The process of cellular respiration begins with molecules of _______ and ends with the production of ______

A rocket started from the rest and goes straight up with an acceleration of 4.0 m/s^2 for the first 100 seconds. Then it coasts

Dimas [21]

Answer:

idk

Explanation:

3 0

3 years ago

Read 2 more answers

If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the

Mnenie [13.5K]

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

3 0

3 years ago

Convert 0.0345mW<br> to MW

Rina8888 [55]

Answer:

3.45e-11MV

that is ur answer

6 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(c) v = 0.31m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

4 years ago

Calculate the magnitude of acceleration of a block that has a mads of 2.0 kg and is pulled along a table by a force of 15 N. The

Alexxx [7]

Answer:

5.05 ms⁻²

Explanation:

(check the graph)

when the block is pulled with a force , <em>force of kinetic friction (Fr)</em> appears resisiting the motion.

Fr = μR = μ*mg (where g- gravitational acceleration = 9.8 ms⁻²)

= 0.25 * 2 * 9.8

=4.9 N

According to Newton's second law of motion

we can apply F = ma to the motion assuming the block is moving right

F = ma (a - acceleration)

15 - 4.9 = 2*a

a = 5.05 ms⁻²

5 0

3 years ago

The process of cellular respiration begins with molecules of _______ and ends with the production of _______.

The process of cellular respiration begins with molecules of _ and ends with the production of _.