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3 years ago

The basic building block of life is

1 answer:

8 0

<span>The basic building block of life is molecules, more specifically macromolecules. There are four different macromolecules which could all be described as the building blocks of life, namely carbohydrates, proteins, nucleic acids and proteins. </span>

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NEED HELP ASAP <br> 2. What is an example of a question that cannot be solved using science?

Goryan [66]

Answer:

How did life begin on earth?

Explanation:

this is an example of a question that cannot be solved or why do we sleep too cannot be solved

7 0

3 years ago

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Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The c

m_a_m_a [10]

Answer:

C) W = - 190 J

Explanation:

Notation

Wf = work done by the friction force (unknown)

Ff = force of the friction

d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m

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3 years ago

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A light ray passing from medium 1 to medium 2 is bent away from the perpendicular normal to the boundary surface. What happens t

SVETLANKA909090 [29]

Answer:

The answer is a for Plato users.

Explanation:

Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.

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3 years ago

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A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field

vazorg [7]

The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time

The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:

φ = BA

B = magnetic field strength, A = loop area

The area of the loop A is given by:

A = πr²

r = loop radius

Make a substitution:

φ = B2πr²

Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

Δφ = ΔB2πr²

ΔB = change in magnetic field strength

Make another substitution:

ℰ = |ΔB2πr²/Δt|

Given values:

ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s

Plug in and solve for ℰ:

ℰ = |(-0.20)(2π)(0.50)²/2.5|

ℰ = 0.13V

3 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$