Please help will be marked most brainlist !!!

A diesel engine a. uses only air during the intake stroke. b. is an external-combustion engine. c. uses a spark plug to ignite f

Mariana [72]

Question : A diesel engine?

Logical Answer : a. uses only air during the intake stroke.

3 0

3 years ago

Read 2 more answers

The filament is broken in a light bulb. The circuit is now

Zina [86]

Answer:

It is A. Open Circuit.

Explanation:

Filament heats up and produces the light. If the filament breaks, the current does not flow and the circuit will be open. The ammeter will reach zero when this happens.

3 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou

madreJ [45]

Answer:

Part a)

E = 0

Part b)

$E = 6.77 \times 10^7 N/C$

Part c)

Electric field inside the conductor is again zero

$E = 0$

Part d)

$E = 8.52 \times 10^6 N/C$

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

$E = \frac{kq}{r^2}$

$E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}$

$E = 6.77 \times 10^7 N/C$

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

$E = 0$

Part d)

Now at r = 7 cm

again by Gauss law

$E = \frac{kQ}{r^2}$

$E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}$

$E = 8.52 \times 10^6 N/C$

5 0

3 years ago

Why does a siren have a lower pitch as it moves away from you? A The period of the sound wave is decreased B The amplitude of th

sweet-ann [11.9K]

Answer:

it is letter C

there is a video explaining this

4 0

3 years ago

Read 2 more answers

Use the information below to answer questions

Ulleksa [173]

Answer:

The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁ = 5 × 10⁻⁸ C - q₂ (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

$q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\ = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16} - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8} - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C$

q₁ = 5 × 10⁻⁸ C - q₂

q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C

5 0

3 years ago