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FinnZ [79.3K]
3 years ago
15

A quantity y is to be determined from the equation y=(px)/q^2

Physics
1 answer:
ki77a [65]3 years ago
4 0

Answer:

heya answer option b

Explanation:

please mark me brainliest

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Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can
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A boy and his skateboard have a combined mass of 65 kg. What is the speed of the boy and skateboard if they have a momentum of 2
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Speed (s) = momentum (M) : mass (m)

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What is the net charge on a sphere that has the following? (a) 5.87 106 electrons and 8.11 106 protons C
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Answer:

3.6 * 10^-13 C

Explanation:

The net charge of the sphere will be the sum of the total electron charge and total proton charge. Mathematically,

Q = Qe + Qp

TOTAL ELECTRON CHARGE:

An electron has an electronic charge of -1.6022 * 10^-19C.

Hence, the charge of 5.87 * 10^6 electrons will be:

Qe = - 1.6022 * 10^-19 * 5.87 * 10^6

Qe = - 9.405 * 10^-13 C

TOTAL PROTON CHARGE:

A proton has an electronic charge of 1.6022 * 10^-19. Hence, 8.11 * 10^6 protons will have:

Qp = 1.6022 * 10^-19 * 8.11 * 10^6

Qp = 1.3 * 10^-12 C

=> Q = (-9.405 * 10^-13) + (1.3 * 10^-12)

Q = 3.6 * 10^-13 C

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3 years ago
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Lasers emit light of a certain frequency in one, precise direction. The light that a laser emits can be tuned to have a high fre
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Answer:

An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.

Explanation:

This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.

Currently, the most powerful lasers reach intensities of the order of 1021W cm-2 and powers of Petawatts, PW, in each pulse. This range of intensities has opened the door for lasers to a multitude of disciplines and scientific areas traditionally reserved for accelerators and nuclear reactors, applying as generators of high-energy electron, ion, neutron and photon beams, without the need for expensive infrastructure.

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Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

F = 1.68 N

Finally, the reading scale is:

F_{scale = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

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