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MA_775_DIABLO [31]
2 years ago
12

A typical machine tests the tensile strength of a sheet of material cut into a standard size of 5.00 centimeters wide by 10.0 ce

ntimeters long. The machine consists of one clamp that holds the entire width (5.00 centimeters) so that it hangs vertically. A second clamp is placed on the lower end of the object, to which a variable downward force is applied. The force is slowly increased until the object ruptures, and the breaking force is recorded.
A strip of aluminum foil with a thickness of 15.0 micrometers and matching the size recommendations of the machine is placed in the machine and tested. The force needed to rupture the foil is found to be 233 newtons. What is the tensile strength of the aluminum foil sample?
Physics
1 answer:
Minchanka [31]2 years ago
4 0

Answer:

Explanation:

tensile strength is stress that is needed to break the wire made of the material .

Here force required to break the sheet of material = 233 N

cross sectional area of the foil = breadth x thickness

= 5 x 10⁻² x 15 x 10⁻⁶ m²

= 75 x 10⁻⁸ m²

breaking stress = force / cross sectional area

= 233 / 75 x 10⁻⁸

= 3.1 x 10⁸ Pa .

Tensile strength = 3.1 x 10⁸ Pa .

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the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point
Andreyy89

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We need to calculate the pressure at this point

Using formula of pressure

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v = velocity

Put the value into the formula

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Hence, The pressure at this point is 0.875 mPa

5 0
1 year ago
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
1 year ago
A particle moving in the x direction is being acted upon by a net force F(x)=Cx2, for some constant C. The particle moves from x
elixir [45]

Answer:

Change in kinetic energy is ( 26CL³)/3

Explanation:

Given :

Net force applied, F(x) = Cx²  ....(1)

Displacement of the particle from xi = L to xf = 3L.

The work-energy theorem states that change in kinetic energy of the particle is equal to the net amount of work is done to displace the particle.

That is,

ΔK = W = ∫F·dx

Substitute equation (1) in the above equation.

ΔK =  ∫Cx²dx

The limit of integration from xi = L to xf = 3L, so

\Delta K=\frac{C}{3}(x_{f} ^{3} - x_{i} ^{3})

Substitute the values of xi and xf in the above equation.

\Delta K=\frac{C}{3}((3L) ^{3} - L ^{3})

\Delta K=\frac{C}{3}\times26L^{3}

5 0
2 years ago
A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0
GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

F=qE

Put the value into the formula

F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

We need to calculate the total charge on the rod

Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}

Put the value into the formula

Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}

Q=47.8\times10^{-9}\ C

Q=47.8\ nC

Hence, The total charge on the rod is 47.8 nC.

6 0
2 years ago
Let A be the last two digits and let B be the sum of the last three digits of your 8-digit student ID. (Example: For 20245347, A
zheka24 [161]

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

v=25.0+47+14

v =86\ urk s/dort

v=86\times\dfrac{58.0}{28.8}

v=173.19\ m/s

Hence, The speed is 173 m/s.

7 0
2 years ago
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