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velikii [3]
3 years ago
13

A cannon fires a ball vertically upward from the Earth’s surface. Which one of the following statements concerning the net force

acting on the ball at the top of its trajectory is correct?a) The net force on the ball is instantaneously equal to zero newtons at the top of the flight path.b) The direction of the net force on the ball changes from upward to downward.c) The net force on the ball is less than the weight, but greater than zero newtons.d) The net force on the ball is greater than the weight of the ball.e) The net force on the ball is equal to the weight of the ball.
Engineering
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

a) The net force on the ball is instantaneously equal to zero newtons at the top of the flight path.

Explanation:

At an instantenous time at the top of the flight path, the upward force due to the Canon explosion on the ball is just equal to the weight of the ball, this will equate the net force on the ball to zero. At this point the velocity of the ball is zero before it decends down to earth under its own weight.

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A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
3 years ago
A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2is subjected to an external tensile
lakkis [162]

Answer:

(a)  The force sustained by the matrix phase is 1802.35 N

(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa

(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively

Explanation:

Find attachment for explanation

8 0
3 years ago
What is engineering?
ANTONII [103]

Answer:

the branch of science and technology concerned with the design, building, and use of engines, machines, and structures. Anything that involves engines, wires, etc. basically

8 0
3 years ago
The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable
Fittoniya [83]

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

   Tray should be located ( 1 to 2 meters below surface )

   max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray :  1m below surface

<u>a) Determine percent removal of particles having a settling velocity of 1m/h</u>

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

<u>b) Determine if  the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray  and the maximum removal efficiency</u>

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

=  percent removal_{above} + percent removal_{below}

= ( d_{a},v_{p} ) . \frac{d_{a} }{depth}  + ( d_{a},v_{p} ) . \frac{depth - d_{a} }{depth}  = 100

hence max removal efficiency ≈ 70%

<u>c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?</u>

The maximum removal will drop as the particle settling velocity = 0.5 m/h

7 0
3 years ago
Can space debris take out a whole state
vodomira [7]
Yes depends how big the debris is
5 0
3 years ago
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