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algol [13]
3 years ago
11

In a water park, people walk a distance of 20 meters up an inclined ramp before getting on a water slide. If the spot from where

they start sliding down is located 7 meters above the water’s surface, calculate the angle of the ramp with the horizontal.
Physics
2 answers:
Elenna [48]3 years ago
8 0
Thank you for posting your Physics question here. I hope the answer helps.  Upon calculating the ramp with the horizontal the answer is 20.49 Deg. Below is the solution:

Y = 7 m. 
<span>r = 20 m. </span>

<span>sinA = Y/r = 7/20 = 0.35. </span>
<span>A = 20.49 Deg.</span>
Elena L [17]3 years ago
7 0

Answer:

E.  20°

Explanation:

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When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What ma
Lina20 [59]

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm

3 0
3 years ago
Three forces act on a statue. Force F⃗ 1F→1 (magnitude 45.0 NN) points in the +x-direction, Force F⃗ 2F→2 (magnitude 105 NN) poi
Trava [24]

Answer:

Resultant force = (232.93î + 246.10j) N

x-component of the resultant force = (+232.93î) N

y-component of the resultant force = (+246.1j) N

Explanation:

The net external force on the statue is equal to the resultant force on the statue.

And the resuphant force is a vector sum of all the other forces acting on the statue.

Force 1 = (45î) N

Force 2 = (105j) N

Force 3 = (235cos 36.9°)î + (235 sin 36.9°)j = (187.93î + 141.10j) N

Resultant force = (Force 1) + (Force 2) + (Force 3)

Resultant force = 45î + 105j + (187.93î + 141.10j) = (232.93î + 246.10j) N

Hope this helps!!!

5 0
3 years ago
A wire in a circuit carries a current of 0.9 A.
ioda

Answer:

45coulombs

Explanation:

Using your equation current=0.9 & time=50secs multiply and your answer is 45. Hope the answer is good enough for u

7 0
3 years ago
Two vehicles of mass M and 2M are moving in the same direction on a highway. Both drivers apply their brakes at the same time an
JulsSmile [24]

Answer:

The mass is inversely proportional to the acceleration so the acceleration a1 is twice that acceleration a2

a_{1},a_{2}=\frac{a_{1}}{2}

Explanation:

The force of friction and the kinetic force make the law of mass in moving so

F=m*a

F=f_{m}+f_{k}

f_{k1}=u*m1*g

f_{k2}=u*2*m1*g

f_{m}=\frac{1}{2}*m1*v^2

The forces are the same however at the moment to determinate the acceleration

a_{1}=\frac{F}{m}

a_{1}=\frac{F}{2*m}

\frac{F}{m} are constant because they make the same motion however the difference of mass make the acceleration difference

3 0
3 years ago
Roemer's experiment failed because of his ignorance about the what
miskamm [114]
There are no choices available, however through my research this could be the choices for your question: 

a) the speed of light
b) the dimensions of the earth´s orbit
c) the number of Saturn´s moons
<span>d) the use of the telescope
</span>
The most probable answer would be his miscalculations about the dimensions of the Earth's orbit which is why his experiment ultimately failed.
3 0
3 years ago
Read 2 more answers
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