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anygoal [31]
3 years ago
5

How much time does it take a dropped object to fall 180 m on Earth?

Physics
1 answer:
rusak2 [61]3 years ago
7 0

Answer:

6s

Explanation:

Assume it is dropped from rest and the gravitational acceleration is 10

By the equation of motion under constant acceleration:

s=ut+\frac{1}{2} at^2

180 = (0)t+10(t^2)/2

t = 6 or -6 (rejected)

t = 6 s

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One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th
lozanna [386]

Answer:

k=105.0359\times 10^4\,W.m^{-1}.K^{-1}

Explanation:

Given:

temperature at the hotter end, T_H=100^{\circ}C

temperature at the cooler end, T_C=0^{\circ}C

length of rod through which the heat travels, dx=0.7\,m

cross-sectional area of rod, A=1.1\times 10^{-4}\,cm^2

mass of ice melted at zero degree Celsius, m=8.7\times 10^{-3}\,kg

time taken for the melting of ice, t=15\times60=900\,s

thermal conductivity k=?

By Fourier's Law of conduction we have:

\dot{Q}=k.A.\frac{dT}{dx}......................................(1)

where:

\dot{Q}=rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice,  L = 334000\,J.kg^{-1}

Q=m.L

Q=8.7\times 10^{-3}\times 334000

Q=2905.8\,J

Now the heat rate:

\dot{Q}=\frac{Q}{t}

\dot{Q}=\frac{2905.8}{900}

\dot{Q}=3.2287\,W

Now using eq,(1)

3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}

k=205.4606\,W.m^{-1}.K^{-1}

8 0
3 years ago
When we touch things, how close do atoms get together? Or does anything touch anything else?? Full answer Please 50 - 100 words
kkurt [141]
Well, im pretty sure that when we do touch eachother, the atoms themselves are touching. idk if this is what ur looking for but hope this helps.
3 0
3 years ago
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Which of the following weighing balances performs measurement in a closed compartment with no air currents to disturb measuremen
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Out of the following choices given, hydraulic balances performs measurement in a closed compartment with no air currents to disturb measurement. The correct answer is C. 
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Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from
qwelly [4]
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0
3 years ago
A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).
yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

f = n\frac{v}{2L}

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

f = n\frac{v}{2L}\\L = \frac{nv}{2f}

The radius between the two frequencies would be 4 to 5,

\frac{528Hz}{660Hz}= \frac{4}{5}

4:5

Therefore the frequencies are in the ratio of natural numbers.  That is

4f = 528\\f = \frac{528}{4}\\f = 132Hz

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m

Therefore the length of the pipe in the second harmonic is 2.6m

7 0
3 years ago
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