Which kind of mirror can produce real images? A)concave B)convex C)flat

You can burn fat by all of the following Question 2 options:

Katen [24]

Answer:

D

Explanation:

6 0

2 years ago

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A small, uncharged metal sphere is placed near a larger, negatively charged sphere. Which diagram best represents the charge dis

Rudik [331]

Answer:

I'm sorry I don't know what is the answer

3 0

3 years ago

A real image is four times as far from a lens as is theobject.

saveliy_v [14]

Answer:

1.25 focal lengths

Explanation:

The lens equation states that:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the object distance

q is the image distance

In this problem, the image is 4 times as far from the lens as is the object: this means that

$q=4p$

If we substitute this into the lens equation and we rearrange it, we get

$\frac{1}{f}=\frac{1}{p}+\frac{1}{4p}=\frac{4+1}{4p}=\frac{5}{4p}\\p=\frac{5}{4}f=1.25 f$

so, the object distance measured in focal lengths is

1.25 focal lenghts

3 0

3 years ago

Stewart James, Calculus, Section 6.4, Page 449, Problem 5

faltersainse [42]

The total work done is
W=60 plus 120
=180J

For the graphs please find the attached image

4 0

3 years ago

An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0

Mazyrski [523]

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

$\theta=0.260t^2$

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

$\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t$

Then, you evaluate for t=2.40 s:

$\omega=0.52(2.40)=1.248\frac{rad}{s}$

(b) The linear velocity is calculated by using the following formula:

$v=\omega r$

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

$v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}$

(c) The tangential acceleration is:

$a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}$

(d) The radial acceleration is:

$a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}$

4 0

3 years ago