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3 years ago

What is the shape of the earth's orbit around the sun?

Physics

2 answers:

Daniel [21]3 years ago

7 0

<span>............D. Elliptical</span>

valkas [14]3 years ago

7 0

It's orbit is elliptical

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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

3 years ago

To prevent accidental electric shock, most electrical equipment is A. painted. B. isolated. C. coated. D. grounded.

Nesterboy [21]

D. It is Grounded below ground

4 0

3 years ago

Read 2 more answers

The difference in a brass rod and steel rod is same as 12cm at all temperature .Find the length of the rod at 0°C.

egoroff_w [7]

Answer:

I think its 12 cm itself. Because in the question it is said that the length is same in all temperature

5 0

2 years ago

for the primitive yo-yo in fig, to find the downward acceleration of the cylinder and the tension in the string. you can take th

mixas84 [53]

The downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

<h3>Downward acceleration of the cylinder</h3>

The downward acceleration of the solid cylinder is determined from the principle of conservation of angular momentum as shown below;

Iα = Tr

where;

I is moment of inertia of the solid cylinder
α is angular acceleration of the cylinder
T is tension in the string
r is length of the string

α = Tr/I

$\frac{a}{R} = \frac{Tr}{I} \\\\\frac{a}{R} = \frac{Tr}{\frac{1}{2} MR^2}\\\\\frac{a}{R} =\frac{2Tr}{MR^2} \\\\a = \frac{2Tr}{MR}$

where;

a is the downward acceleration of the solid cylinder
R is radius of the cylinder

Thus, the downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

7 0

1 year ago

A jet airliner moving initially at 548 mph

sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction

$v_2 = 343 mph$ in a direction $67^{\circ}$ north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

$v_{1x}= 548 mph$

$v_{1y}=0$

$v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph$

$v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph$

So, the components of the resultant velocity in the two directions are

$v_{x}=548 mph+134 mph=682 mph$

$v_{y}=0 mph+315.7 mph=315.7 mph$

So the new speed of the aircraft is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph$

3 0

3 years ago