•What is the gravitational potential energy of a girl

2. In your own words, what is direct plagiarism?

alekssr [168]

Answer:

copying another writer's work with no attempt to acknowledge that the material was found in external source is considered as a direct plagiarism.

7 0

3 years ago

Read 2 more answers

A clamp-type measuring instrument operates on the principle of

Hitman42 [59]

<em>A clamp-type measuring instrument operates on the principle of; </em>

A. induction

8 0

3 years ago

A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal

kogti [31]

Answer:

The answer to your question is 636.6 ft

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

sin 19.5 = 212.5/hyp

solve for hyp

hyp = 212.5 / sin 19.5

Result

hyp = 212.5/ 0.333

hyp = 636.6 ft

4 0

3 years ago

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin

Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0

3 years ago

Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp

In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0

3 years ago