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Shalnov [3]
3 years ago
13

Plz help, its an emergency, giving lots of points for it. and srry if its in the wrong subject, not really sure? I just don't un

derstand it at all, it's about atoms and all that stuff.

Physics
2 answers:
Nonamiya [84]3 years ago
8 0
What is the question being asked?
exis [7]3 years ago
4 0
Anything I can help with
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I just want to shift to another country to learn like London so who can tell me the progress​
scoray [572]

Answer:

by shift do you mean travel

Explanation:

3 0
2 years ago
How are frequency and wave period related?
dedylja [7]
-- The unit of frequency is "per second"  (Hz), which is [reciprocal time].

-- The unit of period is "second", which is [time].

Do you see where this is going ?

'Frequency' and 'period' are reciprocals of each other.

For any wave ...

Period  =  (1) / (frequency) .

Frequency  =  (1) / (period) .
6 0
3 years ago
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
3 years ago
an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm)
Juli2301 [7.4K]

Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

To Find :

the image height (in cm).

Solution :

By lens formula :

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Here, u = - 100 cm

f =  150 cm

\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm

Now, magnification is given by :

m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m

Therefore, the image height is 3 m or 300 cm.

5 0
2 years ago
A car travels a distance of 200m in 20s. The engine of the car provides a driving force of 1000N. What is the power output of th
sp2606 [1]

Answer:

10000W

Explanation:

work=1000*200=200000

then

P=work/time=200000/20=10000W

6 0
3 years ago
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