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kirza4 [7]
2 years ago
15

Please Its urgent I need your HELP!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
ozzi2 years ago
4 0

Raising the temperature results in the radiator giving off photons of high-energy ultraviolet light. As heat is added, the radiator emits photons across a wide range of visible-light frequencies

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Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground
Papessa [141]

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that sin(betha)=(leg opposite)/(hypotenuse)

In this case we will need to find β, this way:

betha=sin^-1((15ft/20ft))

Then β=48,6°  

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

betha=(48,6°)*(2pirads/360°)

then we find β=0,85rads

7 0
3 years ago
A scientist who wants to study the affects of fertilizer on plants sets up an experiment. Plant A gets no fertilizer, Plant B ge
Solnce55 [7]

Answer: Plant A

Explanation: The control group is participants who do not receive the experimental treatment.

8 0
3 years ago
A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a
yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N

- Force B (259 N)

B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N

So the x- and y- component of the total force acting on the block are:

R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N

7 0
3 years ago
What single property was the most important in Jesseca’s material?
aleksandrvk [35]

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation:

Plato Answer

7 0
3 years ago
Read 2 more answers
A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec
kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
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