How do you convert 3.5 * 10^4 to standard form

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out

blondinia [14]

Answer:

$\theta = 20.98 degree$

Explanation:

As we know that the speed of the sound is given as

$v = 332 + 0.6 t$

now at t = 273 k = 0 degree

$v = 332 m/s$

so we have

$a sin\theta = N\lambda$

$a sin\theta = N(\frac{v_1}{f})$

now when temperature is changed to 313 K we have

$t = 313 - 273 = 40 degree$

now we have

$v = 332 + (0.6)(40)$

$v_2 = 356 m/s$

$a sin\theta' = N(\frac{v_2}{f})$

now from two equations we have

$\frac{sin19.5}{sin\theta} = \frac{332}{356}$

so we have

$sin\theta = 0.358$

$\theta = 20.98 degree$

7 0

3 years ago

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andreev551 [17]

Answer:

Polarization occurs when an electric field distorts the negative cloud of electrons around positive atomic nuclei in a direction opposite the field. Polarization P in its quantitative meaning is the amount of dipole moment p per unit volume V of a polarized material, P = p/V.

Explanation:

8 0

3 years ago

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A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t

nexus9112 [7]

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m$

So, the hayride will cover a distance of 68.18 m.

6 0

4 years ago

A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t

HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0

4 years ago

A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping. His resultant displacement was

Serhud [2]

<u>Answer:</u>

Resultant displacement is 4 km north.

<u>Explanation:</u>

Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping,

So initial displacement, 15 kilometers north = 15 j km

Second displacement, 11 kilometers south = -11 j km

Total displacement = 15 j -11 j = 4 j km

Total displacement is 4 km north.

4 0

4 years ago

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