3) The energy resulting from motion is called

A tiger is running across a field. What would happen if the hydrogen ion pumps into the tigers mitochondria stopped functioning

galben [10]

Answer:

The tiger would not be able to produce glucose causing it to stop running

Explanation:

Since the mitochondria is in charge of producing ATP the tiger would not be able to use any glucose causing it to not be able to run.

3 0

3 years ago

Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

Sati [7]

Answer:

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$ , $z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$ , $z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

$z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]$

Where angles are measured in radians and k represents an integer between $0$ and $n - 1$ .

The magnitude of the complex number is $27$ and the equivalent angular value is $1.817\pi$ . The set of cubic roots are, respectively:

k = 0

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$

k = 1

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$

k = 2

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

5 0

3 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

7 0

3 years ago

Joey drives along the 110 South freeway and notices a mile marker that reads 260 miles. He drives until he reaches the 150 mile

Nostrana [21]

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150 = 110 miles

Displacement along the 110 North freeway = 150 - 175 = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

8 0

3 years ago

(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim

jeka94

Answer:

Two identical spheres are released from a device at time t = 0 from the same ... Sphere A has no initial velocity and falls straight down. ... (b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time. ... Which ball has the greater vertical velocity

Explanation:

4 0

2 years ago