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3 years ago

Which statement best describes the difference between acceleration and velocity?

Physics

1 answer:

Sloan [31]3 years ago

8 0

C. is the answer because acceleration is the change in velocity in time while velocity is speed with a direction

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Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the

ikadub [295]

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $\frac{velocity}{2 *length}$

velocity = $\sqrt{\frac{tension}{mass per unit length} }$

mass per unit length= $\frac{3.5}{1000*1.22}$ =0.00427 $\frac{kg}{m}$

Now calculating velocity v= $\sqrt{\frac{255}{0.00427} }$

=244.3 $\frac{m}{sec}$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $\frac{244.3}{2 *0.7}$ =174.5 hz

6 0

3 years ago

Friction is necessary when you are on a bike to stay

zzz [600]

Answer:

yes friction is needed hope this helps might of been to long tho

7 0

3 years ago

As demonstrated by the Doppler Effect, why does sound increase in pitch as a sound source approaches you?

scZoUnD [109]

I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.

3 0

3 years ago

Read 2 more answers

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

netineya [11]

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

6 0

3 years ago