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3 years ago

6

Water has a specific heat of 4 J/g C. He heats 30 g of water in the microwave for 1 minute and finds that the temperature of the

water has increased by 100 C. What change in temperature will the chef observe when he heats 90 g of water in the same microwave for the same amount of time? A. An increase of 100 °C B. An increase of 150 °C C. An increase of 33 °C D. An increase of 50 °C

1 answer:

Bogdan [553]3 years ago

7 0

It’s been a long time since I last learned this but I think it’s C.
(30x4x100)/4x90 = 100/3

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In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea

Vikentia [17]

For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).

fringe = (delta t) / (λ/c)

We can find (delta t) with the equation:

delta t = [v^2(L1+L2)]/c^3

Derivation of this formula can be found in your physics text book. From here we find (delta t):

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.

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3 years ago

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

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3 years ago

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

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$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

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3 years ago

An object is thrown off a cliff with a horizontal speed of 10 m/sec. After 3 seconds the object hits the ground. Find the height

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Answer:

223

Explanation:

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Answer:

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