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aliina [53]
3 years ago
5

If an object moving at 5 m/s accelerates for 30 seconds at a rate of 2

Physics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

V=u + at

V= final velocity

U= Initial velocity

a= acceleration

t= time taken

V= 5 + 30*2

V=5+60

V=65m/s

Explanation:

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In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release
Scilla [17]

Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

Explanation:

When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}

where:

r = distance between the charges

\epsilon_0= permittivity of free space

By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as  the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.

7 0
3 years ago
After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A pol
vfiekz [6]

Answer:

18.03 s

Explanation:

We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.

x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2

The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:

x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2

Equalizing these two equations and solving for t:

(55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s

6 0
3 years ago
Which of these steps would most likely be part of a lab procedure?
Snezhnost [94]

Answer:

These are a part of lab procedures:

1. Write a hypothesis to answer a question.

2. Write a title at the top of a completed lab report.

3. Record the time to complete a chemical reaction.

These are NOT a part of lab procedures:

1. Create a question on the cause of a chemical reaction.

8 0
3 years ago
How does the strong nuclear force hold the nucleus of an atom together?
erastovalidia [21]

The strong nuclear force holds the nucleus of an atom together.  

Somehow, it overcomes the electrical force of repulsion between protons in the nucleus, which all have the same charge but still stay close together somehow. (b)

3 0
3 years ago
Read 2 more answers
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