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3 years ago

An egg is dropped from a building that is 61 m high.

Physics

1 answer:

Allisa [31]3 years ago

3 0

Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

time = 3.5 s

Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

<u>Initial Velocity = 0 m/s</u>

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.6 m/s

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Answer:

Paying for employees seminars and workshops related to their careers

Explanation:

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3 years ago

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

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Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

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3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$