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3 years ago

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A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates at 1.5m/s2? A. 51.5 N B. 4

8.5 N C. 75 N D. 33.3 N

1 answer:

const2013 [10]3 years ago

3 0

Use the Second Law of Newton, which states that Net Force = mass times acceleration

F = m * a

F = 50 kg * 1.5 m/s^2 = 75 N

Answer: option C

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A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration

kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

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2 years ago

32. (FR) A mass moving at 25 m/s slides along a rough horizontal surface. The coefficient of friction is 0.30. (A) Use force and

scoray [572]

Answer:

A)s = 104.16 m

b)s= 104.16 m

Explanation:

Given that

u = 25 m/s

μ = 0.3

The friction force will act opposite to the direction of motion.

Fr= μ m g

Fr= - m a

a=acceleration

μ m g = - m a

a= - μ g

a= - 0.3 x 10 m/s² ( take g= 10 m/s²)

a= - 3 m/s²

The final speed of the mass is zero ,v= 0

We know that

v² = u² +2 a s

s=distance

0² = 25² - 2 x 3 x s

625 = 6 s

s = 104.16 m

By using energy conservation

Work done by all the forces =Change in the kinetic energy

$- Fr.s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Negative sign because force act opposite to the displacement.

$- \mu\ m\ g \ s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$-\mu\ g \ s=\dfrac{1}{2}v^2-\dfrac{1}{2}u^2$

$-0.3\times 10\times \ s=\dfrac{1}{2}\times 0^2-\dfrac{1}{2}\times 25^2$

- 3 x 2 x s = - 625

s= 104.16 m

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2 years ago

Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.

MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>

It is given that, the infinite well having a width of 0.050 mm.
We have the expression for energy of electron in an infinite well as,

$E_n=\frac{n^2h^2}{8mL^2}$

where;

$m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1$

Thus, the lowest energy of electron in an infinite well is,

$E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J$

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

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9 months ago

For a beam of light, the direction of polarization is defined as

evablogger [386]

<span>The direction of the electric field's vibration</span>

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3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago