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3 years ago

Grop 17 of the periodic table contains the ?

1 answer:

8 0

Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and (Ts). Astatine and are radioactive elements with very short half-lives and thus do not occur naturally.

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A race car circles 10 times around a circular 8.0-km track in 20 min.

Otrada [13]

Find the average speed and the average velocity.

Average speed = distance / time

distance = 10 x 8000 m = 80,000 m
time = 20 min * 60 s/min = 1200 s

Average speed = 80,000 m / 1200 s = 66.67 m/s

Average velocity = displacement / time

Given that the race car made complete circles the final poin is the same initial point, then its displacement is zero and the average velocity is zero too.

6 0

3 years ago

Read 2 more answers

How much energy will be transferred to your eardrum while listening to this sound for 1.0 min?

Phoenix [80]

So E = 2x10^-3W/m^2*(π*(3.0x10^-3m)^2)*1min*60s... = 3.4x10^-6J

5 0

3 years ago

If the magnitude of the electric field in air exceeds roughly 3 âœ• 106 n/c, the air breaks down and a spark forms. for a two-di

Vlad1618 [11]

Answer: 39.8 μC

Explanation:

The magnitude of the electric field generated by a capacitor is given by:

$E = \frac{V}{d}$

d is the distance between the plates.

For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

$C =\frac{\epsilon_o A }{d}$

where A is the area of the plate and ε₀ is the absolute permittivity.

substituting, we get

$E = \frac{Q}{\epsilon_o A}$

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.

radius of the plates of the capacitor, r = 69 cm = 0.69 m

Area of the plates, A = πr² = 1.5 m²

Thus, the maximum charge that can be placed on disks without a spark is:

Q = E×ε₀×A

⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.

8 0

3 years ago

The electric field of a sinusoidal electromagnetic wave obeys the equation E=(375V/m)cos[(1.99×107rad/m)x+(5.97×1015rad/s)t].a.

andrey2020 [161]

Answer:

A) max = Eo = 375 V/m, B) B = 125 10⁻⁸ T , C) f = 9.50 101⁴ Hz, D) λ = 3.158 10⁻⁷ m, E) T = 1.05 10⁻¹⁵ s , F) invisible for humans G) v = c = 3 10⁸ m/s

Explanation:

The expression given for the electric field is

E = 375 cos (1.99 107x + 5.97 105t)

The general formula for the electric field of a transverse traveler is

E = Eo cos (kx-wt)

Where k is the wave number and w the angular velocity

A) The amplitudes as electric is

Emax = Eo = 375 V / m

B) the electric and magnetic field are related

E / B = c

B = E / c

B = 375/3 108

B = 125 10⁻⁸ T

C) angular velocity and frequency is related

.w = 2π f

f = w / 2π

f = 5.97 10¹⁵ / 2π

f = 9.50 101⁴ Hz

D) the speed of light has the formula

c = λ f

λ = c / f

λ = 3 10⁸ / 9.50 10¹⁴

λ = 3.158 10⁻⁷ m

E) The period

T = 1 / f

T = 1 / 9.5 10¹⁴

T = 1.05 10⁻¹⁵ s

F) let's reduce the wavelength nm

λ = 3.158 10⁻⁷ m (10⁹nm / 1m)

λ = 3.158 10²nm = 315.3 nm

The visible radiation range is between 400nm and 700nm. This radiation is ultraviolet and is invisible humans

G) All electromagnetic radiation has a speed at the speed of light (c)

v = c = 3 10⁸ m/s

5 0

3 years ago

Three Carnot engines operate between the following temperature limits.

Kipish [7]

Answer:c>a>b

Explanation:

Given

All the engine extracts same amount of heat(Q) from High-temperature reservoir

For a) 400 and 500 K

$\eta _{engine}=1-\frac{T_L}{T_H}$

$\eta _{engine}=\frac{work\ supplied}{heat\ supplied}$

$1-\frac{400}{500}=\frac{W_a}{Q}$

$W_a=\frac{Q}{5}$

For b)500 K and 600K

$1-\frac{500}{600}=\frac{W_b}{Q}$

$W-b=\frac{Q}{6}$

For c) 400 K and 600 K

$1-\frac{400}{600}=\frac{W_c}{Q}$

$W_c=\frac{2Q}{3}$

So c will give the highest amount of work

c>a>b

6 0

3 years ago