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Whitepunk [10]
3 years ago
12

What caused day and night on a planet

Physics
1 answer:
Volgvan3 years ago
7 0

Answer:

The rotation of a planet around it's sun

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A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2
Shtirlitz [24]
99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
 = (0²-27.5²)/(2x50.0)
 =-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
4 0
3 years ago
What location on earth experiences the most change the number of daylight hours?
CaHeK987 [17]
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3 0
3 years ago
Which statement describes Redi’s experiment, which helped disprove spontaneous generation?
Kay [80]

Answer:

He examined covered and uncovered meat to determine that maggots came from eggs.

In other other words A

5 0
3 years ago
Read 2 more answers
A 30 ohm resistor and a 20 ohm resistor are
Reptile [31]

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

  • I = V/Rt........... Equation 1

Where:

  • I = Electric current passing through the 30 ohms resistor
  • V = Voltage
  • Rt = Total or effective resistance of the resistors.

From the question,

Given:

  • V = 100 volts
  • Rt = (30+20) ohms (since both resistors are connected in series)
  • Rt = 50 ohms

Substitute these values into equation 1

  • I = 100/50
  • I = 2 A

Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

8 0
2 years ago
Two resistors R1 = 3 Ω and R2 = 6 Ω are connected in parallel. What is the net resistance in the circuit?​
gtnhenbr [62]

Answer:

"2Ω" is the net resistance in the circuit.

Explanation:

The given resistors are:

R1 = 3Ω

R2 = 6Ω

The net resistance will be:

⇒  \frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}

On substituting the values, we get

⇒  \frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}

On taking L.C.M, we get

⇒  \frac{1}{R_{net}} =\frac{2+1}{6}

⇒  \frac{1}{R_{net}} =\frac{3}{6}

⇒  \frac{1}{R_{net}} =\frac{1}{2}

On applying cross-multiplication, we get

⇒ R_{net}=2 \Omega

3 0
3 years ago
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