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DIA [1.3K]
3 years ago
7

How fast would you say the earth rotates near the equator?

Physics
1 answer:
PtichkaEL [24]3 years ago
8 0
460 meters per second, or about 1,000miles per hour.
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PLS HELP!!! WILL GIVE BRAINLIEST
grandymaker [24]

Answer:

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2 years ago
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A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer
Mrac [35]

Explanation:

The given data is as follows.

     m_{1} = 57 kg,        m_{2} = 79 kg

      l_{1} = 6.5 m,         l_{2} = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1} = 0

     57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5 = 0

                     T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}    

  57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5 = 0

                            T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.  

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3 years ago
A wave can be described as
rewona [7]
A wave can be described as the disturbance of particles in an area. Think about it this way: particles (matter) carry energy. For all the laws of physics to work, this energy must be "traded" somehow. This happens by miniscule vibrations in the particles, which are apparent disturbances. This creates a wave, and therefore a wave is, indeed, a disturbance.<span />
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When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C
Alex_Xolod [135]

Answer:

26.466cm³/min

Explanation:

Given:

Volume 'V'= 320cm³

P= 95kPa

dP/dt = -11 kPa/minute

pressure P and volume V are related by the equation

PV^{1.4}=C

we need to find dV/dt, so we will differentiate the above equation

V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt}  = \frac{d[C]}{dt}

\frac{dP}{dt} V^{1.4} + P(1.4)V^{0.4} \frac{dV}{dt} =0

lets solve for dV/dt, we will have

\frac{dV}{dt} =\frac{-\frac{dP}{dt} V^{1.4} }{P(1.4)V^{0.4} } \\\frac{dV}{dt} =- \frac{-\frac{dP}{dt} V}{P(1.4)}

\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}  (plugged in all the values at the instant)

\frac{dV}{dt} = 26.466

Therefore, the volume increasing at the rate of 26.466cm³/min at this instant

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3 years ago
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Katen [24]
The answer is b/ cope a small section word-for-word
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3 years ago
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