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allsm [11]
3 years ago
11

The _____ is about determining if you are the way you are because of your

Physics
1 answer:
REY [17]3 years ago
3 0

Answer: nature versus nurture debate

Explanation:

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Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo
bagirrra123 [75]
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span> =>v¹ = 2.25 m/s (East) </span>
5 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
If a bullet travels at 593.0 m/s, what is its speed in miles per hour?
Ksenya-84 [330]

We have 1 \; mile = 1609.34\; meters. So, 1 \; meter = \frac{1}{1609.34} \;mile.

1 \; hour = 3600 \; s. So 1 \; s = \frac{1}{3600}  \; hour.

Thus we can convert the units of the given quantity.

That is,

593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\&#10;593\;m/s=1,326.51\;miles/hour.

The quantity is converted to the required units.


7 0
3 years ago
According to the law of reflection, the angle of incidence is equal to the angle of
Darina [25.2K]

Before going to answer this question first we have to understand reflection and laws of reflection.

Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .

Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.

Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]

The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]

There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.

There are two laws of reflection.

FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.

SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r

Hence the correct answer will be angle of reflection.

                                         

3 0
3 years ago
Read 2 more answers
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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