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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.

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a powerboat accelerates along a straight path from 0 km/hr to 99.8 km/hr in 10.0 s.Find the average acceleration of the boat in

Alex777 [14]

(27.72-0)/10
= 2.772 m/s2

3 0

3 years ago

A straight bar magnet is initially 4 cm long, with the north pole on the right and the south pole on the left. if you cut the ma

Greeley [361]

The right half will be a new bar magnet of 2cm with north pole on the right side and south pole on teh left.

6 0

3 years ago

A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

And now we can re-arrange the eq.(1) to solve for the resistivity:

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

5 0

3 years ago

Look at the circuit diagram. Which of these components is part of the circuit?

tangare [24]

I think the answer is c AC power source
Hope this help you?

4 0

3 years ago

Why is the escape speed for a spacecraft independent of the spacecraft's mass?

scoray [572]

In order to escape the gravitational pull of our planet, any object must have an escape velocity of 7 km/s or more, anything lower than that will be slowed down by the pull of gravity, and will eventually returned to the surface of our planet. It is independent of mass, any lighter or heavier object must attain the required escaped velocity to reach space.

7 0

3 years ago