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denis23 [38]
4 years ago
9

What are the change in internal energy if 250 j of hear is added to a system of 80 j of work is done by the system

Physics
1 answer:
Sphinxa [80]4 years ago
5 0

Answer:

Explanation:

dU= dq+w

dU is change in internal energy of the system

dq is the amount heat added or released by the system which be positive or negative respectivelý

And w is the amount of work done by the system or on the system which will be positive or negative respectively.

Hence,

dU= 250+80= 330 J

The change will be positive

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What are the magnitude and direction of a uniform electric field perpendicular to the ground that is able to suspend a particle
Oksanka [162]

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the magnitude and direction of the uniform electric field is 1846.38 N/C and UPWARD respectively

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F = mg and F = Eq

so

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7 0
3 years ago
While visiting the Albert Michelson exhibit at Clark University, you notice that a chandelier (which looks remarkably like a sim
aksik [14]

Answers:

a) 0.144 Hz

b) 0.904 rad/s

c) 11.818 m

d)9.77 m/s^{2}

Explanation:

The rest of the question is written below:

a) Calculate the frequency of oscillation (in Hertz) of the chandelier

b) Calculate the angular frequency \omega of the chandelier in radians/ second

c) Determine the length L in meters of the chandelier

d) That evening, while hanging out in JJ. Thompson's House O' Blues, you notice that (coincidentally) there is a chandelier identical in every way to the one at the Michelson exhibit except this one swings back and forth 0.01 seconds slower, so the period is T+0.01 s. Determine the acceleration due to gravity in m/s^{2} at the club.

a) The frequency f has an inverse relation with the period T:

f=\frac{1}(T} (1)

Where T=6.9 s

f=\frac{1}(6.9s}=0.144 Hz (2)

b) The angular frequency \omega is given by:

\omega=2\pi f=\frac{2 \pi}{T} (3)

\omega=2\pi (0.144 Hz) (4)

\omega=0.904 rad/s (5)

c) Another expression for the period is:

T=2 \pi \sqrt{\frac{L}{g}} (6)

Where:

L is the length of the pendulum

g=9.8 m/s^{2} is the mean acceleration due gravity

Isolating L:

L=\frac{T^{2} g}{4 \pi^{2}} (7)

L=\frac{(6.9)^{2} (9.8 m/s^{2})}{4 \pi^{2}} (8)

L=11.818 m (9)

d) In this case the period of the pendulum is T_{p}=T+0.01 s. So, we will use equation (7) with this period and find g:

g=\frac{4 \pi^{2}L}{(T+0.01 s)^{2}} (10)

g=\frac{4 \pi^{2}(11.818 m)}{(6.9 s+0.01 s)^{2}} (11)

g=9.77 m/s^{2} (12) This is the acceleration due gravity at the place, which is near the mean value of 9.8 m/s^{2}

5 0
3 years ago
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