All categories

2 years ago

The graph of an object's position over time is a horizontal line and y is not equal to 0. What must be true abou

Physics

1 answer:

frozen [14]2 years ago

7 0

Answer:D: the velocity is zero

Explanation:

You might be interested in

CI. Design challenge: Your goal is to build a compound microscope than can at least double the size of the object. We will const

lilavasa [31]

Answer:

2. 17.7cm

3. -7 that is magnification

Explanation:

See attached handwritten document

6 0

2 years ago

Which statement about an atom is correct?(1 point)

alexdok [17]

Answer:

The electron has a negative charge and is found outside of the nucleus.

Explanation:

1.) Neutrons have no charge, not a negative charge.

2.) Neutrons are found inside the nucleus, not outside.

3.) Protons have a positive charge.

5 0

2 years ago

Read 2 more answers

A skater increases her speed uniformly from 2.0 meters per second to 7.0 meters per second over a distance of 12 meters. the mag

AVprozaik [17]

<span>The magnitude of her acceleration as she travels this 12 meters is 1.875m/s^2</span>

6 0

3 years ago

A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

$v = \sqrt{\frac{2GM}{R}}$

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

$v_o = \sqrt{\frac{GM}{R+h}}$

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

$T = \frac{2\pi(R+h)}{v_o}$

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

3 years ago

A person pushes on a box with a force of 500N. If the box does not move, what is the force of static friction on the box?

Alexxx [7]

Answer:

Explanation:

4 0

2 years ago