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Viktor [21]
3 years ago
9

Getyour pointsssssssssssssssssssssss

Physics
2 answers:
vladimir2022 [97]3 years ago
8 0

Answer:

hi and thankss

Explanation:

Maurinko [17]3 years ago
3 0

Answer:

thankkkk youuuu soo muchhh

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You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the
kondor19780726 [428]

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

Initial speed of a euro, u = 2 m/s

It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s

Let t is the time taken by the euro to hit the ground. It can be calculated as :

v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s

Hence, it will take 3.29 seconds to hit the ground.

4 0
3 years ago
Explain how you think humans have disrupted the nitrogen cycle HELP QUICK DUE APRIL 15TH ANYBODY
mina [271]

Answer:

Scientists have determined that humans are disrupting the nitrogen cycle by altering the amount of nitrogen that is stored in the biosphere. The chief culprit is fossil fuel combustion, which releases nitric oxides into the air that combine with other elements to form smog and acid rain.

5 0
2 years ago
A gas occupies a volume of 1.0 m3 in a cylinder at a pressure of 120kPa. A piston compresses the gas until the volume is 0.25m3,
Hoochie [10]

Answer:

Approximately 480\; \rm kPa, assuming that this gas is an ideal gas.

Explanation:

  • Let V(\text{Initial}) and P(\text{Initial}) denote the volume and pressure of this gas before the compression.
  • Let V(\text{Final}) and P(\text{Final}) denote the volume and pressure of this gas after the compression.

By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:

\displaystyle \frac{P(\text{Final})}{P(\text{Initial})} = \frac{V(\text{Initial})}{V(\text{Final})}.

Note that in Boyle's Law, P is inversely proportional to V. Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.

For this particular question:

  • V(\text{initial}) = 1.0\; \rm m^3.
  • P(\text{Initial}) = 120\; \rm kPa.
  • V(\text{final}) = 0.25\; \rm m^3.
  • The pressure after compression, P(\text{Final}), needs to be found.

Rearrange the equation to obtain:

\displaystyle P(\text{Final}) = \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial}).

Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.

\begin{aligned}P(\text{Final}) &= \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})\\ &= \frac{1.0\; \rm m^{3}}{0.25\; \rm m^{3}} \times 120\; \rm kPa = 480\; \rm kPa\end{aligned}.

4 0
3 years ago
1. A strobe pattern is made of marks on a piece of paper recorded every 0.1 s. Two adjacent marks near the middle of the pattern
Korvikt [17]
Speed = distance / time

3.4cm / 0.1s = 34 cm/sec

7 0
2 years ago
During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f
dezoksy [38]

Answer:

F= 2.3733 x10^{20} N

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

M_ {sun} = 1.99 * 10^{30} Kg

M_ {Earth} = 5.98 * 10 ^ {24} Kg

M_ {Moon} = 7.36 * 10 ^{22} Kg

Average distances

\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m

\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m

Gravitational constant

G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}

The formula of the Gravitational Force between the Moon and the Earth would be,

F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}

F = 1.9908 * 10 ^{20} N

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}

F_2 = 4.3641*10^-{ 20} N

This force is in the direction of the Sun

The net force must be

F_ {net} = F_2-F

F_ {net} = 2.3733*10^{20} N

This in the direction of the Sun.

8 0
2 years ago
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