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3 years ago

PLEASE HELP

Physics

1 answer:

maxonik [38]3 years ago

5 0

Answer:

Flashlight

Explanation:

Flashlight use batteries and batteries are dc

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Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun

fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

That is given by as relation:

$\alpha+\rho+\tau=1$

where:

$\alpha=$ absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

$\rho=$ reflectivity is defined as the ratio of reflected radiation to the total irradiation

$\tau=$ transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

6 0

3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$