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Arte-miy333 [17]

4 years ago

10

4. We have 4 identical strain gauges of the same initial resistance (R) and the same gauge factor (GF). They will be used as R1,

R2, R3, and R4 in a Wheatstone bridge. We want to measure the tensile strain of a rod subjected to a tensile force. What is the maximum bridge constant (kappa) that we can reach

1 answer:

Luba_88 [7]4 years ago

3 0

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Which substance is a compound Li, N2, H2, CO2?

kupik [55]

Carbon dioxide is a compound (CO2). The other are just considered molecules since they are bonded to themselves. Lithium (Li) is an element since it's on its own and hasn't formed a bond yet.

6 0

4 years ago

Read 2 more answers

Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05

Maru [420]

Answer:

The answer to your question is: F = 50 N

Explanation:

Data

mass = 1000 kg

acceleration = 0.05m/s2

F = ?

Formula

F = m x a

Substitution

F = 1000 kg x 0.05 m/s2 = 50 kgm/s2 = 50 N

Mike is applying a force of 50 N to the car.

3 0

4 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

4 years ago

.... is increasing the temperatures of earths oceans and atmosphere, leading to more intense storms of all types, including hurr

kirza4 [7]

I think with the increasing of temperatures of earths oceans and atmosphere is causing climate change and global warming. When the ocean heats up a lot of sea animals die and with the hot temps in the atmosphere is causing rare hot tempatures in some places in the world.
It can cause some intense storms but it mostly heats up the world and cause a lot of problems.

I hope i answered your question right.

5 0

4 years ago

A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i

konstantin123 [22]

Answer:

$T_f=5854.76$ °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, $E= 1.10\times 10^6 J$

heat capacity of the hiker, $c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}$

initial body temperature of hiker, $T_i=36.6 \degree C$

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

$W=63\times 9.8\times 828$

W= 511207.2 J

∵Hike eats 2 energy bars= $2\times 1.1\times 10^{6} J$

Energy produced $= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

$25\% of E= 511207.2$

$\Rightarrow E= 511207.2\times \frac{100}{25}$

$E=2044828.8 J$

&

Amount of total energy (E) converted into heat(Q) is:

$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$ .................(1)

where:

$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

$T_f\approx 5854.76$ °C

4 0

3 years ago