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3 years ago

Someone help me with this ASAP!!

Physics

2 answers:

docker41 [41]3 years ago

8 0

It does help in protecting us from sun rays so it is a

alexandr402 [8]3 years ago

3 0

It protects us from the sun so the answer would be A

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If a car used 260,000 W of power to complete a race in 15 s, how much work did the car do?

suter [353]

Answer: 3.9 MW

Explanation:

1 W = 1 J/s

260000 J/s (15 s) = 3,900,000 = 3.9 MW

3 0

3 years ago

Another term for electromotive force is _____. voltage current resistance power

Ray Of Light [21]

Voltage because I said so

3 0

3 years ago

If a car rolls gently off a vertical cliff, how long does it take to reach 90km/hr.

Naddik [55]

90 km/h : 3.6 = 25 m/s. If you know that on earth g = 9.81 m/s^2, then all you have to do is divide the speed by g. 25/9.81 = 2.548 seconds

At least, if by 'gently rolls off a vertical cliff' means that your starting velocity equals zero.

5 0

3 years ago

The bogus pipeline is an instrument that was developed to measure attitudes that are otherwise difficult to measure because ____

VMariaS [17]

Answer:

of motivations to give a socially desirable response

Explanation:

The bogus pipeline is nothing but a polygraph machine which is fake.

With the help of a polygraph test it can be determined whether a person is telling the truth. This is done by measuring physiological indicators of a human being such as blood pressure, respiration, pulse, and skin conductivity.

The bogus pipeline appears to be like a real polygraph machine. It is assumed that people will be truthful as they believe that the machine is real. Nowadays, it is used to measure attitude towards a certain stimulus.

6 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago