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Semenov [28]
3 years ago
6

Pls help with science!!!!!​

Physics
1 answer:
Andrews [41]3 years ago
4 0

Answer:

Your answer is C~

Explanation:

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Explain the change of frequency of the wave if the tension of the string is increased​
jek_recluse [69]

Answer:

Increasing the tension on a string increases the speed of a wave, which increases the frequency (for a given length). Pressing the finger at different places changes the length of string, which changes the wavelength of standing wave, affecting the frequency.

Explanation:

8 0
2 years ago
The gamma ray released by each decay carries 140kev of energy. find the total energy e released by decays in the 2 hours.
olya-2409 [2.1K]
First, we would need to know the decaying isotope.
Next, we use the decay formula
A = Ao e^(-kt)
After determining the remaining amount after two hours, the decay reaction can be used to determine the number of gamma rays released. If the given is in terms of mole, then the total energy is
E = 140n KeV where n is the number of moles of gamma rays released
8 0
3 years ago
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An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0
3 years ago
Which planet is smaller than EarthÍs core?
Juliette [100K]
The Earth's core is estimated to be bigger than both Mars and Mercury. However, I would say that Mercury is the safest choice. Mercury is also known to be smaller than Jupiter's Moon!
3 0
3 years ago
Can someone help me please
nlexa [21]

Answer:

b

Explanation:

8 0
2 years ago
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