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1 month ago

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Help asap please!! An aluminum block of mass 12.00 kg is heated from 20 C to 118 C. If the specific heat of aluminum is 913 J-1

kg K-1 then how much energy is required?
A. 10.956 kJ
B. 1073.688 kJ
C. 64.640 kJ
D. 7456.167 kJ

1 answer:

Sav [3.1K]1 month ago

7 0

Q = mCΔT, in which Q = energy required, m = mass of the block, C = specific heat, ΔT = temperature change.

Utilizing the values provided;

Q = 12*913*(118-20) = 1073688 J = 1073.688 kJ.

The correct option is B.

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Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

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Using the balance equation

Horizontal component

$T\cos\theta=mg$

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Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

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$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

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2 months ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

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A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

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where

k represents Coulomb's constant

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By inserting the values into the formula, we arrive at

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B) Attractive

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- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

$q_2 = Ne$

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$q_1 = -Ne$

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1 month ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

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Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

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It is known that

$P_1 = \frac{nRT_1}{v}$

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n and v are constant on both sides. Therefore we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

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let the final pressure be P and the temperature $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

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1 month ago

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Velocity = 71 meters per minute (MPM)
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2 months ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

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Answer:

1)

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2)

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Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

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$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

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Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

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$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

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$y=7.39\ m$

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2 months ago