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bagirrra123 [75]
3 years ago
8

Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop.

B speed up. C keep moving at the same speed. D follow a circular path.
Physics
1 answer:
andrew11 [14]3 years ago
5 0

Answer:

A slow down and stop

Explanation:

When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.

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dolphi86 [110]

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3 0
3 years ago
A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
wolverine [178]
Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.

Let
a =  average acceleration,
t = time taken to attain final speed.

Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²

Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s

Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).
8 0
3 years ago
The haystack maker tractor lifted 120 kg of grass to a height of 5 meters in 6 seconds. Calculate the power of the haystack make
Gelneren [198K]

Answer:

1000 Joules per second.

Explanation:

To lift that much grass 5 meters higher than it was before the tractor did work that is the product of the (force needed to overdo gravity) and (displacement):

W = mg\Delta h=120kg\cdot 10\frac{m}{s^2}\cdot 5m = 6000J

Power is work over the amount of time:

P = \frac{W}{\Delta t}=\frac{6000J}{6 s} = 1000 \frac{J}{s}

4 0
3 years ago
A liquid of density 1288 kg/m3 flows with speed 2.88m/s into a pipe of diameter 0.24 m. The diameter of the pipe decreases to 0.
Tju [1.3M]

Answer:

66.35m/s

Explanation:

Para resolver el ejercicio es necesario la aplicación de las ecuaciones de continuidad, que expresan que

A_1V_1 =A_2 V_2

From our given data we can lower than:

R_i = \frac{0.24}{2} = 0.12m

R_f = \frac{0.05}{2} = 0.025m

So using the continuity equation we have

A_1V_1 =A_2 V_2

V_2 = \frac{A_1V_1}{A_2}

V_2 = \frac{(\pi(0.12^2))(2.88)}{(\pi (0.25)^2)}

V_2 = 66.35m/s

Therefore the velocity at the exit end is  66.35m/s

8 0
3 years ago
Calculate the density of a material that has a mass of 52.457 and a volume of 13.5
Arte-miy333 [17]
The density is about 3.88. You just have to divide the mass and the volume. You can check this on a calculator, too. Hope this helped
7 0
3 years ago
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