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3 years ago

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a motorcycle is trying to leap across the canyon by driving horizontally off a cliff 38 m/s. Ignoring air resistance, find the s

peed with which the cycle strikes the ground on the other side

1 answer:

vaieri [72.5K]3 years ago

3 0

Answer:

46.15m/s

Explanation:

We are given that

Initial speed,u=38 m/s

Height of cliff,h=70 m

We have to find the speed with which the cycle strikes the ground at a height 35 m on the other side.

h'=35 m

Using conservation law of energy

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2+mgh'$

$\frac{1}{2}u^2+gh=\frac{1}{2}v^2+gh'$

Where $g=9.8m/s^2$

Substitute the values

$\frac{1}{2}(38)^2+9.8\times 70=\frac{1}{2}v^2+9.8\times 35$

$\frac{1}{2}v^2=\frac{1}{2}(38)^2+9.8\times 70-9.8\times 35=1065$

$v=\sqrt{2\times 1065}$

$v=46.15m/s$

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Answer:

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Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

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here we know that

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now we will have

$d = \frac{(10 + 0)}{2}\times 1.2$

$d = 5 \times 1.2$

$d = 6.0 m$

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3 years ago

If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Alex

Answer:

0.015 m/s2

Explanation:

Using Newtons 2nd law.

F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.

So substitute the values in SI units.

m = $\frac{3}{1000}$ kg

Therefore F = 0.003×5 = 0.015 m/s2

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3 years ago

Singly charged gas ions are accelerated from rest through a voltage of 10.3 V. At what temperature (in K) will the average kinet

Natasha_Volkova [10]

Answer:

Temperature of the gas molecules is 7.96 x 10⁴ K

Explanation:

Given :

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Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = $\frac{3}{2}kT$

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

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Substitute the suitable values in the above equation.

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T = 7.96 x 10⁴ K

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3 years ago

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

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Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

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$\frac{10620.75}{4186} =2.537 Kcal$

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

x ---> 100%

$x=\frac{2.537*100}{20} =12.685$

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

$\frac{9000Kcals}{12.685Kcals} =709.5 times$

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

$Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\$

$P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]$

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spayn [35]

C. increased stirring i think

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3 years ago

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