All categories

3 years ago

A 440-g cylinder of brass is heated to 97.0 degree Celsius and placed in a 2 points

Physics

1 answer:

nydimaria [60]3 years ago

4 0

Answer:

Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .

Explanation:

Given :

Mass of brass, m₁ = 440 g

Temperature of brass, T₁ = 97° C

Mass of water, m₂ = 350 g

Temperature of water, T₂ = 23° C

Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹

Equilibrium temperature, T = 31° C

Let C₁ be the specific heat of brass.

Heat loss by brass = Heat gain by water

m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )

Substitute the suitable values in above equation.

440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)

C₁ = $\frac{11704}{29040}$

C₁ = 0.40 J g⁻¹ °C⁻¹

You might be interested in

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

6 0

3 years ago

A rectangular piece of cardboard, whose area is 352 square centimeters, is made into an open box by cutting a 2-centimeter sq

Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

$xy=352$

$y=\dfrac{352}{x}$ -------- (1)

The volume of the rectangular piece

$2(x-4)(\dfrac{352}{x}-4)=432$

$x^2-38x+352=0$

$(x-16)(x-22)=0$

x=16,22

Put the value of x in the equation (I)

For x = 16

$y=\dfrac{352}{16}=22$

For x = 22

$y=\dfrac{352}{22}=16$

Dimension of cardboard is 22 m by 16 m

8 0

4 years ago

A 18Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

Andrew [12]

Answer:

2Ω

Explanation:

If a 18Ω resistance is cut into three equal parts each of the resistance will be 18Ω/3 = 6Ω

Equivalent ratio in parallel is expressed as:

1/R = 1/6 + 1/6 + 1/6

1/R = 3/6

Cross multiply

3R = 6

R = 6/3

R = 2Ω

Hence the required equivalent resistance is 2Ω

5 0

3 years ago

Which process produces the energy radiated by the star when it becomes a main sequence star?

inysia [295]

The process that produces the energy radiated by stars is nuclear fusion in the core.
For a star on the main sequence, it's the fusion of hydrogen nuclei into helium.

8 0

4 years ago

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0

3 years ago