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Artist 52 [7]
2 years ago
14

Suppose we have a radar dish that generates a strong signal that travels out to hit an asteroid 10^9 kilometres away.

Physics
1 answer:
tigry1 [53]2 years ago
4 0

Answer:

me when the c um won't stop

Explanation:

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If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the
agasfer [191]

Answer:

Output power of the circuit is 3 Watt.

Given:

loss in decibles = 3 dB

Input power = 6 Watt

To find:

Output power = ?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss = 0.5 ratio

Output power is given by,

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watt

Thus, output power of the circuit is 3 Watt.

4 0
2 years ago
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Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain
mel-nik [20]
The surrounding air will become warm when water vapor condenses. The vapors when become water will give away latent heat they have, we know that latent heat is required for the object to change states, so, the latent heat the water vapor had when it became water vapor from water will be given out when it again becomes water.
8 0
2 years ago
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
2 years ago
A nervous squirrel gets startled and runs 5.0\,\text m5.0m5, point, 0, space, m leftward to a nearby tree. The squirrel runs for
mel-nik [20]

Answer:

−5.0  

Explanation:

6 0
2 years ago
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What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
2 years ago
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