Una rueda gira con una frecuencia de 530 rpm. Determina la velocidad angular, el periodo y la frecuencia.

n April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.57 km mark at a

True [87]

Answer:

$0.084\ m/s^2$

Explanation:

<u>Given:</u>

Length of the race = 10 km
Distance traveled by Steve in 25 min = 7.57 km
Time interval for the constant acceleration = 60 s
Initial velocity of Steve = 0 m/s

<u>Assume:</u>

u = initial velocity
v = final velocity
t = time interval
a = constant acceleration

Since the person has the instantaneous velocity at 25 min equivalent to the average velocity of the person during this time interval. So, let us find out the average velocity of the person till 25 min time interval.

$v_{avg} = \dfrac{7.57\ km}{25\ min}\\\Rightarrow v_{avg} = \dfrac{7.57\times 1000\ m}{25\times 60\ s}\\\Rightarrow v_{avg} = 5.047\ m/s$

Since the person moves with a constant acceleration for the first 60 s and then moves with a constant velocity after this instant. This means the final velocity of the person at the end of 60 s is 5.047 m/s.

$\therefore v = 5.047\ m/s\\u = 0\ m/s\\t = 60\ s\\$

Now using the equation of constant acceleration, we have

$v = u +at\\\Rightarrow a = \dfrac{v-u}{t}\\\Rightarrow a = \dfrac{5.047-0}{60}\\\Rightarrow a =0.084\ m/s^2$

Hence, the acceleration of Steve in the 60 s interval is $0.084\ m/s^2$ .

3 0

2 years ago

Bill is throwing a football at four targets and attempting to knock them over. Which of the following targets will be hardest fo

Naddika [18.5K]

100 g lead target. Lead has a very high density which means even a very small volume and weight quite a lot. So because the lead target weighs only 100 g it's going to be small in size compared to other targets which are all made of lesser dense materials than lead.

7 0

3 years ago

Accuracy of physical balance is dash

Iteru [2.4K]

Answer:

the answer for this is my.

6 0

2 years ago

Speed is a component of skill related fitness. what does speed enable you to do

AveGali [126]

To complete a task in a short amount of time/get from A to B in the quickest amount of time

Hope helps!

7 0

3 years ago

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$

Where $c_p$ represent the specific heat for the water and this value from tables we know that $c_p =1 \frac{cal}{gr C}$ for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

8 0

3 years ago

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