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3 years ago

I need help thanks :)))))))))

Physics

1 answer:

dsp733 years ago

6 0

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end. It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one. It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

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Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs

Anestetic [448]

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

Lines per centimetre = 7000 /cm = 700000 /m

Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

λ is the wavelength of the incident light

d is the distance separating the centres of 2 adjacent slits

θ is the angle at which diffraction occurs

From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

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You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

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Answer:

Explanation:KE = 18 J

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An engineer is designing a runway for a witch. Several brooms will use the runway and the engineer must design it so that it is

jonny [76]

Answer:

1170 m

Explanation:

Given:

a = 3.30 m/s²

v₀ = 0 m/s

v = 88.0 m/s

x₀ = 0 m

Find:

v² = v₀² + 2a(x - x₀)

(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)

x = 1173.33 m

Rounded to 3 sig-figs, the runway must be at least 1170 meters long.

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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

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Answer:

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Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

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