All categories

3 years ago

This nutrient can be hidden in food and has twice as many calories as other nutrients

Physics

1 answer:

vfiekz [6]3 years ago

6 0

Protein has 4 cals carbs have 4 too. Fat however has 9.

You might be interested in

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

Resistance of a light bulb with 0.33 A of current flowing from a 12V battery?

Sergio039 [100]

Resistance = (voltage) / (current)

Resistance = (12v) / (0.33 A)

Resistance = (12/0.33) ohms

Resistance = 36.4 ohms

8 0

3 years ago

choose the correct anwer 9: the range of the gravitional force is given by A: 10_2m B:10_15m C: infinite D: 10_10m

frosja888 [35]

Answer:

The answer is C

Explanation:

The magnitude of the gravitational force depends inversely on the square of the radial distance between the centers of the two masses. Thus, essentially, the force can only fall to zero, when the denominator that is r becomes infinite.

6 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

What happens to gravity when someone jumps up?

Alla [95]

Answer:

The direct answer to the question as written is as follows: nothing happens to gravity when someone jumps up - gravity continues exerting a force on the body of that particular someone proportional to (mass of someone) x (mass of Earth) / (distance squared). What you might be asking, however, is what is the net force acting on the body of someone jumping up. At the moment of someone jumping up there is an upward acceleration, i.e., an upward-directed force which counteracts the gravitational force - this is the net force ( a result of the jump force minus gravity). From that moment on, only gravity acts on the body. The someone moves upward gradually decelerating to the downward gravitational acceleration until they reaches the peak of the jump (zero velocity). Then, back to Earth.

5 0

3 years ago