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Naily [24]
3 years ago
14

If it took Ms. Oldman 37 seconds to life a 400 N student up 15 m, how much power did she use?

Physics
1 answer:
Rom4ik [11]3 years ago
7 0
The answer is 162 W.
Hope this helps!!:)
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2. A Formula One race car speeds up from rest to 27.8 m/s in a distance of 25 meters.
Nadya [2.5K]

Answer:

Correct answer:   a = 15.46 m/s²

Explanation:

The formula for accelerated movement with the given data is:

V² - V₀² = 2 · a · d     where the initial velocity  V₀ and the final V  

Since the initial velocity V₀ is zero, the formula is:

V² = 2 · a · d  => a = V² / (2 · d) = 27.8² / (2 · 25) = 772.84 / 50 = 15.46 m/s²

a = 15.46 m/s²

God is with you!!!

4 0
3 years ago
A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 5.00s what is the speed of the mail
grandymaker [24]
The acceleration of gravity (on Earth) is 9.8 m/s² downward.

This means that every falling object gains 9.8 m/s more downward speed
every second that it falls.

In 5 seconds of falling, it gains (5 x 9.8 m/s) = 49 m/s of downward speed.

If it was already descending at 2.0 m/s at the beginning of the 5 sec,
then at the end of the 5 sec it would be descending at

                                 (2 m/s  +  49 m/s)  =  51 m/s .
7 0
3 years ago
What is the main cause of the seasons
irinina [24]
The main cause of the seasons is earths orbit since each time were in a new season the planet is facing the earth a different way which causes winter etc..
3 0
3 years ago
Anyone want to get a conversation going again? lol
marshall27 [118]

Hello

i  am Marie

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7 0
3 years ago
Read 2 more answers
You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t
Nataly [62]

Answer:

Approximately \rm 2.0\; V.

Approximately \rm 30 \; mA. (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here R_1= 10\; \Omega and R_2= 100\; \Omega are connected in parallel. Their effective resistance would be equal to

\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega.

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be 20\; \rm mA = 0.020\; \rm A. That should also be the current through the effective \displaystyle \rm \frac{10}{11} \; \Omega resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V.

The voltage drop across the entire circuit would equal to

  • the voltage drop across the resistors, plus
  • the voltage drop across the LED.

In this case, that value would be equal to 1.83 + 0.182 \approx 2.0\; \rm V. That's the voltage that needs to be supplied to the circuit to achieve a current of 20\; \rm mA through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega.

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega.

Again, the current in a serial circuit is the same in all appliances.

\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA.

7 0
3 years ago
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