If a person can't accompdate and the glasses is +2.50D. At which distance will the person see cleary

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get

$\sin{\theta} = \dfrac{F}{mg}$

or

$\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)$

$\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]$

$\;\;\;=42.9°$

6 0

2 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from

krek1111 [17]

Answer:

$E=35921.96N/C$

Explanation:

From the question we are told that:

Radius $r=0.321mm$

Charge Density $\mu=0.100$

Distance $d= 5.00 cm$

Generally the equation for electric field is mathematically given by

$E=\frac{mu}{2\pi E_0r}$

$E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}$

$E=35921.96N/C$

4 0

3 years ago

The persistence of vision for normal eye is

attashe74 [19]

Answer:

The answer is 1/16

Explanation:

1. Persistence of vision refers to the optical illusion that occurs when visual perception of an object does not cease for some time after the rays of light proceeding from it have ceased to enter the eye. 2. The persistence of vision for normal eye is 1/16 if a second.

6 0

3 years ago

Read 2 more answers

A publication which contains news on current events

german

Answer:

A newspaper is a serial publication which contains news on current events of special or general interest.

3 0

2 years ago

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