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2 years ago

Which term is defined as the force applied to the object divided by the mass of the object?

Physics

1 answer:

Leokris [45]2 years ago

7 0

Answer:

acceleration

Explanation:

From the question: "FORCE... divided by the MASS..." is translated to a math equation as "F / m", where "F" is the force and "m" is the mass.

Now, the second Newton's law says "F = m . a ", where a is acceleration.

If you move m to the left side, clearly you have "F / m", and "a" is alone to the right. Math equation:

a = F / m

acceleration = force / mass

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To what temperature must you raise a silver wire (c = 0.0038), originally at 20.0°C, to double its resistance, neglecting any ch

balu736 [363]

Answer:

$T=283^{\circ}C$

Explanation:

Given a material with temperature coefficient of resistance <em>c</em>, the equation that relates the resistance $R_0$ at temperature $T_0$ and the resistance $R$ at temperature $T$ is

$\frac{R-R_0}{R_0}=c(T-T_0)$

We want to double our resistance, so $R=2R_0$ , thus having:

$\frac{2R_0-R_0}{R_0}=\frac{R_0}{R_0}=1=c(T-T_0)$

For this T must be:

$1=cT-cT_0$

$T=\frac{1+cT_0}{c}$

which for our values means (with $T=20^{\circ}C=293^{\circ}K$ , remember to write temperature in S.I., and that for silver $c=0.0038^{\circ}K^{-1}$ ):

$T=\frac{1+(0.0038^{\circ}K^{-1})(293^{\circ}K)}{(0.0038^{\circ}K^{-1})}=556^{\circ}K=283^{\circ}C$

3 0

2 years ago

Farah claps his 600times in 60seconds. what is the friquency and periodb

rosijanka [135]

Answer:

$\huge\boxed{\sf Frequency = 10\ Hz}$

$\huge\boxed{\sf Time\ Period = 0.1\ secs}$

Explanation:

<u>Given:</u>

No. of times = 600 times

Time = 60 seconds

<u>Required:</u>

Frequency = f = ?

Time Period = T = ?

<u>Solution:</u>

<h2>Frequency = No. of Times / Time</h2>

Frequency = 600 / 60

<u>Frequency = 10 Hz</u>

<h2>Time Period = 1 / Frequency</h2>

Time Period = 1 / 10

<u>Time Period = 0.1 secs</u>

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3 years ago

A proton accelerates from rest in a uniform electric field of 680 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativis

Ksivusya [100]

Answer:

Acceleration, $a=6.51\times 10^{10}\ m/s^2$

Explanation:

Given that,

Electric field, E = 680 N/C

Speed of the proton, v = 1.3 Mm/s

We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :

$qE=ma$

a and m are the acceleration and mass of the proton.

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}$

$a=6.51\times 10^{10}\ m/s^2$

So, the acceleration of the proton is $a=6.51\times 10^{10}\ m/s^2$ . Hence, this is the required solution.

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