Ask question

Ask question

All categories

2 years ago

10

Alguien me puede ayudar en mi examen de fisica¿ porfavor es este https://forms.office.com/Pages/ResponsePage.aspx?id=AmBpC6bWZkK

JKBXtjQZEEGFFgCLKpPRNlrCyVE_kApNUQjZRQ01NQlVSM1hFWFZTRDFFMkQ5TjU2RiQlQCNjPTEu

1 answer:

Reptile [31]2 years ago

4 0

Answer:

Chicken Feet Or Butt Whole

Explanation:

Call Me Girls!

You might be interested in

Humans protect themselves from natural disasters by all of the following except

jeyben [28]

I'm pretty sure it's sunscreen

4 0

2 years ago

A crate is being lifted into a truck. If it is moved with a 2470 N force and 4850 J of work is done, then how far is the crate b

olchik [2.2K]

Answer:1.96m

Explanation:

Work=4850J

Force=2470N

Distance =work ➗ force

Distance =4850 ➗ 2470

Distance =1.96m

7 0

2 years ago

A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth

Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

$W=mg$

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center

The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

$800 N= W=mg=m \frac{GM}{R^2}$ (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

$200 N = W' = mg' = m \frac{GM}{(R+h)^2}$ (2)

If we divide eq.(1) by eq.(2), we get

$\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}$

$4=\frac{(R+h)^2}{R^2}$

By solving the equation, we find:

$4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0$

which has two solutions:

$h=-3R$ --> negative solution, we can ignore it

$h=R$ --> this is our solution

Since the Earth's radius is $R=6.4\cdot 10^6 m$ , the person should be at $h=R=6.4\cdot 10^6 m=6400 km$ above Earth's surface.

5 0

3 years ago

What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt

ipn [44]

Hope this helps you!

5 0

3 years ago

Read 2 more answers

Ship A is 32 miles north of ship B and is sailing due south at 16 mph. Ship B is sailing due east at 12 mph. At what rate is the

Zielflug [23.3K]

Answer:

$\dfrac{dz}{dt} =-5.6\ mile/h$

Explanation:

distance between ship A and B = 32 mile

Ship A velocity in south, dx/dt = -16 mph

Ship B is sailing toward east with speed, dy/st = 12 mph

time = 1 hour

rate of change of distance between them = ?

x be the distance travel after t time

X = 32 + x

Let distance between them be z

now, using Pythagoras theorem to calculate distance between ships after 1 hours

z² = x² + y²

z² = (32 + x)² + 12²

z² = (32 - 16)² + 12²

z = √400

z = 20 miles

now, calculation of rate of change of distnace

z² = (32 + x)² + y²

differentiating both side w.r.t. time

$2 z \dfrac{dz}{dt} = 2(32+x)\dfrac{dx}{dt} + 2 y\dfrac{dy}{dt}$

$z \dfrac{dz}{dt} =(32-16)\dfrac{dx}{dt} +y\dfrac{dy}{dt}$

$20\times \dfrac{dz}{dt} =16\times (-16) +12\times 12$

$\dfrac{dz}{dt} =\dfrac{-112}{20}$

$\dfrac{dz}{dt} =-5.6\ mile/h$

hence, the rate is the distance between them changing at the end of 1 hour is equal to $\dfrac{dz}{dt} =-5.6\ mile/h$

7 0

3 years ago