Find the acceleration if a 32.5 N force is used on an object that has a mass of 128.6 kg.

What are the reactants of cellular respiration? Check all that apply. carbon dioxide energy glucose oxygen water

Rufina [12.5K]

Answer:

Glucose and Oxygen

Explanation:

Cellular respiration is the process whereby cells derives energy by the use of glucose and oxygen.

Organisms that use cellular respiration to produce their energy are known as heterotophs. They derive the glucose from food materials obtained from plant sources. They use the oxygen from the environment to liberate energy from the glucose obtained from feeding on plant materials.

Cellular respiration can be simply expressed as shown below:

GLUCOSE + OXYGEN → CO₂ + H₂O + ATP

The reactants are glucose and oxygen.

The products are CO₂, water and ATP

7 0

2 years ago

Read 2 more answers

The electric current leaves the battery passes through the it then travels through the next through the and finally padses throu

miss Akunina [59]

The answers are:

bulb, motor, buzzer and swtich.

As seen in the picture attached,

The electric current leaves the battery passes through the bulb. It then travels through the motor, next through the buzzer and finally passes through the switch before returning to the battery.

8 0

3 years ago

In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to

dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, $a=\dfrac{v-u}{t}$ ....(1)

The third equation of kinematics is as follows :

$v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s$

Hence, it will take 25 seconds to increase the speed.

6 0

3 years ago

An 80- kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15m/s . Sort the fol

torisob [31]

Answer:

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

Explanation:

Given that,

Mass of the quarterback, m = 80 kg

Mass of the football, m' = 0.43 kg

Speed of the football, v' = 15 m/s

We need to sort the following quantities as known or unknown.

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

6 0

3 years ago

Read 2 more answers

Please help me i dont get it

Dima020 [189]

The answers are:

39:

a - 1 mile

b - 0 miles

40:

$a=-4\frac{m}{s^{2} }$

41:

$Displacement=3.60 miles\\Distance=5 miles$

Why?

Solving 39:

To answer the questions, we need to remember that distance and displacement are different things. Distance refers to the total "ground" covered during motion, while displacement refers to how far is the object/body from its starting point.

So,

39:

a - the total distance traveled corresponds to the length of the track which is 1 mile.

b - The displacement is equal to 0 because they finished at the starting point, the distance between the starting and the finishing point is equal to 0.

40:

We can solve the problem using one of the given equations:

$a=\frac{v_f-v_i}{t}$

Since we know all the information, we just need to substitute it into the equation:

$a=\frac{v_f-v_i}{t}\\\\a=\frac{2\frac{m}{s} -16\frac{m}{s} }{3.5s}=\frac{-14\frac{m}{s} }{3.5s}\\\\a=-4\frac{m}{s^{2} }$

So, the car's acceleration was -4m/s2 (the car was reducing its speed)

41:

We can solve the problem using one of the given equations (Pythagorean Theorem):

$c^{2}=a^{2}+b^{2}$

$Displacement=\sqrt{(3mi(North))^{2}+(2miles(East))^{2}}\\\\Displacement=\sqrt{9+4}=\sqrt{13mi^{2}}=3.60miles$

The distance will be:

$Distance=3mi+2mi=5miles$

Have a nice day!

3 0

2 years ago