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3 years ago

A plane has a take off speed of 1 point

Physics

1 answer:

oksian1 [2.3K]3 years ago

4 0

Answer:

Acceleration = 1.85m/s²

Explanation:

Given the following data;

Final velocity = 300km/h to m/s = 300*1000/3600 = 83.33m/s

Time = 45 seconds

Since the plane started from rest, initial velocity is equal to 0m/s.

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Where,

a is acceleration measured in $ms^{-2}$

v and u is final and initial velocity respectively, measured in $ms^{-1}$

t is time measured in seconds.

Substituting into the equation, we have;

Acceleration = (83.33 - 0)/45

Acceleration = 83.33/45

Acceleration = 1.85m/s²

Therefore, the acceleration of the plane is 1.85m/s².

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Nikitich [7]

<h2>Answer: B. Gravitational potential energy </h2>

Explanation:

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In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy $U_{p}$ will be:

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3 years ago

Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a

raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

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3 years ago

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3 years ago

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The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

natita [175]

Answer:

3.258 m/s

Explanation:

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x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

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As the energy in the system is conserved we have

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