The moon’s relative motion causes

Marx believed that workers had a false consciousness and should develop a true consciousness instead. What does this mean?

Artyom0805 [142]

Answer:

That means they should actually pay attention.

Explanation:

8 0

4 years ago

Read 2 more answers

To leave the gravitational pull of the Earth, and explore other planets, satellites must have at least:

Alexxx [7]

Answer:

To explore the other planets, the satellite must have the velocity more than the escape velocity.

Explanation:

The minimum velocity required by any object to escape from the earth gravitational pull is called the escape velocity.

The escape velocity for any planet depends on the mass of planet and radius of planet. It does not depends on the mass of object. The escape velocity is same for any mass for a particular planet.

So, to explore the other planets, the satellite must have the velocity more than the escape velocity.

7 0

3 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

3 years ago

Phosphorus-32, a radioactive isotope of phosphorus-31 (atomic number 15), undergoes a form of radioactive decay whereby a neutro

dimulka [17.4K]

Answer:

The product of the decay its Sulfur-32

Explanation:

Phosphorus-32 ( lets write it $_{15}^{32}P$ , where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:

$_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}$

where $e^-$ its the electron, and $\bar{\nu_e}$ the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).

So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:

$_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}$ .

$_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}$ .

The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its

$_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}$ .

and the product of such decay its Sulfur-32

5 0

4 years ago

A 90. 0-kg ice hockey player hits a 0. 150-kg puck, giving the puck a velocity of 45. 0 m/s. If both are initially at rest and i

Mice21 [21]

The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>

The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

$m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s$

The time taken for the puck to reach 15 m is calculated as follows;

$t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s$

The distance traveled by the hockey player at the calculated time is;

$d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m$

Learn more about conservation of linear momentum here: brainly.com/question/7538238

4 0

2 years ago