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RideAnS [48]
3 years ago
9

Suppose a treadmill has an average acceleration of 4.7x10^-3 m/s. a)how much does its speed change after 5min? b)if the treadmil

l's speed 1.7 m/s, what will its final speed be?
Physics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

a)Change in the speed  = 1.41 m/s

b)The final speed will be 3.11 m/s

Explanation:

Given that

Acceleration ,a= 4.7 x 10⁻³ m/s²

a)

We know that

v= u + a t

v=final speed ,u=initial speed

t= time ,a= acceleration

Change in the speed

v- u = a t

t= 5 min  = 5 x 60 s = 300 s

v- u = 4.7 x 10⁻³ x 5 x 60 m/s

v-u = 1.41 m/s

Change in the speed  = 1.41 m/s

b)

Given that

u= 1.7 m/s

v-u = 1.41 m/s

v= 1.7 + 1.41 m/s

v=3.11 m/s

The final speed will be 3.11 m/s

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Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in t
Over [174]
What are the choices ? 

Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation.  A few of them might be . . .

-- Kevin will have no trouble getting back in time for dinner.

-- Kevin will have no time to enjoy the scenery along the way.

-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.

           -- Speed = (distance covered) / (time to cover the distance) .

If time to cover the distance is zero, then speed is huge (infinite).

           -- Kinetic energy = (1/2) (mass) (speed)² .

If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.

         -- Mass = (Kevin's rest-mass) / √(1 - v²/c²)

-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
     if he ever reached Planet-Y, nobody could see him anyway.
8 0
3 years ago
Read 2 more answers
How long will it take a plane to fly 1256km<br> if it travels 500km/hr?
expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

d is the distance

v is the velocity

From the question we have

t =  \frac{1256}{500}  \\  = 2.512

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0
3 years ago
What is produced as the result of unequal warming of the earth's surface?
Soloha48 [4]
Unusual precipitation patterns 
7 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
As you can see, my cousin has a lot of hair. He uses an 1800 W blow dryer and it takes him
maw [93]

Power = 1800W (or 1.8KW by dividing by 1000)

Time = 3 hours

Power = energy/ time

1.8KW = energy/ 3

x3

5.4Kw/h= energy

(5.4KJ or 5400J used)

$0.15 Kw/h

$0.15 X 5.4 = 0.81

Thus, cost $0.81

Hope this helps!

5 0
2 years ago
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