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3 years ago

What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

2 answers:

4 0

Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
= (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
= (9.1×10∧-31 × 1,153,846.154∧2) ÷2
= (1.21154×10∧-18) ÷ 2
= 6.05769×10∧-19 J

Tomtit [17]3 years ago

3 0

Answer:

K = 6.02 × 10⁻¹⁹ J

Explanation:

The momentum (p) of an electron is its mass (m) times its speed (v).

p = m × v

v = p / m = (1.05 × 10⁻²⁴ kg.m/s) / 9.11 × 10⁻³¹ kg = 1.15 × 10⁶ m/s

We can find the kinetic energy (K) using the following expression.

K = 1/2 × m × v²

K = 1/2 × 9.11 × 10⁻³¹ kg × (1.15 × 10⁶ m/s)²

K = 6.02 × 10⁻¹⁹ J

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Svetlanka [38]

60kg would be the answer

8 0

2 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

stira [4]

Answer:

See explanation

Explanation:

We have a mass $m$ revolving around an axis with an angular speed $\omega$ , the distance from the axis is $r$ . We are given:

$\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]$

and also the formula which states that the kinetic rotational energy of a body is:

$K =\frac{1}{2}I\omega^2$ .

Now we use the kinetic energy formula

$K =\frac{1}{2}mv^2$

where $v$ is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

$v=\omega r$

After replacing in the previous equation we get:

$K =\frac{1}{2}m(\omega r)^2$

now we have the following:

$K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2$

therefore:

$mr^2=I$

then the moment of inertia will be:

$I = 13*(0.5)^2=3.25 [Kg*m^2]$

3 0

3 years ago

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

$\frac{v_o}{r_o} = \frac{v_1}{r_1}$

v1 = $\frac{r_1}{r_o} \ \ v_o$

let's calculate

v₁ = $\frac{1.50}{1.00} \ \ 25.0$

v₁ = 37.5 cm / s

4 0

3 years ago

A. How long does it take light to travel through a 3.0-mm-thick piece of window glass?

hodyreva [135]

Answer:

a) $1.517\times10^{-11}$ s

b) 3.41 mm

Explanation:

We take the speed of light, c = $3.0\times10^8$ m/s and the refractive index of glass as 1.517.

Speed = distance/time

Time = distance/speed

Refractive index, n = speed of light in vacuum / speed of light in medium

$n=\dfrac{c}{s}$

$s=\dfrac{c}{n}$

$t=\dfrac{d}{c/n}$

$t=\dfrac{dn}{c}$

$t=\dfrac{3\times10^{-3}\times1.517}{3.0\times10^8}$

$t=1.517\times10^{-11}$

We take the refractive index of water as 1.333.

Speed in water = speed in vacuum / refractive index of water

Distance = speed * time

$d=s\times t$

$d=\dfrac{c}{n_w}\times \dfrac{3\times10^{-3}\times1.517}{c}$

$d=\dfrac{3\times10^{-3}\times 1.517}{1.333}$

d = 3.41 mm

6 0

3 years ago

What are the size and direction of the force that is the sum of a force of 3 n acting south and a force of 6 n acting north?

V125BC [204]

The resultant force is 3 N north.

4 0

3 years ago