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3 years ago

How do you find the rest mass (kg) of a 3.1 eV electron?

Physics

1 answer:

Scilla [17]3 years ago

3 0

Answer:

Explanation:

The rest energy of any substance is defined by the Einstein's mass energy equivalence relation. Thus the rest mass of a electron is 9.11x10^-31 kg. The speed of light is 299,792,458 m/s. Thus multiplying the square of speed of light with the rest mass of electron gives the rest energy of the electron.

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1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?

Anettt [7]

The man is holding the suitcase at the same height above the surface of earth. So the gravitation potential energy remains the same.

<span>work done is force * displacement = weight * 0 = 0</span>

7 0

3 years ago

9. During an egg toss, a catcher must cushion the egg by maximizing the time it takes to stop the

Lorico [155]

Answer:

the impulse experienced by the egg is 0.053 kgm/s.

Explanation:

Given;

mass of the egg, m = 60 g = 0.06 kg

initial velocity of the egg, u = 6 m/s

height moved by the egg, h = 50 cm = 0.5 m

Determine the final velocity of the egg as it moves upward;

v² = u² + 2(-g)h

v² = u² - 2gh

where;

v is the final velocity

-g is negative acceleration due gravity as it moves upward

v² = 6² - 2(9.8 x 0.5)

v² = 26.2

v = √26.2

v = 5.12 m/s

The impulse applied to the egg is the change in linear momentum;

J = ΔP

ΔP = mu - mv

ΔP = m(u - v)

ΔP = 0.06(6 - 5.12)

ΔP = 0.053 kgm/s

Therefore, the impulse experienced by the egg is 0.053 kgm/s.

8 0

3 years ago

What is kinetic friction?

Lunna [17]

Answer:

Kinetic friction is an inhibitory force that is present when a body or object begins to move. It should be noted that this force is in the opposite direction to the movement of the body that slides with respect to a surface and parallel to that surface.

In addition, this friction varies according to the surface characteristics and the material properties of the bodies that slide in this surface. Being quantified by the coefficient of kinetic friction $\mu_{K}$ , which is dimensionless.

4 0

4 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago

In which situation does the energy constantly change from gravitational potential energy to kinetic energy?

Anarel [89]

when you drop a ball and let it fall to the ground

4 0

3 years ago