What is the weight on earth of a girl with a mass of 23kg?

Describe how we know the plates once formed a supercontinent and how we know this.

Lady_Fox [76]

Answer:

<h3>How did scientists “discover” Pangea and other supercontinents of the past?</h3>

Nowadays, they can study the geologic record and use radioactive dating, seismic surveys, and other technologies to construct maps of how the world looked at various points in Earth's history.

3 0

2 years ago

Consider an unknown charge that is released from rest at a particular location in an electric field so that it has some initial

sineoko [7]

Answer:

b)

Explanation:

If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.

For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)

For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)

4 0

3 years ago

What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes

bagirrra123 [75]

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

q is magnitude of master charge
r is distance of separation
k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

E is electric field
q is the test charge

F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

8 0

1 year ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

Which explanation is based on scientific evidence?

Yuki888 [10]

Answer:

Scientific evidence is evidence that serves to either support or counter a scientific theory or hypothesis. Such evidence is expected to be empirical evidence and interpretable in accordance with scientific method.

5 0

3 years ago