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3 years ago

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A cell phone weighing 80 grams is flying through the air at 15 m/s. What is its kinetic energy

1 answer:

Contact [7]3 years ago

8 0

Answer:

The kinetic energy of the cell phone is 9J

Explanation:

The kinetic energy is the energy possessed by a body by virtue of motion.

The kinetic energy is expressed as

KE= 1/2m(v)²

Given data

Mass of cell phone m= 80g--to kg=80/1000= 0.08kg

Velocity of cell phone v= 15m/s

Substituting our given data we have

KE= 1/2*0.08(15)²

KE= (0.08*225)/2

KE=18/2

KE= 9J

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A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen

Sveta_85 [38]

Answer:

$v_x=34 m/s$

$v_y=53.9\ m/s$

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

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vx=v

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The horizontal component of the velocity is always the same:

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3 years ago

Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the

Dafna11 [192]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Heat capacity of body 1 :

$\qquad \sf \dashrightarrow \:m_1s_1$

Heat capacity of body 2 :

$\qquad \sf \dashrightarrow \:m_2s_2$

it's given that, the the head capacities of both the objects are equal. I.e

$\qquad \sf \dashrightarrow \:m_1s_1 = m_2s_2$

$\qquad \sf \dashrightarrow \:m_1 = \dfrac{m_2s_2}{s_1}$

Now, consider specific heat of composite body be s'

According to given relation :

$\qquad \sf \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }$

[ since, $m_2s_2 = m_1s_1$ ]

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}$

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$\qquad \sf \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2}$

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gavmur [86]

Answer:

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b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

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$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

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