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3 years ago

Two factors that can be used to evaluate ______ are life expectancy and quality of life.

Physics

1 answer:

Olin [163]3 years ago

8 0

Two factors that can be used to evaluate health are life expectancy and quality of life

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A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th

s344n2d4d5 [400]

Answer:

15 cm

Explanation:

$h_{o}$ = Diameter of the coin = 15 mm

$h_{i}$ = Diameter of the image of coin = 5 mm

$d_{o}$ = distance of the coin from mirror = 15 cm

$d_{i}$ = distance of the image of coin from mirror = ?

Using the equation

$\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}$

$\frac{d_{i}}{15} = \frac{- (5)}{15}$

$d_{i}$ = - 5 cm

$R$ = radius of curvature

Using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}$

$\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}$

$R$ = - 15 cm

6 0

2 years ago

A professor drives off with his car (mass 870 kg), but forgot to take his coffee mug (mass 0.47 kg) off the roof. The coefficien

SSSSS [86.1K]

Answer: $6.86 m/s^2$

Explanation:

Given

mas of car=870 kg

coffee mug mass=0.47 kg

coefficient of static friction between mug and roof $\mu _s= 0.7$

Coefficient of kinetic Friction $\mu _k=0.5$

maximum car acceleration is $\mu \times g$

here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

$a_{max}=0.7\times 9.8=6.86 m/s^2$

8 0

3 years ago

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$