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3 years ago

plz i need help fast................a motion along a straight line with a constant increase in velocity is ?

Physics

2 answers:

Monica [59]3 years ago

8 0

Answer:

uniform acceleration

Explanation:

The definition for uniform acceleration is:

if an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration is said to be uniform.

Hope this helps.

Good Luck

Effectus [21]3 years ago

5 0

Answer:

Uniform Accleration.

Is the answer.

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A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period o

Jobisdone [24]

Answer:

Explanation:

The amplitude is extraneous information for this question.

Period (T) is simply a measure of time (how long it takes to complete a full cycle).

The frequency (f), is the reciprocal (how many cycles it can complete per unit of time).

The frequency reported is 12Hz. Hz is a unit meaning "cycles per second"

Thus 12cycles = 1second.

To find the number of seconds per cycle, simply create the other unit fraction and simplify:

$T=\frac{1second}{12cycles}\\T=0.8\bar{3}\frac{seconds}{cycle}\\$

$T=0.83seconds$

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1 year ago

A car accelerates at a rate of 3 m/s^2. If it's original speed is 8 m/s, how many seconds will it take the car to reach a final

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2 years ago

The star Betelgeuse is about 600 light-years away. If it explodes tonight,

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Answer:

Explanation:

It has to travel 600 light years before we would be able to observe the explosion.

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3 years ago

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Yuri [45]

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

= $4.59\times 10^{-19} \ J$

or,

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

= $2.87 \ eV$

(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

2 years ago

plz i need help fast................a motion along a straight line with a constant increase in velocity is ?​

plz i need help fast................a motion along a straight line with a constant increase in velocity is ?