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sleet_krkn [62]
3 years ago
9

HELP ME ILL GIVE BRAINLIST ​(please explain how to do it)

Physics
1 answer:
Alenkasestr [34]3 years ago
3 0

Answer:

I honestly don't know(sorry)

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Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in
Svetradugi [14.3K]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

5 0
3 years ago
HELP ASAP
kipiarov [429]

Answer:

This can be part of your paragraph.

Explanation:

From the cornea, the light passes through the pupil. The iris, or the colored part of your eye, controls the amount of light passing through. From there, it then hits the lens. This is the clear structure inside the eye that focuses light rays onto the retina.

4 0
2 years ago
the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67​
baherus [9]

Answer:

\theta_c = 36.78^o

Explanation:

The relationship between the refractive index and the critical angle is given as follows:

\eta = \frac{1}{Sin\ \theta_c} \\\\Sin\ \theta_c = \frac{1}{\eta}\\\\\theta_c = Sin^{-1}(\frac{1}{\eta} )

where,

η = refractive index = 1.67

θc = critical angle =?

Therefore,

\theta_c = Sin^{-1}(\frac{1}{1.67} )

\theta_c = 36.78^o

4 0
3 years ago
Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun
fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

\alpha+\rho+\tau=1

where:

\alpha= absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

\rho= reflectivity is defined as the ratio of reflected radiation to the total irradiation

\tau= transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

6 0
3 years ago
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0
3 years ago
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