Of the seven chambers in the masque of the red death, which room is the least popular, and why?

An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist

Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0

3 years ago

A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 de

jeyben [28]

Answer:

$c_{e1}$ = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded Qh = m1 ce1 ( $T_{h}$ - $T_{f}$ )

Heat absorbed Qc = m2 ce2 ( $T_{f}$ - T₀)

Body 1 is metal and body 2 is water . Where m are the masses of the two bodies, ce their specific heat and T the temperatures

Qh = Qc

m₁ $c_{e1}$ ( $T_{h}$ - $T_{f}$ ) = m₂ $c_{e2}$ ( $T_{f}$ - T₀)

we clear the specific heat of the metal

$c_{e1}$ = m₂ $c_{e2}$ ( $T_{f}$ - T₀) / (m₁ ( $T_{h}$ - $T_{f}$ ))

$c_{e1}$ = 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

$c_{e1}$ = 209.2 (9.36) / (75 78.85)

$c_{e1}$ = 1958.11 / 5913.75

$c_{e1}$ = 0.331 J / g ° C

5 0

3 years ago

a square loop of wire side 3.0 cm carries 3.0 A of current. A uniform magnetic field of magnitude 0.67 T makes an angle of 37 de

Varvara68 [4.7K]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

The root-mean-square value of an alternating current of S0Hz frequency is 10 ampere. The time taken by the alternating current i

stealth61 [152]

Answer:

(D ) $5\times10^{-3} sec$ and $14.14 amp.$

Explanation:

Given that, the rms value of current is 10 ampere and its frequency is 50 Hz.

Now maximum value of current can be calculated as,

$I_{max}=\sqrt{2}I_{rms}\\I_{max}=10\times \sqrt{2} \\I_{max}=14.14 amp.$

The time period is defined as,

$T=\frac{1}{f}\\ T=\frac{1}{50}\\ T=0.02 sec$

Now the time taken by alternating current going from zero to maximum value is,

$t=\frac{T}{4}\\ t=\frac{0.02}{4}\\ t=5\times 10^{-3}sec$

Therefore, the time taken by alternating current to reach maximum value is $t=5\times 10^{-3}sec$ and the peak value of current is $14.14 amp.$ .

7 0

3 years ago

Which best describes protein

Elena L [17]

Answer:

It is nutrients you get from food

7 0

3 years ago

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