What is the acceleration of a skier that goes from 2.50 m/s to 14.5 m/s while traveling 505 m down a slope?

When 224-nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.84

OLga [1]

Answer:

3.71 eV

Explanation:

λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m

c = speed of electromagnetic wave = 3 x 10⁸ m/s

V₀ = stopping potential = 1.84 volts

W₀ = Work function of the metal = ?

Using the equation

$\frac{hc}{\lambda } = eV_{o} + W_{o}$

$\frac{(6.63\times 10^{-34})(3\times 10^{8})}{224\times 10^{-9} } = (1.6\times 10^{-19})(1.84) + W_{o}$

$W_{o}$ = 5.94 x 10⁻¹⁹

$W_{o}$ = 3.71 eV

7 0

3 years ago

HELLO ONCE MORE FRIENDS . PLEASE HELP ME.

tatuchka [14]

Answer:

Do u want to talk?

Explanation:

5 0

3 years ago

Read 2 more answers

In order for work to happen you MUST have?

Nataly_w [17]

Explanation:

it requires energy

hope it helps

7 0

3 years ago

You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Obje

Over [174]

Answer: a. m = 7.7 kg

b. V = 435.52 in³

c. m = 1927 kg

d. V = 335.37 cm³

e. m = 3 kg

Explanation: <u>Density</u> is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

$\rho=\frac{m}{V}$

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

$\rho=\frac{m}{V}$

$m=\rho.V$

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

$m=\rho.V$

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as $V=\frac{4}{3}.\pi.r^{3}$

Diameter is twice the radius, then r = 4.31 cm.

Volume is

$V=\frac{4}{3}.\pi.(4.31)^{3}$

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. $m=\rho.V$

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

4 0

3 years ago

A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge

Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π× $10^{-7}$ ×0.024 / 2π(1.20× $10^{-2}$ ) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 × $10^{-7}$ × 1.09861 × 120

current = 52.73 × $10^{-6}$ A

so here current in loops is 52.73 μA

8 0

3 years ago