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3 years ago

Need some help !!!!!!!!!

Physics

2 answers:

Yuri [45]3 years ago

4 0

Answer:

B. electromagnetic waves

Oduvanchick [21]3 years ago

4 0

The waves that travel through empty fence are electromagnetic:))) I hope this helps!!!

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Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy

Anestetic [448]

Answer:

To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Explanation:

The general formula of kinetic energy is given as follows:

$K.E = \frac{1}{2}mv^{2}$

where,

K.E = Kinetic Energy

m = mass of the object

v = speed of the object

So, for the marble and rock to have same kinetic energy, we can write:

$K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}$

<u>Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.</u>

4 0

3 years ago

A closed, rigid tank is filled with water. Initially, the tank holds 9.9 ft3 of saturated vapor and 0.1 ft3 of saturated liquid,

Galina-37 [17]

Answer:

a) x₁ = 0.058

b) T₂ = 416.02°F

c) q = 5047.39 Btu

Explanation:

see the attached file

7 0

4 years ago

After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calcul

Fittoniya [83]

Answer:

Part a: <em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

Part b: <em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Part c: <em>The value of Young's modulus at given point is 172 GPa.</em>

Part d: <em>The percentage elongation is 18.55%.</em>

Part e: The percentage reduction in area is 15.81%

Explanation:

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

$\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}$

Here F are different values of the load

Now Strain is given as

$\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\$

So the curve is plotted and is attached.

<em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

<em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Young's Modulus is given as

$E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa$

<em>The value of Young's modulus at given point is 172 GPa.</em>

The percentage elongation is given as

$Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\$

So the percentage elongation is 18.55%

The reduction in area is given as

$Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%$

So the reduction in area is 15.81%

6 0

3 years ago

A brick is dropped from rest from a height of 4.9 m. How long does it take for the brick to

noname [10]

Answer:

t=1s

Explanation:

Using the formula

s=ut+1/2at^2

4.9 = (0)(t) + 1/2(9.8)(t^2)

4.9 = 4.9(t^2)

t^2 = 1

t = 1

8 0

3 years ago

Convert 100°c into Faherenite and kelvin in steps.

wel

Answer:

100°c = 373.15 K

100°C=212°F

Explanation:

Given the 100°C

To convert Celsius to Kelvin, we need the following equation.

°C + 273.15 = K

100°C + 273.15 = K

373.15 = K

Therefore, 100°c = 373.15 K

Celsius To Fahrenhite:

F = 9/5C + 32

=9/5(100)+32

= (180) + 32

= 212°

Therefore,

100°C=212°F

3 0

3 years ago