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2 years ago

Which of the following is NOT a way that a machine makes work easier to do?

Physics

1 answer:

puteri [66]2 years ago

5 0

Answer:

Explanation:

ur changing the way it move not to make it easier

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A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

5 0

3 years ago

Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

brilliants [131]

Answer:

Speed will be 30810 rpm

Explanation:

We have given diameter of the tire d = 24 inch

So radius $r=\frac{d}{2}=\frac{24}{2}=12imch$

We have given linear velocity v = 35 mph

We know that linear velocity is given by $v=\omega r$

$35=\omega \times 12$

$\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min$

As we know that 1 mile = 63360 inch and 1 hour = 60 min

3 0

3 years ago

The suspension system mounts the car's wheels<br> solid on the frame. True or false?

nadezda [96]

Suspect and mounts are a solid frame. True.

3 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

To make sure you understand how to use the equation, suppose that there are 1000 habitable planets in our galaxy, that 1 in 10 h

krok68 [10]

Answer:

Explanation:

Number of habitable planets = 1000

Fraction of planet with life = 1/10

Fraction of planet with life and civilization (before) = 1/4

Fraction of planet with life and civilization (now) =1/5

Therefore multiplying we have:

1000×1/10×1/4×1/5 = 5

5 0

2 years ago