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Veseljchak [2.6K]

2 years ago

7

PLS HELP!!! Response: Develop a unique example (not posted on the discussion board by anyone else) and calculate the resultant d

isplacement between two points when there are two legs or distinct parts to the trip. Include the displacement of each leg of the trip as well as the resultant displacement of the entire trip. Don't forget: include both the direction and the magnitude as part of the displacement. Research the actual distance between these two points on the globe. Compare the resultant displacements and account for the error. Don't forget: include both the direction and the magnitude as part of the displacement.

1 answer:

Jlenok [28]2 years ago

8 0

Answer:

he is the first person can be 3in and a 6man is the best not a man of the hartford

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100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:

love history [14]

Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

6 0

3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

3 0

3 years ago

A 946.4 kg ( 2083 lb ) car is moving at 14.8 m / s ( 33.0 mph ) . Calculate the magnitude of its momentum.

IgorLugansk [536]

Answer:

14009. 72 kgms^-1

Explanation:

Momentum is the product of an objects mass and velocity

7 0

3 years ago

How does a heat pump resemble a refrigeration system?

REY [17]

A heat pump is a device that is capable of transferring heat energy from a source of heat to what is known as the heat sink. It also moves thermal energy in the opposite direction of a spontaneous heat transfer through heat absorption from a cold space and releasing it to a warmer space.

When a heat pump is being utilized for heating, it employs the same principle with that of the refrigeration cycle used by an air conditioner or a refrigerator, but in the opposite direction since it releases heat into a conditioned space rather than the surrounding environment. Moreover, heat pump resembles much as refrigeration since it has the same components with the latter except for the presence of a reverse valve.

3 0

3 years ago

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0

3 years ago