Can someone answer this question quickly??

A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagati

Margaret [11]

Answer:

$P=2.57\times 10^{-7}\ N/m^2$

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

$P=\dfrac{2I}{c}$

Where

c is speed of light

Putting all the values, we get :

$P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2$

So, the radiation pressure is $2.57\times 10^{-7}\ N/m^2$ .

3 0

3 years ago

The process of achieving fitness in the human body is almost instant and can occur overnight. True or False

777dan777 [17]

This statement is false nothing in the human body can occur overnight

3 0

3 years ago

I need help with two more Physics problems

sesenic [268]

Answer:

c. They hit at the same time

b. BGS

Explanation:

A marble dropped (initial vertical velocity is 0) will land at the same time as a marble launched horizontally (initial vertical velocity is 0) from the same height.

Boat S has a net speed of 5 m/s (10 − 5).

Boat B has a net speed of 15 m/s (10 + 5).

Boat G has a net speed of ≈11.2 m/s (√(10² + 5²)).

8 0

3 years ago

Read 2 more answers

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

3 years ago

Select all that apply. There MIGHT be more than one.

Stolb23 [73]

I believe the answer to your question is “Lithosphere plate boundaries”

The planet Earth is covered by a layer formed by land and rocks called the earth's crust or lithosphere. This crust is not smooth and uniform, but rather irregular and composed of tectonic plates, also called lithosphere plates. These plates are not fixed as they are under the magma (high temperature molten rock).

Hope this helps!:)

4 0

3 years ago