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AysviL [449]
3 years ago
13

Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?

Physics
1 answer:
KengaRu [80]3 years ago
7 0

Answer:

incomplete question, resistor must be there

Explanation:

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Answer:

no it is not C they are tying to get point the answer is, B: In codominance and incomplete dominance, neither of the alleles are dominant

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2 years ago
3. What is the relationship among wave speed, wavelength, and frequency?
Rus_ich [418]

Answer:

Wave speed is related to both wavelength and wave frequency. Wavelength is the distance between two corresponding points on adjacent waves. Wave frequency is the number of waves that pass a fixed point in a given amount of time. This equation shows how the three factors are related:

Speed = Wavelength x Wave Frequency

Explanation:

4 0
2 years ago
A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for
mash [69]

Answer:

Only a backward force is acting, no forward force.

Explanation:

  • Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
  • As it would move at a constant speed, no net force would be acting on it.
  • So, if it is gradually slowing, there must  be a net force producing an acceleration in a direction opposite to the movement.
  • This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
  • As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
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8 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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2 years ago
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I say it helped then because TrueType had room
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