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vesna_86 [32]
3 years ago
5

Technology applies the principles of science to particular problems.

Physics
2 answers:
dlinn [17]3 years ago
5 0

Answer:

According to these scientists, engineers and philosophers, applying science in technology consists of applying (basic) scientific laws: science produces basic scientific laws that explain and predict natural phenomena, and technology applies those laws in designing technological artefacts by filling them out at ...

Explanation:

yess it is true..

mojhsa [17]3 years ago
4 0

Answer:

true

Explanation:

true,due to technology has prograing effects

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3/360 is 120. your velocity is 120

Explanation:

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Please help with vectors (will give BRAINLIEST answer)
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)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app
masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

F \propto \frac{1}{r}

F = \frac{k}{r} (Eq. 1)

Where:

F - Force, measured in newtons.

k - Proportionality ratio, measured in newton-meters.

r - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

F_{A}\cdot r_{A} = F_{B}\cdot r_{B}

F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}(Eq. 2)

Where:

F_{A}, F_{B} - Initial and final forces, measured in newtons.

r_{A}, r_{B} - Initial and final distances, measured in meters.

If we know that F_{A} = 5\,N, r_{A} = 0.9\,m and r_{B} = 0.35\,m, then final force is:

F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)

F_{B} = 12.857\,N

A force of 12.857 newtons must be applied to open the door.

3 0
3 years ago
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6 0
3 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0
2 years ago
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