Ask question

Ask question

All categories

2 years ago

14

Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v

iew distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?

1 answer:

butalik [34]2 years ago

5 0

Answer:

$u=34cm$

Explanation:

From the question we are told that:

Far point is $V=34 cm$

Near point is $u=17 cm$

Therefore

Focal Length

$f=-34cm$

Generally the equation for the Lens is mathematically given by

$\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$

$\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}$

$u=34cm$

You might be interested in

Please help me out as soon as you can. Really need help.

olga nikolaevna [1]

The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

mark me brainliest

4 0

2 years ago

Why is a warm, tropical cumulus cloud more likely to produce precipitation than a cold, stratus cloud?

Lelu [443]

Explanation:

Warm,tropical cumulus cloud more likely to produce precipitation because they have high liquid content, strong and consistent updraft, these clouds are very thick. Moreover, they have large range of cloud droplet sizes. Whereas, stratus cloud have somewhat opposite characteristics, hence tropical cumulus clouds produce more precipitate.

7 0

2 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

2 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

timama [110]

Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

$2.9=0.5g(t_1^2-t_2^2)$

$5.8=g(t_1+t_2)(t_1-t_2))$

and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

$2t_2+0.134=4.416$

$t_2=0.5\times 4.282$

$t_2=2.141\ s$

substitute the value of $t_2$ in equation 2

$h=0.5\times 9.8\times (2.141)^2$

$h=22.46\ m$

8 0

2 years ago

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec

olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

$h=u't+\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}\times 9.8\times (4.3)^2$

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

5 0

2 years ago