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erastovalidia [21]
3 years ago
11

Is a light on potential or kinetic?

Physics
1 answer:
Basile [38]3 years ago
7 0

its the the first one u said

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Reading glasses use what property of light waves
tresset_1 [31]
Visible range of electromagnetic radiation
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3 years ago
Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins aroun
Juliette [100K]

Answer:

a.) L = 2.64 kgm^2/s

b.) V = 4.4 m/s

Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.

So,

Radius r = 0.6 m

Mass M = 2 kg

Velocity V = 1.1 m/s

Angular momentum L can be expressed as;

L = MVr

Substitute all the parameters into the formula

L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1

the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1

b. If she pulls her arms into 0.15 m,

New radius = 0.15 m

Using the same formula again

L = 2( MVr)

2.64 = 2( 2 × V × 0.15 )

1.32 = 0.3 V

V = 1.32/0.3

V = 4.4 m/s

Her new linear speed will be 4.4 m/s

4 0
3 years ago
How does the law of conservation of energy apply to machines?
allsm [11]
Energy can be changed from one form to another, but it cannot be created or destroyed. ... This principle is referred to as the first law of thermodynamics or the law of energy conservation. The law applies to all systems both large and small, and, again, it states that energy cannot be created or destroyed.
4 0
3 years ago
Read 2 more answers
A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to
nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket V=10\ gallon

Time taken to fill the bucket t=50\ s

so volume flow rate is \dot{V}=\frac{10}{50}=0.2\ gal/s

1 gal is equivalent to 0.133\ ft^3

\dot{V}=0.0267\ ft^3/s

mass flow rate \dot{m}=\rho \times \dot{V}

\dot{m}=62.4\times 0.0267

\dot{m}=1.668\ lbs

(b)Average velocity through nozzle exit

\dot{V}=Av_{avg}

v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}

v_{avg}=49.51\ ft/s

8 0
4 years ago
An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of
Aloiza [94]

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

From the question, we know that

435 nm is given as the wavelength of the wave, at the same time, we also know that the amplitude of the electric field, E(max) has been given to be 2.7*10^-3 V/m

a)

To find the frequency of the wave, we would be applying this formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

b) again, to find the amplitude of the magnetic field, we would use this relation

E(max) = B(max) * c, magnetic field amplitude, B(max) =

B(max) = E(max)/c

B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

c) and lastly,

1T = 1 (V.s/m^2)

6 0
3 years ago
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