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Sliva [168]
3 years ago
5

To increase power, it is necessary to

Physics
2 answers:
gavmur [86]3 years ago
5 0

Answer:

Increase the work being done or decrease the time in which the work is being completed.

Explanation:

yuradex [85]3 years ago
3 0
In getting the power in physics, there are factors to be considered, the work and the time. The formula is Power = work / time. If we base on this formula, I think the answer to this problem is letter B. There will be an increase of power if your time will decrease and your work will increase. Much force is exerted during the process to finish a work in a short period of time. 
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What rock type would most likely be found in the location marked by the black rectangle in the diagram below?
pychu [463]

There is no diagram below so I can't answer the question

5 0
3 years ago
Read 2 more answers
Three charges are on a line. A positive charge is on the far left labeled q Subscript 1 baseline positive 6 Coulombs. The second
tiny-mole [99]

Answer:

–1.1 × 10^11 N

Explanation:

I did the assignment

4 0
3 years ago
the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes
slamgirl [31]

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

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5 0
3 years ago
A moving object must have which type of energy
xxMikexx [17]

Answer:

Kinetic Energy

Explanation:

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7 0
4 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

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or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

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(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
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