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2 years ago

When are tides highest

Physics

2 answers:

N76 [4]2 years ago

7 0

The highest tides are on full moons

Lunna [17]2 years ago

7 0

The Proxigean Spring Tide is a rare, unusually high tide. This very high tide occurs when the moon is both unusually close to the Earth and in the New Moon phase

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A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate t

vivado [14]

Answer:

The force that would be applied on the rope just to start moving the wagon is 122 N

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

μ is the coefficient of friction ( μs, in this case, static friction);

m is mass of the object and;

g is the acceleration due to gravity( a constant equal to 9.81 m/ $s^{2}$ )

from the equation we are provide with;

μs = 0.25

m = 50 kg

g = 9.81 m/ $s^{2}$

F =?

Using equation 1

F = 0.25 x 50 kg x 9.81 m/ $s^{2}$

F = 122.63 N

Therefore a force of 122 N must be applied to the rope just to start the wagon.

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3 years ago

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Plz help!!!!! Best answer will get B R A I N L I E S T!!!

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They spill out chemicals into our air that may be bad for our lungs

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3 years ago

Is a electron a positively charged, negatively charged, or a neutral particle

Effectus [21]

An electron is negatively charged.

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3 years ago

What kind of energy is in a rock at the edge of a cliff.

soldi70 [24.7K]

Answer:

kinetic energy

Explanation:

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2 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

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2 years ago