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Rudik [331]
3 years ago
10

The force that pulls planets torwards the sun is called​

Physics
1 answer:
Studentka2010 [4]3 years ago
7 0

Answer:

gravity

Explanation:

Gravity pulls the planets toward the Sun. Gravity pulls the moon toward Earth. Gravity pulls us toward the Earth. Gravity is a force. Inertia.

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Solved, can someone check it over! ​
kakasveta [241]

Answer:

its good no need to change anything :))

4 0
2 years ago
Which feature of a heating curve indicates a change of state?
Andreas93 [3]
A "heating curve" is a graph that shows the temperature of the substance
against the amount of heat you put into it.

For most of the graph, as you'd expect, the temperature goes up as you
add heat, and it goes down as you take heat away.  BUT ... While the
substance is changing state, its temperature doesn't change even though
you're putting heat in or taking heat out.

So that part of the graph is a horizontal line. 
7 0
3 years ago
Read 2 more answers
How much work is done when you push a crate horizontally with 130
Harrizon [31]
Work= Force in the direction of displacement*displacement.

You know the force in the direction of displacement (horizontally) and the displacement. So,

W=130*11=1430 

Therefore, the work done is 1,430 Joules


3 0
3 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
2 years ago
Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

8 0
1 year ago
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