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3 years ago

What energy transformations take place when you eat a taco and then go for a jog

Physics

1 answer:

sesenic [268]3 years ago

6 0

Answer:

Chemical to kinetic and thermal.

Explanation:

You would eat the food (chemical) than you would jog and move around (kinetic). While running your body would also give off heat (thermal).Than your body would sweat to cool itself down.

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An "energy bar" contains 26 g of carbohydrates. For the steps and strategies involved in solving a similar problem, you may view

masha68 [24]

To solve this problem we will apply the definition of Power and Speed. In turn, we will consider that one gram of carbohydrate, according to numerous scientific studies, contributes around 17kJ of energy. Therefore, if this were true, the total energy of 26 grams would be

$E = (26)(17000) = 4.42*10^5J$

Power can be described as the amount of energy applied at a given time, that is,

$P = \frac{E}{t} \rightarrow t = \frac{E}{P}$

$t = \frac{4.42*10^5}{380}$

$t = 1.16*10^3s$

The speed is described as the distance traveled in a certain time, and its units in international system is m / s, converting and replacing we will have

$v = 5km/h(\frac{1000m}{1km})(\frac{1h}{3600s})$

$v = 1.388m/s$

Now,

$v = \frac{d}{t} \rightarrow d = vt$

The distance is,

$d = vt$

$d = (1.388)(1.16*10^3)$

$d = 1610.08m$

Therefore the distance walked is 1610.08m

7 0

3 years ago

A proton and an electron are moving in the +x direction in a magnetic field in the +z

Cloud [144]

Answer:

Explanation:

A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction

V = v•i

Magnetic field Is the positive z direction

B = B•k

A. For proton.

Proton has a positive charge of q

Direction of force on proton

Force is given as

F = q(v×B)

F = q( v•i × B•k)

F = qvB (i×k)

From vectors i×k = -j

F = -qvB •j

Then, for the positive charge, the force will act in the negative direction of the y-axis

B. For electron

Electron has a negative of -q

Direction of force on proton

Force is given as

F = q(v×B)

F = -q( v•i × B•k)

F = -qvB (i×k)

From vectors i×k = -j

F = --qvB •j

F = qvB •j

Then, for the negative charge, the force will act in the positive direction of the y-axis

5 0

3 years ago

A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of

Tresset [83]

Answer: 2.92 s

Explanation:

Given

Mass of ball is $m=3.9\ kg$

The initial velocity of the ball is $u=-3.5\ m/s$

Velocity after the rebound is $v=15.9\ m/s$

Force during the contact is $F=25.9\ N$

We know, change in momentum is Impulse

$\Rightarrow F\cdot \Delta t=m(\Delta v)$

$\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s$

Thus, the force is applied for 2.92 s

4 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface: