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3 years ago

7

Circle O has a circumference of approximately 44 in. Circle O with diameter d is shown. What is the approximate length of the di

ameter, d?

2 answers:

Zanzabum3 years ago

7 0

The answer is 14.001

GrogVix [38]3 years ago

5 0

Answer:

14.001 in

Explanation:

Circumference = pi * diameter

Circumference / pi = diameter

44 inches/pi = 14.001 inches

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astra-53 [7]

Venus moves faster than earth because it move with a higher speed as compared to earth.

<h3>Is Venus move faster than earth?</h3>

Venus move with the speed of 35.02 km/s while on the other hand, the earth moves with the speed of 29.78 km/s. So due to more speed of Venus, the Venus completes its revolution earlier than earth.

So we can conclude that Venus moves faster than earth because it move with a higher speed as compared to earth.

Learn more about Venus here: brainly.com/question/2323914

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5 0

2 years ago

Which is true of the greenhouse effect?

Aleksandr [31]

Answer:

c is correct option thanks to brainly

3 0

3 years ago

A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a

Zolol [24]

Answer:

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$ (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

$E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L$

For small oscillations, the equation can be re-arranged into the following form:

$E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L$

Where:

$\theta = \frac{A}{L^{2}}$ , measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is $4\theta$ and the total energy of the pendulum is:

$E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L$

The factor of change is:

$\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}$

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$

3 0

3 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

7 0

3 years ago

A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy

GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0

2 years ago