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natta225 [31]
3 years ago
11

Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied

halfway between points A and B), and a moment M=49 ft-lbs. Provide your answer in lbs and pay attention to your signs.
Physics
1 answer:
liraira [26]3 years ago
5 0

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

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A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
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Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

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A hiker is walking down a straight road traveling due west. Her journey can be thought of as occurring in 3 consecutive parts: t
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Answer:

Velocity in the first part= 0.75 m/s

Velocity in the second part=  1.1 m/s

Velocity in the third part= 0.92 m/s

Explanation:

Conceptual analysis : The velocity ​​is defined by the following formula:

v= d/t

Where:

v : velocity in The speed in meters over second (m/s)

d : distance in meters (m)

t : time in seconds (s)

Problem development

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velocity in the second part=  1.1 m/s

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3 years ago
Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to t
snow_lady [41]

Answer:

a) 3.632 m/s

b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}..........(1)

m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s

Substituting the above values into equation (1) and make V2 the subject of the formula:

0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\

V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}..................(2)

Using the law of conservation of kinetic energy:

0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\

2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}.......(3)

Substitute equation (2) into equation (3)

2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8})  ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2}  }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s

Substituting V_{1} into equation(2)

V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s

8 0
3 years ago
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