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laiz [17]
3 years ago
9

A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating

Styrofoam beaker containing 1.00 kg of water at room temperature (20.0∘C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 20.0∘C.
(a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal?
(b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain.
(c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.
Physics
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

electeicity is caused by the movement of. from one atom to the next

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The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.4 V is across it at a frequ
zzz [600]

Answer:

The value of the inductance is 1.364 mH.

Explanation:

Given;

amplitude current, I₀ = 200 mA = 0.2 A

amplitude voltage, V₀ = 2.4 V

frequency of the wave, f = 1400 Hz

The inductive reactance is calculated;

X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms

The inductive reactance is calculated as;

X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}

where;

L is the inductance

L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH

Therefore, the value of the inductance is 1.364 mH.

7 0
3 years ago
Large amounts of water that quickly infiltrate soil on a sloped surface can
sergey [27]
A is the correct choice
7 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
bezimeni [28]

solution:

y = v0t + ½at²

1150 = 79t + ½3.9t²

0 = 3.9t² + 158t - 2300

from quadratic equations and eliminating the negative answer

t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)

t = 11.37 s to engine cut-off

the velocity at that time is

v = v0 + at

v = 79 + 3.9(11.37)

v = 123.3 m/s

it rises for an additional time

v = gt

t = v/g

t = 123.3 / 9.8

t = 12.59 s

gaining more altitude

y = ½vt

y = 123.3(12.59) /2

y = 776 m

for a peak height of

y = 776 + 1150


5 0
3 years ago
Read 2 more answers
Find the minimum kinetic energy in MeV necessary for an α particle to touch a 39 19 K nucleus that is initially at rest, assumin
SIZIF [17.4K]

Answer:

KE = 9.6MeV

Explanation:

Given the relationships we understand that,

q_1 = 19e, q_2 = 2e

The Potential at point p, is given by the following formula

V_ {p} = \frac {Kq_1} {r}

According to the graphic designed, you have,

V_ {p} = \frac {Kq_1} {(r_1 + r_2)}

V_ {p} = \frac {(9 * 10 ^ 9) (19 * 1.6 * 10 ^ {- 15})} {(3.7 + 2) * 10 ^{- 15}}}

V_ {p} = 4.8 * 10 ^ 6V

The kinetic energy of the particle would be given by

KE = q_2 * V_ {p} = 2e * 4.8 * 10 ^ 6V

KE = 9.6MeV

4 0
3 years ago
Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th
Evgen [1.6K]

Answer:

\omega=0.24\ rad.s^{-1}

Explanation:

Given:

mass of person, m=36\ kg

mass of merry go-round, M=300\ kg

radius of merry go-round, R=2\ m

velocity of the person running, v=4\ m.s^{-1}

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

I=M.R^2

I=300\times 2^2

I=1200\ kg.m^{-2}

<u>Moment of inertia of the person considering as a point mass:</u>

I_p=m.R^2

I_p=36\times 2^2

I_p=144\ kg.m^2

<u>Now according to the conservation of angular momentum:</u>

I.\omega=I_p.\omega_p

where:

\omega = angular velocity of the merry-go-round

\omega_p= angular velocity of the person running

1200\times \omega=144\times \frac{v}{R}

\omega=\frac{144}{1200} \times \frac{4}{2}

\omega=0.24\ rad.s^{-1}

4 0
3 years ago
Read 2 more answers
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