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Dmitriy789 [7]
3 years ago
8

What is the resultant velocity of a plane that is traveled at 245m/s, north and encounters a tailwind of 55m/s?

Physics
1 answer:
lesya [120]3 years ago
6 0

For this problem, you should use Pythagorean Theorem.

The formula for this is a^2 + b^2 = c^2.

The plane is moving north, and then moves to the right. Therefore, we need to find the hypotenuse, or resultant velocity. Plugging it into the equation we get, (245 m/s)^2 + (55 m/s)^2 = c^2

Solving for this, you will get 250 m/s.


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the apparent weight of a body wholly immersed in water is 32N and its weight in 96N and calculate volume of the body
ArbitrLikvidat [17]

Answer:

0.0065 m³

Explanation:

Apparent weight = weight − buoyancy

32 N = 96 N − (1000 kg/m³) (9.8 m/s²) V

V = 0.0065 m³

3 0
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Besides providing food and lumber, some trees are valuable natural resources because they contain substances that help make ____
kondaur [170]
Medicines would be your answer


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Select the properties that apply to superconductors.
makkiz [27]
made from pure metals . . . no;
they've been made from all kinds of weird compounds and alloys.
 
conduct electricity with zero resistance . . . yes;
that's why they're called "superconductors".

produce a strong magnetic field . . . possible, but not because it's a superconductor;
just like any other conductor, the magnetic field depends on the current that's flowing in the conductor.

no loss of energy in the transfer of electricity . . .
there's no loss of energy in the current flowing in the superconductor;
but if you tried to transfer the current out of the superconductor into
something else, then there would be some loss.
7 0
3 years ago
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Is there a frame of reference one can go into that seems to eliminate gravity as Newton described it?
andriy [413]

Answer:

Yes such a frame exists: a free-fall (free-float frame) frame. This frame of reference is subject only to gravity and no forces such as electromagnetic forces or nuclear forces.

3 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
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