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NikAS [45]
3 years ago
12

Like the moon's orbit, Earth's orbit is not a perfect circle, but an oval How do you think this might affect Earth's tides?

Physics
1 answer:
ExtremeBDS [4]3 years ago
8 0
During the parts in the orbit where the moon is farthest away from the earth the tides will be low. Whereas during the parts where the moon is closer to the earth the tides will be higher.

Good Luck!
Please correct me if i'm wrong
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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Where,

m_1 = mass of ball

m_2 = mass of the person

u_1 = Velocity of ball before collision

u_2 = Velocity of the person before collision

v_1 = velocity of ball afer collision

v_2= velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

v_1 = v_2

m_1u_1 + m_2u_2 = (m_1+m_2)v_2

Re-arrenge to find v_2,

v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}

Our values are,

m_1= 0.425kg

u_1= 12m/s

m_2= 68.5kg

u_2= 0m/s

Substituting,

v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}

v_2 = 0.074m/s

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0
3 years ago
You use a lever to lift a heavy tree branch you apply a force of 50 n and the lever lifts the branch
valentinak56 [21]

1.8 is the mechanical advantage of the lever.

<h3>Definition of mechanical advantage</h3>

The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.

The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.

Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.

In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.

Mechanical advantage= 90/50 =1.8

Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?

To learn more about the Mechanical advantage visit the link

brainly.com/question/16617083

#SPJ4

5 0
1 year ago
A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of
levacccp [35]

\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{DSS} = 10mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS(off)} = -6V

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS} =   {?}

\:  \:  \:

<u>Let's Slove :</u><u> </u>

  • \tt \large  I_{D} = I_{(DSS)}  (1 -   \frac {V_{GS}}{V_{GS(off)}} )^{2}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times  { - 6}

\:  \:

  • \underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}
3 0
1 year ago
You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds
german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

f_{beat}=\frac{1}{T}

Then the frequency for the previous period given (2sec) is

f_{beat}=\frac{1}{2}

f_{beat} = 0.5Hz

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz

<em>Hence option A is incorrect.</em>

We can do this process for 254Hz as f_1 and 258 Hz for f_2 , then

f_{beat} =|254Hz-258Hz|

f_{beat} = 4Hz

<em>Hence option B is incorrect. </em>

We can also do this process for 255Hz as  f_1 and 257 Hz for f_2 , then

f_{beat} =|255Hz-257Hz|

f_{beat} = 2Hz

<em>Hence option C is incorrect. </em>

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz

<em>Hence option D is incorrect. </em>

We can also do this process for 255.75Hz as f_1 and 256.25 Hz for f_2, then

f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz

<em>Hence option E is incorrect. </em>

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0
3 years ago
A baseball pitcher throws a 0.14 kg ball toward a batter who is 18 m away.
Olin [163]

Answer:

A. 112 J

Explanation:

KE = ½mv² = ½(0.14)40² = 112 J

5 0
3 years ago
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