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Novosadov [1.4K]
3 years ago
10

After training for a marathon, you would like to complete your first attempt of the 26.22 mile race in under 4.800 hours. At wha

t pace is the slowest you could run and still finish by your goal time (provide your answer in miles/s)?
Physics
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

<em>1.52 x 10^-3 miles per sec</em>

<em></em>

Explanation:

distance to be covered = 26.22 miles

targeted time = 4.8 hours

4.8 hours in seconds = 4.8 x 60 x 60 = 17280 sec

Speed = distance/time

speed = 26.22/17280 = <em>1.52 x 10^-3 miles per sec</em>

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Which blood component stops the bleeding after a cut?
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1. Is there a relationship between the volume of water displaced and the total volume of the block that has anything to do with
strojnjashka [21]

Answer:

The volume of the block is equal to the volume of water displaced by the block.

Explanation:

Volume refers to the amount of space occupied by a given object (in this case the block). When an object such as the block is immersed in water, it displaces its own volume of water. This volume of water displaced is equal to the volume of the block. Hence we can write;

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3 years ago
plz i need help fast................a motion along a straight line with a constant increase in velocity is ?​
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Answer:

uniform acceleration

Explanation:

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Hope this helps.

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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A   (N/m²)

P1=P2

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

F_{2} = \frac{F_{1}*A_{2}  }{A_{1}}  Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

F₂ = 10000 N

W₂= F₂= 10000 N

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3 years ago
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