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3 years ago

What type of plate tectonic activity do you think is being shown in the picture below?

Physics

1 answer:

blondinia [14]3 years ago

3 0

Answer:

Convergent boundaries,over time,a rift valley will form.

Explanation:

The picture shows 2 plates moving apart,and a shallow "valley"is also noticeable.

Mark me brainliest if I helped:D

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Carlos is making phosphorous trichloride using the equation below. He adds 15 g of phosphorus.

Degger [83]

Given:

The balanced chemical reaction of the synthesis of phosphorus trichloride:

2P + 3Cl2 ===> 2PCl3

Initial amount of phosphorus = 15 grams

The amount of product produced from 15 grams of phosphorus:

15 grams / 31 g/mol * (2/2) = 66.46 grams PCl3

The amount of chlorine is 44.31 grams, nearest to 45 grams.

8 0

3 years ago

Read 2 more answers

Which fact was most likely discovered between the time of Mendeleev’s table and the time of Moseley’s table that helped Moseley

igomit [66]

Answer:

The number of protons in an atom is different than the atom's total mass.

Explanation:

Mendeleev had presented the first structured periodic table in 1869. But, in 1913, Henry Moseley developed the modern periodic table. The crucial difference between the two periodic tables is that Mendeleev's periodic table is generated on the basis of the atomic masses of chemical elements while the chemical element's atomic numbers were the main focus of Moseley's periodic table. There were just 56 chemical elements in Mendeleev's periodic table. On the other hand, 74 chemical elements are presented in Moseley's periodic table.

7 0

3 years ago

A mom pushes her 19.3 kg daughter on the swing. If she gives her an initial velocity of 7.5 m/s at the bottom of the swing and t

kiruha [24]

Answer:

3.17333333333? I hope I get it right

Explanation:

..................hello

3 0

3 years ago

Help me pleaseee, it’s due today: Two people push on a large gate as shown on the view from above in the diagram. If the moment

Sladkaya [172]

Answer:

the angular acceleration of the gate is approximately 1.61 $\frac{rad}{s^2}$

Explanation:

Recall the formula that connects the net torque with the moment of inertia of a rotating object about its axis of rotation, and the angular acceleration (similar to Newton's second law with net force, mass, and linear acceleration):

$\sum \tau_1=I\,\alpha$

In our case, both forces contribute to the same direction of torque, so we can add their torques up and get the net torque on the gate:

$\tau_{net}=(20*2+30*3.5) \,N\,m=145\,\,N\,m$

Now we use this value to obtain the angular acceleration by using the given moment of inertia of the rotating gate:

$\sum \tau_1=I\,\alpha\\145\,\,N\.m=(90\,\,kg\,m^2)\,\alpha\\\alpha= \frac{145}{90} \frac{rad}{s^2} = 1.61\, \frac{rad}{s^2}$

4 0

3 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

n200080 [17]

a) $8.93\cdot 10^{-30} N$ , downward

b) $1.76\cdot 10^{-17} N$ , upward

c) $1.20\cdot 10^{-17} N$ , downward

Explanation:

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

$F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N$

And the direction is downward, towards the Earth's centre.

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$E=110 N/C$ is the strength of the electric field

Substituting,

$F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N$

And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

$F=qvB$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$v=1.50\cdot 10^6 m/s$ is the velocity

$B=50.0\mu T = 50.0 \cdot 10^{-6} T$ is the strength of the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N$

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

5 0

3 years ago