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3 years ago

Which of the following is a part of the geocentric model?

2 answers:

6 0

In astronomy, the geocentric model is a model that depicts the Earth at the center of the Universe, with the Sun and the other planets revolving around it.

Therefore, the correct answer is
<span>"All objects in the universe revolve around Earth. "
</span>which corresponds to the geocentric model.

vazorg [7]3 years ago

5 0

The correct answer is C where all objects in the universe revolve around earth.

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a battery of 5v and internal resistance of 5 ohms is connected to a resistor of 20 ohms. calculate the value of terminal voltage

Alborosie

Answer:

A bird flies 75m in 15s.Calculate its speed

7 0

1 year ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

8 0

3 years ago

In every trench over 4 feet (1.2 m) deep, there must be an exit every

ss7ja [257]

Explanation:

25 ft I believe that this is the best answer

5 0

2 years ago

Jason is taking a physical exam and has to do push ups for sixty seconds. He does 30 pushups in the sixty seconds. Each push up

Alexandra [31]

Answer: 2 seconds is the unit rate.

Explanation:

We know that Jason does 30 pushups in 60 seconds at a constant rate, and we know that each push up takes 2 seconds.

then, we can write this as 30 pushups/60 seconds = (1/2) pushups per second.

Here, two seconds represents the time needed to do one pushup, is the unit rate (this means that we need 2 seconds to have a unit "one pushup")

8 0

3 years ago

A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 m above the ground,

katrin [286]

Answer:

0.405 seconds

Explanation:

Consider the amount of time it takes the block to fall from 53 m up to 14 m above the ground; then consider the amount of time it takes the block to fall from 53 m up to 2 m above the ground.

First, d = (1/2) gt^2 or t= ( 2 d / g)^1/2

= ( 2 × 39 / 9.8)^1/2 = 2.8212 seconds

Then, to fall from 53 down to 2 meters...

d = (1/2) gt^2 or t= ( 2 d / g)^1/2

= ( 2 * 51/ 9.8 )^1/2 = 3.2262 seconds

So the amount of time it takes for the block to fall from 14 m upto 2 m above the ground

3.2262 - 2.8212 = 0.405 seconds

this is how much time there is from when the man sees the block until it hits him. Not much time...

5 0

2 years ago