All you have to do is find the description for a b c and d words and find the meaning 8 9 10 11 and 12

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago

9. True or false. Training at high altitudes can also increase the amount of

Westkost [7]

False because when the training higher there is less oxygen

5 0

2 years ago

what helps differentiate between the sound of a fire truck, an ambulance and an 18 wheeler? imbre form dynamics sound

DENIUS [597]

Dynamics sound helps differentiate between the sound of a fire truck, an ambulance and an 18-wheeler.

<h3 /><h3>What is dynamics sound?</h3>

Elements allude to the din or delicateness of music. Elements offer a method for showing articulation in printed music. They help to drive the profound substance of music through volume and force. Elements can likewise be shown at the large-scale level for a piece of music in general. This may be just a single time toward the beginning, or a few times all through in the event that the din changes during various segments. Static elements are melodic directions that advise us to play the music at a specific volume that doesn't change. As such, don't get stronger or calmer, play each note at a similar volume as the final remaining one.

Learn more about dynamics sound, refer:

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6 0

11 months ago

What is the electric flux passing through a gaussian surface that surrounds a 0.075 c point charge?

Valentin [98]

An 0.075 c point charge's surrounding gaussian surface's electric flux is 8.5 ×10⁹Nm²/C.

<h3>Which is a gaussian surface?</h3>

In three dimensions, the Gaussian surface is referred to as a closed surface where the flux of a vector field may be determined. The gravitational field, the electric field, or the magnetic field are all examples of these vector fields.

<h3>Why is a Gaussian surface drawn?</h3>

We build a fictitious Gaussian surface around the supplied surface when the surface of which an electric field or flux must be established is asymmetrical or the surface area is challenging to obtain, such as the surface area for an infinitely long wire or plane. The Gaussian surface is said to utilize symmetry the best.

When a charge q is surrounded by a gaussian surface, the electric flux that passes through it is

$\phi=q / \epsilon 0$ = 0.075 / ( 8.85 ×10⁻¹²)

= 8.5 ×10⁹Nm²/C.

To know more about Gaussian Surface visit:

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7 0

1 year ago