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2 years ago

If a fly gets his wings cut of, is it still a fly?

Physics

2 answers:

jenyasd209 [6]2 years ago

4 0

Answer:

Explanation:

because if the fly get its wings cut off its obviously not going to be able to fly

svetoff [14.1K]2 years ago

3 0

Answer:

Yes I think... But the joke is it is a walk then.

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A plane has a cruising speed of 250 miles per hour when there is no wind. at this speed, the plane flew 300 miles with the wind

kherson [118]

The answer to your question is 50 miles per hour

4 0

3 years ago

A test charge is placed at a distance of 3.5 meters from a charge of 8.7 × 10-9 coulombs. What is the electric field strength at

zavuch27 [327]

In order to calculate the electric field strength, we may use the formula:
E = kQ/d²
Where Q is the charge and d is the distance between the charge and the test charge. Substituting the values into the equation:
E = (9 x 10⁹)(8.7 x 10⁻⁹) / (3.5²)
E = 6.39 Newtons per coulomb

Therefore, the answer is 6.4 Newtons/coulomb

8 0

3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

3 0

2 years ago

You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.

Blizzard [7]

Answer:

Remains same

Explanation:

$T$ = Time period of oscillation

$m$ = mass

$k$ = spring constant

Time period of oscillation is given as

$T = 2\pi \sqrt{\frac{m}{k} }$

we know that as we move from earth to moon, the value of spring constant "k" and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0

3 years ago

4. What is the electric field strength 1.4 nm from a charge of 4.7 cC?

pentagon [3]

The electric field strength is $2.16\cdot 10^{26} N/C$