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3 years ago

Does anyone know the answer to this?

Physics

1 answer:

sweet-ann [11.9K]3 years ago

5 0

Disclaimer: I just answered this, here is the answer again!
*Used copy paste from my own answer as it is a repeated question, no copied work*

3. A
The relation between V and I at constant R is;V=IR, so it is a direct linear relation.
4. A
This is another direct linear relation as P=IV.
5. D
The relation between P, R, and V is P=, so P is inversely proportional to R.
6.B
The relation between P,I, and R is , so P is directly proportional to the square of I.

Please note that y:x relations are always straight lines while relations are parabolic lines.

Hope this helps!

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If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic

harkovskaia [24]

Answer:

the thermal energy generated in the loop = $6.64*10^{-4} \ W$

Explanation:

Given that;

The length of the copper wire L = 0.614 m

Radius of the loop r = $\frac{L}{2 \pi}$

r = $\frac{0.614}{2 \pi}$

r = 0.0977 m

However , the area of the loop is :

$A_L = \pi r^2$

$A_L = \pi (0.0977)^2$

$A_L = 0.02999 \ m^2$

Change in the magnetic field is $\frac{dB}{dt}= 0.0914 \ T/s$

Then the induced emf e = $A_L \frac{dB}{dt}$

e = $0.02999 * 0.0914$

e = 2.74 × 10⁻³ V

resistivity of the copper wire $\rho = 1.69* 10^{-8}$ Ω m

diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³ m

Thus, the resistance of the wire R = $\frac {\rho L}{\pi r^2}$

R = $\frac{(1.69*10^{-8})(0.614)}{ \pi (0.54*10^{-3})^2}$

R = 1.13× 10⁻² Ω

Finally, the thermal energy generated in the loop (i.e the power) = $\frac{e^2}{R}$

= $\frac{(2.74*10^{-3})^2}{1.13*10^{-2}}$

= $6.64*10^{-4} \ W$

8 0

3 years ago

A friend says that the reason one's hair stands out while touching a charged Van de Graaff generator is simply that the hair str

Llana [10]

Answer:

Explanation:

Yes I agree with the statement .

When a person who is perfectly insulated from the earth , touches a Van de Graaff , his body acquires charge . when the hair acquires it, it stands out due to mutual repulsion . It is to be noted here that at pointed areas on a surface , there is larger accumulation of charge. Accumulation of charge is greater at hair tops .

It is also a general observation that when a bird sits on high tension wire , his feather stands out due to the same reason.

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3 years ago

The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in me

Kamila [148]

Answer:

(a) $\frac{m}{s^2}$

(b) $\frac{m}{s^3}$

(d) 20 m

(e) 1 m

(f) $0\frac{m}{s}$

(g) $-12\frac{m}{s}$

(h) $-36\frac{m}{s}$

(i) $-72\frac{m}{s}$

(j) $-6\frac{m}{s^2}$

(k) $-18\frac{m}{s^2}$

(l) $-30\frac{m}{s^2}$

(m) $-42\frac{m}{s^2}$

Explanation:

Since x is measured in meters and t in seconds, constants a and b must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean $\frac{m}{s^2}$ for a and $\frac{m}{s^3}$ for b.

We can get the velocity v equation by deriving the position with respect to t, which gives:

$v=6*t-6*t^2$

And the acceleration a equation by deriving again:

$a=6-12*t$

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is v=0, which gives

$6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1$

For t = 0, x = 0 so the maximun position is archieved at 1 second, which gives x = 1 meter.

For obtaining it's displacement r, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

$r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m$