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3 years ago

A substance freezes at -58c therefore substance melts at ?

1 answer:

3 0

The freezing point is the same as the melting point.

If it freezes at -58°C, hence the melting point is also -58°C.

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Select the correct answer What is the importance of writing a draft?

givi [52]

Answer:to revise or edit

anything that can be made in the non draft one

Explanation:

4 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet

ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S $_{y}$ = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326 E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S $_{y}$ = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N and S $_{y}$ > σ

3 0

3 years ago

A coil is wrapped with 332 turns of wire on the perimeter of a circular frame (of radius 30 cm). Each turn has the same area, eq

almond37 [142]

Answer:

$E=84.5V$

Explanation:

From the question we are told that:

Number of Turns $N=332turns$

Radius $r= 30cm$

Field Change $B=56mt-29mt=27mt$

Time $t=63ms$

Generally the equation for Magnetic Field is mathematically given by

$\frac{dB}{dt}=\frac{27*10^{-3}}{29*10^{-3}}$

$\frac{dB}{dt}=0.9T/s$

Generally the Flux at 332 turns is mathematically given by

$\phi=N*A*B$

Generally the equation for Area of coil is mathematically given by

$A=\pi*r^2$

$A=\pi*(r*10^{-2})^2$

Since

$\phi=332*\pi*(\theta*10^{-2})^2*B$

Therefore

$\frac{d \phi}{dt}=332*\pi*(900*10^{-4}*\frac{dB}{dt}$

Generally the equation for emf Magnitude is mathematically given by

$E=\frac{d\phi}{dt}$

$E=332*\pi*(900*10^{-4}*0.9$

$E=84.5V$

4 0

3 years ago

In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizont

EleoNora [17]

Answer:

$K_{2}=7302.4J$

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the conservation of total mechanical energy

$K_{1}+U_{1}=K_{2}+U_{2}$