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2 years ago

14

Which year was pluto no longer considered a planet?.

1 answer:

katrin2010 [14]2 years ago

6 0

2006, i hope this helps

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a. As you coast down a hill on your bicycle, you accelerate at 0.5 m/s2. If the total mass of your body and the bicycle is 80 ki

zalisa [80]

According to Newton’s second law of motion:
F=ma
=80*0.5
=40N

7 0

3 years ago

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

If 6.24×10¹⁸ electron pass through a wire in 1s how many pass through it during a time interval of 2hr, 47min and 10s

levacccp [35]

Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)

3 0

3 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

3 years ago

The greater the amplitude of a wave a. The quieter the sound b. The higher the frequency of the sound c. The dimmer the light in

Stolb23 [73]

Answer:

The greater the amplitude the greater the energy.

(Think of a water wave - which carries greater energy a 1 ft wave or

a 10 ft wave)

8 0

3 years ago