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77julia77 [94]
3 years ago
7

Would anyone pls help me on these questions Thanks

Physics
2 answers:
Karolina [17]3 years ago
5 0
Number 1 is 0, number 2 is moving, number 3 is Joules, number 4 is a thermometer, number 5 is gas, number 6 is more, number 7 is specific heat. Hope this was helpful
Alenkinab [10]3 years ago
3 0
1. 0°C
2. Motion
3. Joules
4. Thermometer
5. Gas
6. Less
7. Temperature
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Which wave has a greater frequency?
DedPeter [7]
The wave diagramed in blue.
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2 years ago
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If a transformer has a primary with 100 V and fifty coils, and a secondary that yields 20 V, how many coils are on the secondary
melomori [17]
Ideally, if all the magnetic of one winding cuts the other winding, and there isn't any loss in the transformer core or the resistance of the wire, then the voltage across each winding is proportional to the number of turns in its coil.

If you apply 100 V to a winding of 50 turns, then a winding that yields 20 volts
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7 0
3 years ago
On the weather map, what does the symbol shown below represent?
Evgen [1.6K]

Answer:

wind speed i think where i live?

8 0
3 years ago
When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amp
Paraphin [41]

Answer:

796.18 Hz

Explanation:

Applying,

Maximum velocity = Amplitude×Angular velocity

Therefore,

V' = A(2πf)............... Equation 1

Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie

make f the subject of the equation

f = V'/2πA................ Equation 2

From the question,

Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,

Constant: 3.14.

Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

6 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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