All categories

4 years ago

Which of the following is the center of the universe according to the geocentric model?

Physics

2 answers:

Katarina [22]4 years ago

6 0

The earth is the center

Dima020 [189]4 years ago

4 0

Earth is the center according to the geocentric model.

You might be interested in

If P = 4.00î + 3.00k, P · Q = 17.0, and P ✕ Q = −6.00ĵ, determine the vector Q.

Butoxors [25]

let the vector Q is given as

$\vec Q = a\hat i + b\hat j + c\hat k$

given that

$P X Q = -6\hat j$

here we know that

$P = 4\hat i + 3 \hatk$

now by above equation

$(4\hat i + 3\hat k) X (a\hat i + b\hat j + c\hat k) = - 6\hat j$

$4b\hat k - 4c\hat j + 3a\hat j - 3b\hat i = - 6\hat j$

so by comparing both sides

b = 0

4c - 3a = 6

also we know that

$a^2 + b^2 + c^2 = 17^2$

$a^2 + 0 + (1.5 + 0.75a)^2 = 289$

by solving above equation

a = 12.85 and c = 11.14

so the vector Q is given as

$Q = 12.85\hat i + 11.14\hat k$

8 0

4 years ago

Can you help me with what kind of wave is shown?

Romashka-Z-Leto [24]

1 ? 2 transverse 3 longitudinal

4 0

4 years ago

How is heat involved in changing the phase of a substance from solid to liquid to gas?

natali 33 [55]

Answer:do it correct and do your best please yourself do what is right let every body know this is ak200342 talking to you.

Explanation:keep your head up.

3 0

3 years ago

Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$

if we solve for the magnitude, we arrive at

$F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}$

Hence the net force on one of the masses is

$F=8.0*10^{-10}N$

8 0

3 years ago

a block accelerates at 3.3 m/s^2 down a plane inclined at an angle of 27°. find the kinetic friction coefficient between the blo

finlep [7]

Answer:

0.13

Explanation:

Draw a free body diagram. There are three forces:

Weight force mg pulling down.

Normal force N pushing perpendicular to the incline.

Friction force Nμ pushing parallel up the incline.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel down the incline:

∑F = ma

mg sin θ − Nμ = ma

mg sin θ − mgμ cos θ = ma

g sin θ − gμ cos θ = a

gμ cos θ = g sin θ − a

μ = (g sin θ − a) / (g cos θ)

Given a = 3.3 m/s² and θ = 27°:

μ = (9.8 m/s² sin 27° − 3.3 m/s²) / (9.8 m/s² cos 27°)

μ = 0.13

3 0

3 years ago