3 years ago

Los rieles de una vía de tren de acero tienen 1500 m de longitud. ¿Qué longitud tendrá cuando

Physics

1 answer:

melomori [17]3 years ago

6 0

Answer:

can you translate the question

into english

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Every magnet must have what at its ends?

slega [8]

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The correct is A.

every magnet have two pole

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2 years ago

A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley c

sergiy2304 [10]

Answer:

Part a)

$a_{hanger} = g - \frac{T}{m}$

Part b)

$a_{cart} = \frac{T}{M}$

Part c)

$a = \frac{mg}{M + m}$

Part d)

$a = 1.35 m/s^2$

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

$mg - T = ma$

so we have

$a_{hanger} = g - \frac{T}{m}$

Part b)

now for car that is rolling on the floor the net force is given as

$F = Ma$

$T = Ma$

$a_{cart} = \frac{T}{M}$

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

$T = Ma$

$mg - Ma = ma$

$a = \frac{mg}{M + m}$

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

$a = \frac{0.50(9.81)}{2.40 + 0.50}$

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3 years ago

A/an ___ is a machine which tells us about the strength and speed of sisemec waves.

svp [43]

The answer is Seismograph

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3 years ago

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A compound machine is sued to lift a car. If you apply a force of 60 N to the machine, it lifts the car with a force of 550 N.

mrs_skeptik [129]

a) MA=550/60=9.17

b) Wu=1150*0.35=402.5 J

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3 years ago

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When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

yarga [219]

Answer:

$\Delta E=1.64*10^{-18}J$

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

$\Delta E=h\nu(1)$

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

$\nu=\frac{c}{\lambda}(2)$

Here c is the speed of light. Replacing (2) in (1):

$\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J$

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3 years ago