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2 years ago

How much time does it take for a car travelling at 5m/s in 21 metres

Physics

2 answers:

Paha777 [63]2 years ago

7 0

Answer:

21/5 seconds or 4.2 seconds

Explanation:

Distance=Speed*Time

21=5 *Time

21/5=Time

4.2=time

Flura [38]2 years ago

5 0

Answer:21/5 seconds

Explanation:

Time =speed÷distance

Time=21/5 seconds

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but I think it's a and f

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2 years ago

Sound travels at about 360 meters/sec. What is the wavelength of sound with a

ollegr [7]

<u>Answer:</u>

0.24 m

<u>Explanation:</u>

Given:

Wave velocity ( v ) = 360 m / sec

Frequency ( f ) = 1500 Hz

We have to calculate wavelength ( λ ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > λ = v / f

Putting values here we get:

= > λ = 360 / 1500 m

= > λ = 36 / 150 m

= > λ = 0.24 m

Hence, wavelength of sound is 0.24 m.

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3 years ago

Explain<br> (0) what is meant by regeneration,<br> (ii) why an analogue signal cannot be regenerated

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Explanation:

<h3>1.) Regeneration is the natural process of replacing or restoring damaged or missing cells, tissues, organs, and even entire body parts to full function in plants and animals.</h3>

2.) When noise is added to analogue signals, it usually sounds like background hiss. Such noise can not be removed so the original clean signal can not be re-created or re-generated.

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2 years ago

Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o

muminat

As per above given data

initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

$\Delta v = v_f - v_i$

$\Delta v = -18.8 - 19.3$

$\Delta v = -38.1 km/s$

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Part b)

Now we need to find acceleration

acceleration is given by formula

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

given that

$\Delta v =- 3.81 * 10^4 m/s$

$\Delta t = 2.07 years = 6.53 * 10^7 s$

now the acceleration is given as

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2 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

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3 years ago