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4 years ago

14

When a craton is exposed at earth's surface, it is called a ________. when a craton is exposed at earth's surface, it is called

a ________. caldera continental shield mid-ocean ridge plate transform fault region?

1 answer:

harina [27]4 years ago

3 0

When a Craton is exposed at earth's surface, it is called a continental shield. Continental shield is any of the large stable areas of low relief in the Earth's crust that are composed of Precambrian Crystalline rocks. Shield areas are regarded as continental nuclei, the observation often being made that most continental shields are bordered by belts of folded rocks of post-Precambrian age.

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Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0

3 years ago

Light strikes a flat mirror with an angle of incidence of 73º. At what angle does it reflect off at?

Montano1993 [528]

Answer:

73º.

Explanation:

4 0

3 years ago

Which one of the following statements concerning the electric dipole moment is false?

baherus [9]

Answer:

A.The dipole moment is a scalar quantity.

Explanation:

We know that

Dipole moment :It measure the polarity of chemical bond in a molecule .

The electric dipole moment is equal to product of any charge (positive or negative ) and the distance between the two charges.

Mathematical representation:

$\mu=q\times d$

Where $\mu$ =Dipole moment

q=Charge on atom or particle

d=Distance between two charged particles

It helps to find out the molecule is polar or non- polar.

When the dipole moment is zero then the molecule is non-polar.

Dipole moment is a vector quantity.

The direction of dipole moment from negative charge to positive.

When unit of charge is C and unit of distance is m.

Then, unit of dipole moment=C-m

Hence, option A is false.

6 0

4 years ago

a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

<em>where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final </em>

<u>From (1):</u>

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

<em>where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision </em>

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

<em>where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission </em>

<u>From (3):</u>

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

<u>Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$ . </u>

Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

<u>From getting (5) into (4) we can obtain the $v_{2f}$ :</u>

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

7 0

4 years ago

What is the energy of a baby who weighs 20 N sitting on a 1.5 m high chair

guajiro [1.7K]

The energy of the baby is gravitational potential energy, and it is equal to the weight of the baby times its height from the ground:
$U=(mg) h$
where
mg=20 N is the weight (the mass times the gravitational acceleration)
h=1.5 m is the height from the ground
If we plug the numbers into the equation, we find
$U=(mg)h=(20 N)(1.5 m)=30 J$

3 0

4 years ago