All categories

3 years ago

Exposure to the Sun's harmful infrared radiation should be kept to a minimum.

1 answer:

7 0

Of course! If it's harmful, then your exposure to it should be kept
to a minimum. That's a no-brainer. But the sun's infrared radiation
is generally less harmful than its ultraviolet radiation is.

You might be interested in

An 81.5-kg man stands on a horizontal surface.

Elza [17]

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10² N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

$P=\dfrac{F}{A}$

$P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2$

P=181.11 x 10² N/m²

4 0

3 years ago

Find the acceleration of the blocks when the system is released. The coefficient of kinetic friction is 0.4, and the mass of eac

grin007 [14]

Answer:

a = 4.9(1 - sinθ - 0.4cosθ)

Explanation:

Really not possible without a complete setup.

I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g

F = ma

mg - mgsinθ - μmgcosθ = (m + m)a

mg(1 - sinθ - μcosθ) = 2ma

½g(1 - sinθ - μcosθ) = a

maximum acceleration is about 2.94 m/s² when θ = 0

acceleration will be zero when θ is greater than about 46.4°

8 0

3 years ago

When light passes through a convex lens, it does this.....

Tpy6a [65]

Explanation:

6. Converge or come together

7. convex

3 0

2 years ago

Read 2 more answers

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago

The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr

tatyana61 [14]

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.

8 0

2 years ago