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3 years ago

In which situation will the lowest resistance occur?

Physics

2 answers:

insens350 [35]3 years ago

8 0

Answer:

thick wire and cold temperatures

Lena [83]3 years ago

6 0

Answer: There may be two types of resistances here:

Air resistance: for example when something is falling down, the air will push the object upwards. As a bigger surface has the object, more air is pushing, so if you want to reduce the resistance yo must reduce the surface area. For example, a sheet of paper will experience more resistance if it is open if you make a small ball with it.

Electric resistance: This is the resistance of conductors against the electrons that try to travel along with it.

there are two equations here.

R = p*L/A, this talks about the volume and density of the conductor, where p is the density, L is the length and A is the area perpendicular to the direction in which the electricity flows.

Then if you reduce the length, and increase the area, the resistance will be reduced.

the other equation is: R(T) = R0( 1 + a*(T - 20°))

Where R0 is the resistance at 20° and a is a coefficient for T = 20°.

Here, you can see that at smaller temperatures, the resistance decreases, so we can conclude that we must increase the area, and decrease the length and temperature.

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An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

3 0

3 years ago

The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a

Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

$J=F.\Delta t$

$J=10^6\ N\times 20\ s$

$J=2\times 10^7\ N-s$

(b) Impulse is also given by the change in momentum i.e.

$J=\Delta p=p_f-p_i$

$J=p_f-p_i$

$p_f=J+p_i$

$p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s$

$p_f=20143000\ kg-m/s$

(c) For new velocity,

$v_f=\dfrac{p_f}{m}$

$v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}$

$v_f=1549.46\ m/s$

Hence, this is the required solution.

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Wewaii [24]

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masya89 [10]

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Alekssandra [29.7K]

Answer is Seismic Wave

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