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3 years ago

13

How do the lens of a light microscope work

1 answer:

Burka [1]3 years ago

8 0

The smallest one is the least powerful one. And the medium size one is the medium powerful one. And the largest one is the most powerful one of them all

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What is the speed of a wave with a wavelength of 3 m and a frequency of .1Hz?

Yuri [45]

We know,
Speed = Frequency * Wavelength
Speed = 3 * 0.1 m/s [hertz = 1/sec.]
So, your final answer is 0.3 m/s

Hope this helps!!

6 0

3 years ago

Read 2 more answers

Please HEEEEELPPPP !!!!!

stepladder [879]

When you rub a balloon on a sweater, for example, some electrons come off and end up on the balloon. The fibers have lost electrons giving them a positive charge. The rubber gained electrons giving it a negative charge. ... The positively charged fibers are now attracted to the negatively charged balloon.

5 0

3 years ago

Help m-e-e people -_-!

xxTIMURxx [149]

Answer:

the answer is

Explanation:For equilibrium

Weight = Tension

mg=T

∴T=4×3.1π=12.4πN (as can be inferred from the question)

Y=

△l/l

T/A

=

1000

0.031

/20

12.4π/π(

1000

2

)

2

=

4×0.031

12.4×20×1000×(1000)

2

=2×10

12

N/m

2

6 0

3 years ago

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

4 0

3 years ago

The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.

Rainbow [258]

Answer: $0.0353\ s^{-1}$

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in $2.59\times 10\ s$

Sample at any time is given by

$N=N_oe^{-\lambda t}$

where, $\lambda=\text{decay constant}$

Put values

$\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10$

Taking natural logarithm both side

$\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}$

8 0

3 years ago