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Tema [17]
3 years ago
12

6. The temperature at which water molecules have the

Physics
1 answer:
nikklg [1K]3 years ago
3 0
The answer is a which is zero
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Difference between hair dryer and heat gun
Bumek [7]
The heat gun<span> obviously wins this round. Master Appliance </span>heat guns<span> can reach temperatures of up to 1,000 Fahrenheit. A handheld </span>blow dryer<span> might reach 131 degrees Fahrenheit. A </span>hair dryer<span> gets hot enough to burn skin, but not hot enough to complete serious tasks like striping paint and removing serious. By the way I got this from google.</span>
5 0
3 years ago
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen
bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

f_{a} =(\frac{v+v_{c} }{v} )f_{s} (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s} (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s

b) Substitute eq. 1 in eq. 2:

f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz

7 0
3 years ago
The number of protons in the nucleus is also the what?
Natasha_Volkova [10]
The number of protons in the nucleus is also called the Atomic Number
4 0
3 years ago
Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a r
garri49 [273]

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

8 0
3 years ago
Find the following answer based on the image.
saul85 [17]

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷)                      [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

5 0
3 years ago
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