All categories

4 years ago

Which of the following describes the energy transformation that takes place when a boy eats an apple and then lifts weights?

Physics

2 answers:

zysi [14]4 years ago

5 0

the correct answer is 1. Chemical energy is converted to mechanical and heat energy.

When the boy eats apple, the chemical energy of the apple first converted in the heat energy in the process of digestion, which maintains our body temperature, than the body of the boy converts the heat energy into mechanical energy.

Therefore the correct option is 1.

mezya [45]4 years ago

3 0

Answer:

Chemical energy is converted to mechanical and heat energy.

Explanation:

You might be interested in

Increasing excitatory signals above threshold levels for neural activation will not affect the intensity of an action potential.

ahrayia [7]

Answer:

Option C

An all-or-none response

Explanation:

Increasing excitatory signals above threshold levels for neural activation will not affect the intensity of an action potential indicates that the reaction of a neuron is an all-or-none response

4 0

3 years ago

The table lists the mass and charge of a proton and a neutron.

Olegator [25]

The gravitational force is much larger than the electrical force for any distance between the particles.

Explanation:

The gravitational force between two particles is given by:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two particles

r is the separation between the two particles

The electric force between two particles is given by:

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the two particles

In this problem, we are comparing the gravitational force and the electric force between a proton and a neutron.

We must immediately notice that the charge of the neutron is zero, since it is electrically neutral:

q = 0

This means that the electric force between the proton and the neutron is always zero, regardless of their distance. And therefore, since their masses is not zero (and the gravitational force is never zero), the correct answer is

The gravitational force is much larger than the electrical force for any distance between the particles.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

About electrical force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

GalinKa [24]

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time $t$ , the horizontal position $x$ and vertical position $y$ of the ball are given respectively by

$x = v_i \cos(\theta_i) t$

$y = v_i \sin(\theta_i) t - \dfrac g2 t^2$

and the horizontal velocity $v_x$ and vertical velocity $v_y$ are

$v_x = v_i \cos(\theta_i)$

$v_y = v_i \sin(\theta_i) - gt$

The ball reaches its maximum height with $v_y=0$ . At this point, the ball has zero vertical velocity. This happens when

$v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g$

which means

$y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g$

At the same time, the ball will have traveled half its horizontal range, so

$x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g$

Solve for $v_i$ and $\theta_i$ :

$\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0$

Since $0^\circ$ , we cannot have $\sin(\theta_i)=0$ , so we're left with (e)

$3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}$

Now,

$\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}$

so it follows that (d)

$R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}$

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

$v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}$

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

$v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}$