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2 years ago

A robot that can move a hand left and right and rotate the hand 360º is said to have

Physics

1 answer:

Luda [366]2 years ago

3 0

You've described two (2) axes of motion.
The third one would have been up-and-down.

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I will fan and give 2 medals!!!!!!

Maksim231197 [3]

Corona

I. Rare gaseous envelope

II. Irregular pearly glow

III. Total solar eclipse

IV. Surrounds darkened moon

V. Caused by diffraction

Solar Prominence

I. loop shape

II. large bright gaseous

III. luminous hydrogen gas

IV. rises above chromospheres

V. anchored to surface (sun’s surface)

Solar Flare

I. Intense High-energy

II. Causes electromagnetic disruptions

III. In solar atmosphere

IV. Varies in brightness

V. Emits radiation

Sunspots

I. Surface dark spots

II. Strong magnetic fields

III. Solar magnetic storms

IV. Visible through telescope

V. Cooler than photospheres

I did this in class I got 100%

7 0

2 years ago

To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from

Natali5045456 [20]

1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.

3 0

2 years ago

Read 2 more answers

When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre wi

mixer [17]

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

5 0

2 years ago

1. Is the relationship between velocity and centripetal force a direct, linear relationship or is it a nonlinear square relation

Svetllana [295]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

7 0

2 years ago

If a roller coaster cart, with a mass of 100 kg, traveled this coaster, how much kinetic energy would it have at point 'E'?

zzz [600]

Answer:

Explanation:

Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.

8 0

2 years ago