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Salsk061
3 months ago
5

The resistivity of iron is 1 x 10 -7 Ωm. The resistance of a iron wire of particular length and thickness is 1 ohm. If the lengt

h and the diameter of wire both are doubled, then the resistivity in Ωm will be
Physics
1 answer:
Maru [3.3K]3 months ago
7 0
2×10⁻⁷ Ωm. From the problem, we have R = 4ρL/πd². Hence, solving for ρ, we get ρ = Rπd²/4L. When both the length and diameter are doubled while keeping the resistance fixed, the new resistivity is derived as ρ' = 2(1×10⁻⁷), resulting in ρ' = 2×10⁻⁷ Ωm.
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The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports
kicyunya [3294]
To counteract a 58 mph crosswind, the western component of the trajectory must be accounted for. Consequently, directing towards the northwest creates a 45-degree angle, aligning with the destination. This triangle's third vertex is located at the destination, with the right angle positioned there. The western aspect of their flight represents the triangle's base, while the vertical side reflects the resultant path, and the hypotenuse indicates the actual distance traveled. Since the 58 mph crosswind was countered by flying in a northwest direction, the distance from the starting point to the destination should equal the westward segment of their journey. The hypotenuse can be determined via the square root of twice the dimension of the identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02

An alternative method:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Thus, they must fly at 82.02 mph.
5 0
2 months ago
Read 2 more answers
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
inna [3103]
1) The projectile's motion follows
,h(t) = 2+22.5 t-4.9 t^2
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
t=2.30 s

3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
t=4.68 s
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s
, carrying a negative sign to denote a downward direction.
8 0
2 months ago
A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?
Keith_Richards [3271]

Calculating the average speed is straightforward by using the formula involving distance and time:

average speed = distance / time

 

Thus, we have:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0
4 months ago
Read 2 more answers
A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
Sav [3153]

Answer:

The electric flux going through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε = 8.8542 * 10^{-12}

Substituting ε = 8.8542 * 10^{-12} and q =7.6\µC; the formula simplifies to

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Thus, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
4 months ago
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