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lyudmila [28]
2 years ago
8

Explain giving an example for;

Physics
1 answer:
Bingel [31]2 years ago
6 0

\boxed{\large{\bold{\red{ANSWER~:) }}}}

  • The motion of a ball falling through the atmosphere, or a model rocket being launched up into the atmosphere are both excellent examples of Newton’s 1st law of motion.

  • Riding a bicycle is a good example of Newton’s 2nd law of motion.In this example, the bicycle is the mass. The leg muscles pushing on the pedals of the bicycle is the force.

  • You hit a wall with a certain amount of force, and the wall returns that same amount of force. This is an example of Newton’s 3rd law of motion
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A body with mass 4 kg moves with a velocity of 108km/h <br> Calculate it kinetic<br> energy
Fed [463]

Explanation:

The answer is in the pic above

4 0
3 years ago
Cynthia forgot to put the fabric softener in the wash. As her socks tumbled in the dryer, they became charged. If a small piece
polet [3.4K]

Answer:

E = 24000 N/C = 24 KN/C

Explanation:

The electric field experienced by a test charge is given by the following formula:

E = \frac{F}{q}\\\\

where,

E = Electric Field = ?

F = Force of attraction = 3 x 10⁻⁶ N

q = Charge on piece of lint = 1.25 x 10⁻¹⁰ C

Therefore, using these values in the equation, we get:

E = \frac{3\ x\ 10^{-6}\ N}{1.25\ x\ ^{-10}\ C}\\\\

<u>E = 24000 N/C = 24 KN/C</u>

6 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
3 years ago
Solve this question
Rudik [331]
Ok


........................
6 0
2 years ago
Pedro e Maria saíram para passear de carro. Eles partiram de São Paulo às 10 h em direção à Braúna, localizada a 500 km da capit
Salsk061 [2.6K]

Answer:

This can be translated to:

"P edro and Maria went for a drive. They left São Paulo at 10 am towards Braúna, located 500 km from the capital. As the journey was long, they made two 15-minute stops for gas and also spent 45 minutes for lunch. When arriving at the final destination, Maria looked at the clock and saw that it was 6 pm. What is the average speed of the trip?"

Ok, the first thing we know is, that for the average speed we can write:

Speed = Distance/time.

First, we know that Distance = 500km.

And for the time we have two possiblities:

The total average speed will decrease because they were stopped a total of:

15min + 15min + 45min = 75min = 1.25 hours

Then if the travel starts at 10am, and ends at 6 pm, the total time that has passed is:

6pm = 18hs

6pm - 10am = 18 - 10 = 8hs

Then the average speed will be:

Speed = 500km/8h = 62.5 km/h.

Now, if we considerate the average speed only when they are moving, the total time that they are moving is:

Total time in travel - time that they where stopped

8h - 1.25h = 6.75h

Then the average speed would be:

Speed = 500km/6.75h = 74.1 km/h

4 0
3 years ago
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