<h2>
Answer: F=32.2m</h2>
Explanation:
According Newton's 2nd Law of Motion the force
is directly proportional to the mass
and to the acceleration
of a body:
(1)
When we talk about the force of gravity on an object (the weight) the constant acceleration is due gravity, this means:
(2)
Substituting (2) in (1):
(3)
This means the equation that best represents the force on an object due to gravity according to its mass, among the given options is:
![F=32.2m](https://tex.z-dn.net/?f=F%3D32.2m)
Answer:
51.28 grams of fat will gain
Explanation:
Correction: 10,000 kj about 2500 cal not 2500 kcal.
Asumption: basal metabolic rate will depend of weight, lenght, age and gender but its estipulated a 2000 cal basal metabolic rate as a standar for the calculation for nutritional value of food.
eating a 2500 cal in one day will left
Consuming about 2,500 calories in a day where you only sit relaxed for 16 hours and the rest of the time was used to sleep will mean a basal metabolism of 2000 cal, so 500 calories were left unused, taking into account that 1 calorie is about 4 kj we have to
500 cal * 4kj / cal = 2000 kj
The amount of energy stored as fat will be given by the following equation:
Unused energy / energy contained in one gram of fat = 2000 kj / (39 kj / g) = 51.28 grams of fat gain
The first blank: HEAT
The second blank: ELECTRICAL
Of my knowlaedge, the suns light rays are so intense that they bounce off of other planets and shine on the face of the moon giving us the ability ti see the moon at night.
Answer:
The compression is
.
Explanation:
A Hooke's law spring compressed has a potential energy
![E_{potential} = \frac{1}{2} k (\Delta x)^2](https://tex.z-dn.net/?f=E_%7Bpotential%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20%28%5CDelta%20x%29%5E2)
where k is the spring constant and
the distance to the equilibrium position.
A mass m moving at speed v has a kinetic energy
.
So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity
. Knowing that the energy is constant.
![\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m%20v_1%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20d%5E2)
If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:
![2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2](https://tex.z-dn.net/?f=%202%20%2A%20%28%5Cfrac%7B1%7D%7B2%7D%20m%20v_1%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20D%5E2)
But, in the left side we can use the previous equation to obtain:
![2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2](https://tex.z-dn.net/?f=%202%20%2A%20%28%5Cfrac%7B1%7D%7B2%7D%20k%20d%5E2%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20D%5E2)
![D^2 = \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}](https://tex.z-dn.net/?f=%20%20D%5E2%20%3D%20%20%5Cfrac%7B2%20%5C%20%28%5Cfrac%7B1%7D%7B2%7D%20k%20d%5E2%29%7D%7B%5Cfrac%7B1%7D%7B2%7D%20k%7D%20)
![D^2 = 2 \ d^2](https://tex.z-dn.net/?f=%20%20D%5E2%20%3D%20%202%20%5C%20%20d%5E2%20)
![D = \sqrt{2 \ d^2}](https://tex.z-dn.net/?f=%20%20D%20%3D%20%20%5Csqrt%7B2%20%5C%20%20d%5E2%7D%20)
![D = \sqrt{2} \ d](https://tex.z-dn.net/?f=%20%20D%20%3D%20%20%5Csqrt%7B2%7D%20%5C%20%20d%20)
And this is the compression we are looking for