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Liono4ka [1.6K]
3 years ago
13

What is the energy change when the temperature of 15.0 grams of solid silver is decreased from 37.3 °C to 20.5 °C ?

Engineering
1 answer:
Gala2k [10]3 years ago
8 0

Answer:

Q=58.716 W

Explanation:

Given that

mass ,m  = 15 g

The decrease in the temperature ,ΔT = 37.3 - 20.5  °C

ΔT = 16.8°C

We know that specific heat of silver ,Cp = 0.233 J/g°C

The energy change of the silver  is given as

Q = m Cp ΔT

Now by putting the values we get

Q= 15 \times 0.233\times 16.8\ W

Q=58.716 W

Therefore the change in the energy will be 58.716 W.

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Seperate real and imaginary parts tan(2x+i3y)
ahrayia [7]

Answer:

First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

3 0
3 years ago
1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame
Vesna [10]
I DONT KNOW OKAY UGHHH
6 0
3 years ago
A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig
Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0
3 years ago
While discussing VIN numbers, Technician A says that the first digit of the VIN identifies the country where the vehicle was man
ruslelena [56]
Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect
4 0
3 years ago
The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current
Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}

\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

If i = 2A and g = 10 cm

\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97

f_{m}=-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N

b) Using the coenergy of the system:

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2}  }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N

8 0
3 years ago
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