Answer:
(d) ATP molecules are produced in the cytosol as glucose is converted into pyruvate.
<span>A boat would
produce the highest concentration of carbon monoxide in the exhaust system.
</span>Carbon monoxide<span> (CO) is a colorless, odorless, and tasteless gas that is
slightly less dense than air. It is toxic to </span>hemoglobic<span> <span>animals (both </span></span>invertebrate<span> <span>and
vertebrate, including humans) when encountered in concentrations above about 35 </span></span>ppm<span>.</span>
The normal force acts to counter the gravitational force, that is the upward direction.
Answer:
we agree with
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
Explanation:
Weight of the ball is given as

so we have


now tension force at the top is given as


Now at the top position by force equation we can say that ball will have two downwards forces
1) Tension force
2) Weight of the ball
so net force on the ball is given as


So we agree with
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
Answer:
The magnetic field along x axis is

The magnetic field along y axis is zero.
The magnetic field along z axis is

Explanation:
Given that,
Length of the current element 
Current in y direction = 5.40 A
Point P located at 
The distance is


We need to calculate the magnetic field
Using Biot-savart law

Put the value into the formula

We need to calculate the value of 



Put the value into the formula of magnetic field


Hence, The magnetic field along x axis is

The magnetic field along y axis is zero.
The magnetic field along z axis is
