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SpyIntel [72]
3 years ago
11

Jasper made a list of the properties of electromagnetic waves. Identify the mistake in the list. Electromagnetic Wave Properties

1. Are made up of an electric field and a magnetic field 2. Are longitudinal waves 3. Can travel through space 4. All travel at the same speed in a vacuum Line 1 has a mistake. Line 2 has a mistake. Line 3 has a mistake. Line 4 has a mistake.
Physics
1 answer:
Nata [24]3 years ago
6 0

Answer:

Statement 2 is wrong

Explanation:

To check the statements in this exercise, let's describe the main properties of electromagnetic waves. Let's describe the characteristics

* they are transverse waves

* formed by the oscillations of the electric and magnetic fields

* the speed of the wave is the speed of light

with these concepts let's review the final statements

1) True. Formed by the oscillation of the two fields

2) False. They are transverse waves

3) True. Can travel by vacuum as they are supported by oscillations of the electric and magnetic fields

4) True. They all have the same speed of light

Statement 2 is wrong

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The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th
Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

        v=\sqrt{\frac{2GM}{R}}

Here we have

   Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

   R = 1 AU = 1.496 × 10¹¹ m

   M = 2.3 × 10³⁰ kg

Substituting

    v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s

Approximate escape speed = 45.3 km/s

6 0
3 years ago
Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t
nlexa [21]

Answer:

Explanation:

Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.

Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

Tsinθ = F ( F is force of repulsion between two charges sphere)

Dividing the two equations

Tanθ = F / mg

tan17 = F / (7.1 x 10⁻³ x 9.8)

F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17  = .41 m )

21.27 x 10⁻³  = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

no of electrons required  = q / charge on a single electron

= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .39375 x 10¹³

3.9375 x 10¹² .

4 0
3 years ago
How to find the time with only the distance or height from the ground
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You use acceleration due to gravity
and 1/2 atsqr=d
therefore 1/2 * 9.8 * tsqr= d
5 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. A
vagabundo [1.1K]

Answer:

Q = 1.2*10⁻¹² C

Explanation:

  • For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

       C = \frac{Q}{V}  (1)

  • For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

       C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d}  (2)

  • So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
  • Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

       Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} =  1.2e-12 C = 1.2 pC  (3)

6 0
3 years ago
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibrati
coldgirl [10]

Answer:

1. The period is 1.74 s.

2. The frequency is 0.57 Hz

Explanation:

1. Determination of the the period.

Spring constant (K) = 30 N/m

Mass (m) = 2.3 Kg

Pi (π) = 3.14

Period (T) =?

The period of the vibration can be obtained as follow:

T = 2π√(m/K)

T = 2 × 3.14 × √(2.3 / 30)

T = 6.28 × √(2.3 / 30)

T = 1.74 s

Thus, the period of the vibration is 1.74 s.

2. Determination of the frequency.

Period (T) = 1.74 s

Frequency (f) =?

The frequency of the vibration can be obtained as follow:

f = 1/T

f = 1/1.74

f = 0.57 Hz

Thus, the frequency of the vibration is 0.57 Hz

4 0
3 years ago
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