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GalinKa [24]
2 years ago
7

Suppose that a car skids 15 meters if it is moving at 50 kilometers per hour Acceleration-Velocity-Position when the brakes are

applied. Assuming that the car always has the same contstant deceleration, how far (in meters) will it skid if it is moving at 100 kilometers per hour when the brakes are applied?.

Physics
1 answer:
julia-pushkina [17]2 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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Human eyes detect only a very small band of the electromagnetic spectrum. However, some animals and insects see in higher and lo
hoa [83]

Answer:

Squids = 450 - 490 nm (Moderate Frequency) (Blue)

Bees = 300 - 650 nm (Lower Frequency Bands)

Frogs = 280 - 580 nm (Very Low Frequency)

Explanation:

All of the above mentioned ranges are compared to that of humans.

I'm just surprised a little bit in the imagination that how these organisms see the world through their unique eyes. On the other hands, they are evolved like this just like we do so that may not be surprising enough. SIKE

3 0
3 years ago
Where are bar magnets the strongest?
Alecsey [184]

Answer:

The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet.

Explanation:

3 0
2 years ago
Read 2 more answers
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
bixtya [17]
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
8 0
3 years ago
If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f
Anna007 [38]

(1) Doubling of the current through the wire will result in doubling of its magnetic  field.

The magnetic field around a wire is a function of the current I and radial distance r

B = \frac{\mu\cdot I }{2\pi r}

(with mu denoting the magnetic permeability of the medium). So, B is directly proportional to I. The field magnitude will double with the doubled current from 5A to 10A

(2) Using the same formula as in (1), we can see that the magnetic field is inversely proportional to the radial distance from the wire. So, a particle at 20cm will experience half the magnitude compared to a particle at 10cm.

(3) Answer

If a particle with a charge q  moves through a magnetic field B with velocity v, it will be acted on by the magnetic force

F_m = B\cdot v\cdot q

So, a particle with charge -2uC will experience a magnetic force of same magnitude but opposite direction (and perpendicular to B) as compared to a particle with a charge of 2uC

3 0
2 years ago
Read 2 more answers
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