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3 years ago

I WILL GIVE BRAINLIEST Which of the following statements is true with regard to transverse and longitudinal waves?

1 answer:

8 0

Answer: Transverse waves have motion perpendicular to velocity, while longitudinal waves have motion parallel to velocity.

Explanation:

Transverse waves are characterized by the fact that the particles of the medium in which they propagate move transversely to the direction of propagation of the wave.

In other words,<u> its displacement is perpendicular to the direction of propagation of the wave</u>, being a good example the circular waves in the water.

On the other hand, Longitudinal waves are characterized by the fact that <u>the oscillation of the particles in the medium is parallel to the direction of propagation of the wave.</u> A good example of this is the sound wave.

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The small metal cylinder has a mass of 0.20 kgkg, the coefficient of static friction between the cylinder and the turntable is 0

deff fn [24]

Answer:

velocity of the metal cylinder = 0.343 m/s

Explanation:

Force putting the metal cylinder is given by

F = mv²/r

But this force will balance the frictional force between the metal cylinder and the turntable

The frictional force is given by

μN = μ × mg = 0.08 × 0.2 × 9.81 = 0.15696 N

r = 0.15 m, m = 0.2 kg,

F = mv²/r = 0.2 v²/(0.15) = 1.3333 v²

1.3333 v² = 0.15696

v² = 0.117

v = 0.343 m/s

8 0

3 years ago

The driver of a 2.0 × 10³ kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car

nignag [31]

Answer:

Explanation:

a = F/m = 7500/2000 = 3.75 m/s²

v² = u² + 2as

s = (v² - u²) / 2a

s = (0² - 45²) / (2(-3.75))

s = 270 m

6 0

2 years ago

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ = 400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction, μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

$v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f$

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

$S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2$

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0

3 years ago

Read 2 more answers

Find the electron and hole mobilities, and the resistivity of intrinsic silicon at 300K. Is intrinsic silicon a semiconductor

tino4ka555 [31]

Answer:

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

Explanation:

Solution:

At 300K:

Let suppose mobility of electron in intrinsic semiconductor = $M_{e}$

Mobility of electron in intrinsic semiconductor is:

$M_{e}$ = 1300 $cm^{2}$ /volt.sec

Let suppose mobility of hole in intrinsic semiconductor = $M_{h}$

$M_{h}$ = 500 $cm^{2}$ /volt.sec

We know that, intrinsic silicon semiconductor has equal number of holes and electrons. So,

At 300 K

Intrinsic Carrier Concentration = 1.5 x $10^{10}$ / $cm^{3}$ = C

And,

Conductivity of intrinsic Silicon is:

σ = C x ( $M_{h}$ + $M_{e}$ ) e

e = 1.6 x $10^{-19}$ C

So, plugging in the values, we get:

σ = C x ( $M_{h}$ + $M_{e}$ ) e

σ = 1.5 x $10^{10}$ x (500 + 1300) x 1.6 x $10^{-19}$

σ = 4.32 x $10^{-6}$

So, now we can find the resistivity.

Resistivity = 1/σ

Resistivity = 1/ 4.32 x $10^{-6}$

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

7 0

2 years ago

(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat

vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have

∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore

ñ = ⅟4(2·i +3·j -√3·k)

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0

3 years ago