3>
  • Orbital velocity can be defined as the minimum velocity required to put the satellite in its orbit around the earth.
  • The expression for orbital velocity near to the surface of earth will be,

                   V_o=\sqrt{gR}

  • Escape velocity can be defined as the minimum velocity with which a body must be projected from the surface of earth, so that it escapes from the gravitational field of earth.
  • The expression for orbital velocity will be,

                    V_e=\sqrt{2gR}

  • If we want to get into the sun, we want to slow down almost completely, so that your speed relative to the sun became almost zero.
  • We need about twice the raw speed to go to the sun than to leave the sun.

Thus, we can conclude that, the speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Learn more about orbital and escape velocity here:

brainly.com/question/28045208

#SPJ4

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Cheetah mothers perform a number of different behaviors. They and their cubs stay in one place for only four days, moving on bef
Mekhanik [1.2K]

Answer:

Cheetah cubs are in danger from predators like lions and hyenas which can track their prey by scent and so the mother and her cubs leave an area when their scent is too strong so that they are not hunted and the cubs survive.

Mother Cheetahs also train their cubs to hunt so that they may get food for themselves which will ensure their survival as well thus showing that both of these practices can impact on reproductive success.

4 0
3 years ago
Identify the independent, dependent, and constant variables for different experiments.
diamong [38]

ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

Independent variable(IV)- what you change during the experiment.

dependent variable(DV)- what you're measuring during an experiment. The dependent variable is DEPENDENT because it's results DEPEND on the independent variable at play.

Constant variables(CV)- things that do not change in order to isolate the tested variables as much as possible.

3 0
3 years ago
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the
Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g=\frac{530}{1000}=0.530 kg

1kg=1000 g

Area=A=935 cm^2=935\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Height,h=4.4 m

C=1

\rho=1.21 kg/m^3

Width of rectangular prism,b=11.6 cm=\frac{11.6}{100}=0.116 m

1 m=100 cm

Length,l=23.2 cm=0.232 m

Area=l\times b=0.116\times 0.232=0.0269 m^2

Terminal velocity,v_t=\sqrt{\frac{2mg}{\rho CA}}

Where g=9.8 m/s^2

Using the formula

v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}

v_t=17.86 m/s

The velocity of person,v=\sqrt{2gh}

Using the formula

v=\sqrt{2\times 9.8\times 4.4}

v=9.29 m/s

4 0
2 years ago
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