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Sati [7]
3 years ago
5

1. What is Newton's first law of motion and when does it occur during this experiment?

Physics
2 answers:
Yanka [14]3 years ago
5 0
Newton’s first law of motion, or inertia is an object at rest stays at rest. I may not be able to answer the second question, I do not know what the experiment is or what it is about.
Anna35 [415]3 years ago
4 0
For example if you would put an object on your car and drive and if you would turn left, the object would move on in the direction the car was driving before it turned left. It keeps its direction.
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What's an easy way to create an interference pattern of waves?
igor_vitrenko [27]

Answer:

B

Explanation:

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Which county in Florida is most in need of safe rooms and hurricane ties?
Murrr4er [49]
Probably Pasco. Lol.
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3 years ago
Read 2 more answers
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
Burning oil and coal adds to the atmosphere.
True [87]

Answer:

carbon dioxide

Explanation:

8 0
3 years ago
I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR =  E- V  

                                                    IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0
3 years ago
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