By what factor will the electric force between two charged objects change if the charge on both objects is quadrupled

Which will a positively charged objects attract

Vitek1552 [10]

Me and you! enjoy your day girllllaa or boy

6 0

4 years ago

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Please answer my question!,

Karolina [17]

Answer:

Hope this helps! Mark as brainliest if liked thanks!

Explanation:

Your reasoning that the shadow is the shortest at mid-day is spot-on!

The wording of the question is the key to the answer. It says that the measurements were made in Summer. So this means that British Summer Time (BST) is being applied. BST is one hour ahead of Greenwich Mean Time and so what looks like 1pm is really 12 noon.

The safest sort of answer is to say that the shadow is shortest when the sun is at its highest point, and in this particular question that is at 1 pm because it is BST.

3 0

3 years ago

A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with t

sergij07 [2.7K]

Answer: $135.29\º$

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

*Note the normal is perpendicular to the plane, with a $90\º$ angle with the surface

And this can be visualized in the figure, where the angle ${\theta}_{1}$ with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium refractive index (the air)

$n_{2}=1.309$ is the second medium refractive index (the ice)

$\theta_{1}=25.5\º$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Now, firstly we have to find $\theta_{2}$ and then, by geometry, find $\alpha$ and $\beta$ which sum the angle between the reflected and the refracted light.

Finding $\theta_{2}$ from (1):

$(1)sin(25.5\º)=(1.309)sin(\theta_{2})$

$sin(\theta_{2})=0.328$

$\theta_{2}=arcsin(0.328)$

$\theta_{2}=19.201\º$ (2)

Remembering that the Normal makes a $90\º$ angle with the surface, we can say:

$90\º=\theta_{1}+\beta$ (3)

Finding $\beta$ :

$\beta=90\º-25.5\º=64.5\º$ (4)

Doing the same with $\theta_{2}$ and $\alpha}$ :

$90\º=\theta_{2}+\alpha$ (5)

Finding $\alpha$ :

$\alpha=90\º-19.201\º=70.79\º$ (6)

Adding both angles (4) and (6):

$\alpha+\beta=70.79\º+64.5\º$ (7)

$\alpha+\beta=135.29\º$ >>>This is the angle between the reflected and the refracted light.

5 0

3 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

3 years ago

Airplanes moves in blank dimensions through. What is Blank

raketka [301]

Blank is something empty lol

6 0

3 years ago

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