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mojhsa [17]
3 years ago
8

A flagpole is a simple device on which a flag can be raised or lowered. Which simple machine is usually part of a flag pole?

Physics
2 answers:
kozerog [31]3 years ago
4 0
The anwser is C it can also be A but C would make the most sense
Talja [164]3 years ago
4 0

Answer:

C, a pulley! A pulley is made of a wheel and rope and in the case of the flagpole, it uses the pulley to raise the flag by pulling the rope.

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A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up
prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,  

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)  

We know that, Net force = m × a

\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}

\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}

\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}

a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

4 0
3 years ago
The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it
Sindrei [870]

Answer with Explanation:

We are given that

Speed ,v_0=107257 kmph

Angular velocity,\omega=7.292\times 10^{-5} rad/s

Radius of earth,r=6371 km=6371000 m

1 km=1000 m

Linear velocity,v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s

Linear velocity,v=464.57\times \frac{18}{5}=1672.46 km/h

Velocity at point A,v_A=-vi+v_0 j=-1672.46 i+107257j kmph

Velocity at point B,v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph

Velocity at point C,v_C=v_0j+vi=1672.46 i+107257j kmph

Velocity at point  D,v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph

3 0
3 years ago
Answers are - <br><br>A 235 N<br><br>B 376 N<br><br>C 271 N<br><br>D 188 N<br><br>E 470 N
amm1812

Answer:

C

Explanation:

8 0
3 years ago
A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the
vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ ln[\frac{P_{1} }{P_{2} }]

   = <em>mRT</em>₁ln[\frac{P_{1} }{P_{2} }]      [∵<em>pV</em> = <em>mRT</em>]

   = (5) (0.287) (272.039) ln[\frac{2.0}{1.0}]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = mc_{v}(T_{2}  - T_{1} ) = 0

From 1st law of thermodynamics

Q = Δ<em>U  </em>+ <em>W</em>

   = 0 + 270.588

   = 270.588 kJ

4 0
3 years ago
Where do earthquake waves originate from? select one:
arsen [322]
The correct answer is C) the focus
6 0
3 years ago
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