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3 years ago

The answer is C, method?

Physics

1 answer:

Arturiano [62]3 years ago

4 0

Answer:

Numbers with more number

Explanation:

am I right or am I right?

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A body starting from rest and moves with uniform acceleration. The velocities of the body are 40 m/s at the 4th second and 60 m/

Sergeeva-Olga [200]

Answer:

GIVEN

u is 40m/s and v=60m/s

time is 4s

therefore acceleration =60-40/4

=20/4=5m/s^2

b)next 2 sec=70m/s

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3 years ago

What human disease can result from exposure to ultraviolet radiation from the sun?

aev [14]

Melanoma skin cancer

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4 years ago

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English schoolteacher who proposed the atomic theory model of matter

vodka [1.7K]

Answer:

Ernest Rutherford.

Explanation:

Ernest Rutherford

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3 years ago

Christopher drives into the city to buy new new hockey equipment. Because of traffic conditions, he averages only 15 mph. On the

alina1380 [7]

Answer:

36 minutes

Explanation:

Distance= Time*Speed

Let time taken to go to city be x hours, therefore, time to travel back is (2-x) hours

The distance to and from the city are equal hence

15x=35(2-x)

15x=70-35x

50x=70 hence x=70/50=1.4 hours

Time to drive home will be 2-x=2-1.4=0.6 hours

0.6*60=36 minutes

Therefore, time to travel back home is 36 minutes

8 0

3 years ago

Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl

tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

$v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s$

For Train 2 Velocity of the Source,

$v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s$

Frequency of Source,

$f_{s}=500\ Hz$

To Find:

Frequency of Observer,

$f_{o}=?$ (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source $v_{s}$ and observer $v_{o}$ , the original frequency of the sound waves $f_{s}$ and the observed frequency of the sound waves $f_{o}$ ,

The Equation is

$f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})$

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

$f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz$

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

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3 years ago