Ask question

Ask question

All categories

3 years ago

15

The model that describes a Sun-centered solar system is called theA.laws of motion.B.geocentric model.C.heliocentric model.D.gen

eral theory of relativity.

1 answer:

artcher [175]3 years ago

7 0

The answer is C the heliocentric model.

You might be interested in

Solar cells change into electricity.

ValentinkaMS [17]

Answer:

yes that is true because if you put the solar thing in the sunlight it gets converted to electricity

3 0

3 years ago

Read 2 more answers

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Burka [1]

Answer:

the force of attraction between the two charges is 4.2 x 10⁹ N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

$F = \frac{Kq_1q_2}{r^2}$

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

$F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N$

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.

5 0

3 years ago

The front and rear sprockets on a bicycle have radii of 8.40 and 4.91 cm, respectively. The angular speed of the front sprocket

Rudik [331]

Explanation:

It is given that,

Radius of the front sprockets, $r_f=8.4\ cm=0.084\ m$

Radius of the rear sprockets, $r_r=4.91\ cm=0.0491\ m$

The angular speed of the front sprocket is 12.3 rad/s, $\omega_f=12.3\ rad/s$

(a) Linear speed of the front sprockets, $v_f=r_f\times \omega$

$v_f=0.084\times 12.3$

$v_f=1.0332\ m/s$

$v_f=103.32\ cm/s$

Linear speed of the rear sprockets, $v_r=r_r\times \omega$

$v_r=0.0491\times 12.3$

$v_r=0.60393\ m/s$

$v_r=60.393\ cm/s$

(b) Let $a_r$ is the centripetal acceleration of the chain as it passes around the rear sprocket.

$a_r=\dfrac{v_r^2}{r_r}$

$a_r=\dfrac{(60.393)^2}{0.0491}$

$a_r=74283.39\ m/s^2$

$a_r=7428339\ cm/s^2$

`Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0

3 years ago

PLEASE HELP MEEEE<br> ITS SIMPLE BUT CONFUSING

avanturin [10]

The density of Liquid T is less than that of water.

So, Liquid T will float on water.

<h3>Answer:</h3>

D. Liquid T

7 0

3 years ago

Read 2 more answers