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alexandr1967 [171]

3 years ago

8

A candle is placed 30 cm in front of a convex mirror with a focal length of 20 cm, as shown in the diagram. What is the distance

from the lens to the image?
A. 60 cm

B. -12 cm

C. -60 cm

D. 12 cm

2 answers:

AnnyKZ [126]3 years ago

7 0

A That is the answer

lora16 [44]3 years ago

5 0

I think the answer is 60 cm.

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A certain type of bird lives in two regions of a state. The distribution of weight for birds of this type in the northern region

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Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.

Explanation: <u>Z-score</u> is how many standard deviations a data is from the population mean or how far a data point is from the mean.

The z-score is calculated by the following:

$z=\frac{x-\mu}{\sigma}$

where

x is the data point

μ is population mean

σ is standard deviation

For the <u>northern</u> <u>region</u> birds:

μ = 10, σ = 3, x = 13

$z=\frac{13-10}{3}$

z = 1

The z-score for birds living in the northern region is 1, which means it is 1 standard deviation <em>above the mean</em>.

For the southern region:

μ = 16, σ = 2.5, x = 13

$z=\frac{13-16}{2.5}$

z = -1.2

The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations <em>below the mean</em>.

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A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t

svp [43]

Answer:

Explanation:

Given

for $\theta=10^{\circ}$

Sphere are $d=40\ cm$

when sphere $d_2=10\ cm$ apart suppose deflection is $\theta _2$

We know

$F=k_t\cdot \theta$

Where F=force between charged particle

$\theta =$ Deflection

$F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta$

$\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1$

thus $\theta \propto \frac{1}{r^2}$

for $\theta _2$

$\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2$

$\theta _2=16\times \theta _1$

$\theta _2=160^{\circ}$

(b)for $10^{\circ}$ deflection Potential $v_1=8\ kV$

Electric Potential is $V=\frac{kQ}{r}$

$Q=\frac{V\cdot r}{k}$

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

$\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}$

$\theta =\frac{V^2r^2}{k\cdot k_t}$

thus $\theta \propto V^2$

therefore

$\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2$

$\frac{10}{\theta _2}=(\frac{8}{4})^2$

$\theta _2=\frac{10}{4}=2.5^{\circ}$

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2 years ago