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RideAnS [48]
3 years ago
5

please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
\frac{total distance}{ total time}

\frac{72m}{36s} = 2m/s

Same thing for the rest 
 2) 5km/h
 3)360km/h

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The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti
ivanzaharov [21]

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So   h * f = KE + Wf = 1.2 + 1.88 = 3.08    incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

7 0
3 years ago
A diagram of a closed circuit with power source on the left labeled 12 V, a resistor on the top labeled 10 Ohms, a resistor on t
zalisa [80]

Answer:

7.5 A

6.0 A

5.0 A

Explanation:

6 0
3 years ago
Read 2 more answers
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
11. A sensitive thermometer is one which
Marta_Voda [28]

Answer:

A. ls sensitive to heat

this is correct answer

I hope it's helpful for you.......

4 0
3 years ago
A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj
Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

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Vx = V cos 35 = 24.6 m / sec     horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0
2 years ago
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