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3 years ago

How much magnification is needed to see a nanometer?.

1 answer:

6 0

20,000,000

To see things at a nanometer, which is a trillionth of a meter, you would need to increase magnification nearly 20,000,000 times those are ionizing atoms

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A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

$f_1 = \dfrac{f_0(u + v)}{u}$ ..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

$f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

$f_2= f_0\dfrac{u+v}{u-v}$

$1275=1240\times \dfrac{340+v}{340-v}$

1.02822(340 - v) = 340 + v

2.02822 v = 340 x 0.028822

2.02822 v = 9.799

v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

5 0

3 years ago

Why are airbags, helmets, and other safety devices important? Physics

Ivenika [448]

Answer:to prevent serious damage when an accident occur

Explanation:to prevent serious damage when accident occur

4 0

3 years ago

Dolphins and other sea creatures can leap to great heights by swimming straight up and exiting the water at a high speed. A 200

scoundrel [369]

Answer:

i don't know sorry

Explanation:

so if a pie is baked

5 0

3 years ago

A marble is thrown horizontally with a speed of 7 m/s. When the marble lands, it hastraveled a horizontal displacement of 30 m.

Alex Ar [27]

Answer:

t = 4.28 s

Explanation:

The horizontal speed of a marble = 7 m/s

The horizontal displacement covered by the marble = 30 m

We need to find the time for which the marble is in air. Let the time is t. Using the formula for speed to find it as follows :

$s=\dfrac{d}{t}\\\\t=\dfrac{d}{s}\\\\t=\dfrac{30\ m}{7\ m/s}\\\\=4.28\ s$

So, it will in the air for 4.28 s.

6 0

3 years ago

A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

Darya [45]

Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

6 0

3 years ago