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3 years ago

HELP PLEASE. WILL GIVE BRAINLIEST!!!

Physics

2 answers:

Troyanec [42]3 years ago

4 0

Answer:

Explanation:

nikitadnepr [17]3 years ago

3 0

Answer:

its A

Explanation:

Just took the exam

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Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As

Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

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An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0

Murrr4er [49]

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

$C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F$

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

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3 years ago

If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1

Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

$I=\frac{Q}{t}$

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

$\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}$

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>

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