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pychu [463]
3 years ago
13

PlS help I really need help.very urgentPLS help

Physics
1 answer:
horsena [70]3 years ago
4 0
Draw a freebody diagram, it will explain it really well

the boat is floating on top of the water, which means that the net acceleration in the y direction must be zero

the boat is not sinking (dominant downwards acceleration/force)
the boat is not flying (dominant upward acceleration/force)

that measn
F_{net_y} =0

now, if you drew the FBD, you only have 2 forces acting on the boat.
the upward bouyancy force on the boat and the downward force due to weight

F_{net_y}= F_{bouyancy} - F_{weight}

since the net force is equal to zero

F_{bouyancy} - F_{weight} = 0

and thus

F_{bouyancy} = F_{weight}
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Which is not an element? Water , Arsenic, sodium, Aluminum
MrRa [10]

Answer:

water

Explanation:

water is not an element, it is a molecule

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4 years ago
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What if The rotational speed of Earth increased ?
dsp73

Answer:

This would happen.

Explanation:

If the earth’s rotation speed increases then the weight of the body decreases. This is because you see a moving body on the rotating earth’s surface itself is in the reference frame. So when the earth rotates, the centripetal force acts towards the centre of rotation.

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Find the force on an object which has the mass of 20kg and an acceleration of 10m/s
Elis [28]

Answer:

200 Newtons

Explanation:

20 kg x 10 m/s squared

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ORIGIN OF MOON AND THE RELATIONSHIP WITH THE EARTH
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The moon's gravity pulls at the Earth, causing predictable rises and falls in sea levels known as tides. To a much smaller extent, tides also occur in lakes, the atmosphere, and within Earth's crust.

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4 years ago
A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w
lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

4 0
3 years ago
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