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Ann
1 day ago
8

Priscilla is driving her car on a busy street and Harvey passes her on his motorcycle. What will happen to the sound from his mo

torcycle after it
passes her car?
O A.
The frequency of the sound will decrease.
OB. The amplitude of the sound will increase.
0
O C. The intensity of the sound will increase.
OD. The wavelength of the sound will decrease.
0
O E.
The velocity of the sound will increase.​
Physics
2 answers:
kicyunya [3.1K]1 day ago
8 0
d
Sav [3K]1 day ago
4 0
E - the sound's velocity will increase.
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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
Ostrovityanka [3070]

Response:

\text{heat loss} = 24864.05 \ W/m^2

Clarification:

If

  • T_1, T_2 represent the temperatures of gases and liquids in Kelvins,
  • t_1 and t_2 denote the thicknesses of the gas layer and steel slab in meters,
  • h_1, h_2 are the convection coefficients for gas and liquid in W/m^2 \cdot K,
  • R_c represents the contact resistance in m^2 \cdot K/W,
  • and k_1, k_2 signify thermal conductivities of gas and steel in W/m \cdot K,

then: part(a):

\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

by employing known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Utilizing the rate equation:

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature is T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Correspondingly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature profile is depicted in the image provided

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29 days ago
A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a
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The sound wave intensity at the sphere's surface is described as follows: B = Bulk modulus. The oscillation amplitude of the sphere can be represented as: Substitute velocity and amplitude into Pmax. The intensity of the sound wave at a distance is determined by:
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7 days ago
A football game begins by flipping a penny to decide which team will get the ball first. The 5.25 g penny has a speed of 3.27 m/
ValentinkaMS [3354]
To start, we first need to determine the kinetic energy of the penny before it strikes the ground. This is calculated using the formula where m equals 5.25 g, which is 0.00525 kg for the penny's mass, and v equals 3.27 m/s for its speed. Replacing the values into the equation provides: When the penny lands, all this kinetic energy transforms into internal energy for both the penny and the ground. If half of this energy goes into the penny's internal energy, the change is determined by a specific formula where m is the penny's mass, Cs is its specific heat capacity (2.03 J/gC), and \Delta T, the change in temperature. To find the last element, the equation will be solved.
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13 days ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
Keith_Richards [3146]

Case A:

A.75 kg 65 N/m 1.2 m

m = weight of the car = 0.75 kg

k = spring's stiffness = 65 N/m

h = elevation of the hill = 1.2 m

x = spring's compression = 0.25 m

Applying the principle of energy conservation from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill equals Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B:

B.60 kg 35 N/m.9 m

m = weight of the car = 0.60 kg

k = spring's stiffness = 35 N/m

h = height of the hill = 0.9 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Top hill energy = Bottom hill energy

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C:

C.55 kg 40 N/m 1.1 m

m = weight of the car = 0.55 kg

k = spring's stiffness = 40 N/m

h = height of the hill = 1.1 m

x = spring's compression = 0.25 m

Using conservation of energy from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill = Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D:

D.84 kg 32 N/m.95 m

m = weight of the car = 0.84 kg

k = spring's stiffness = 32 N/m

h = height of the hill = 0.95 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


thus, the closest result is from case C at 5.1 m/s




7 0
1 month ago
Read 2 more answers
15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If
Keith_Richards [3146]

Answer:

Explanation:

Considering that,

The mass of the first vehicle

M1= 328kg

It is traveling in the positive x direction at a speed of

U1 = 19.1m/s

The speed of the second vehicle

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The mass of the second vehicle

M2 = 790kg

The speed of the second vehicle post-collision

V2 = 15.1 m/s

The speed of the first vehicle following the collision

V1 =?

This represents an elastic collision,

and applying the principle of conservation of momentum

The momentum prior to the collision must equal the momentum afterwards

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The speed of the first vehicle after the collision is 14.04 m/s

5 0
7 days ago
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