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2 years ago

What are the ending materials produced from cell respiration

Physics

2 answers:

Mnenie [13.5K]2 years ago

8 0

Answer:

The products of cellular respiration are carbon dioxide and water

Explanation:

Lunna [17]2 years ago

5 0

Carbon dioxide and water :) hope this helped!

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A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s

zmey [24]

Answer:

The angle is 65.6°.

Explanation:

Given that,

Speed = 2.20 m/s

Distance from the shore= 500 m

Distance from the bottom= 1100 m

Speed of boat = 7.30 m/s

According to figure,

We need to calculate the angle with shore

Using formula of angle

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\tan\theta=\dfrac{500}{1100}$

$\theta=\tan^{-1}(\dfrac{500}{1100})$

$\theta=24.4^{\circ}$

We need to calculate the angle

$\alpha=90-\theta$

Put the value into the formula

$\alpha=90-24.4$

$\alpha=65.6^{\circ}$

Hence, The angle is 65.6°.

5 0

3 years ago

A golfer hits a golf ball with a club head velocity of 94 mph. Mass of golf club head (m): 190 g Mass of golf ball (ms): 46 g Co

MAXImum [283]

Answer:

Velocity of golf ball≅ 12.87mph

Explanation:

Using the theory of conservative momentum to both the club head and the golf ball we have;

Qi1 + Qi2= Qf1 + Qf2

Qi1: initial momentum for the club head

Qf1: final momentum for the club head

Qi2: initial momentum for the golf ball

Qf2: final momentum for the golf ball

momentum (Q) = Mass x velocity

which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.

using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since

0> e=0.83 >1.

Qi1= 190 x 94= 17,860 mph.g

Qf1= 190 x Vf1

Qi2= 46 x 0= 0 mph.g

Qf2= 46x Vf2

Using the value of e to determine Vf2 as the final velocity of the golf ball:

e= $\frac{Vf2- Vf1}{Vi1 - Vi2}$

0.83= $\frac{Vf2 - Vf1}{94 - 0}$

Vf1= (0.83 x 94)+ Vf2

Vf1= 78.02 + Vf2

Qi1 + Qi2= Qf1 + Qf2

17,860 + 0 = 190xVf1 + 46xVf2

17,860= 190x (78.02+Vf2) + 46xVf2

17,860= 14,823+ 190xVf2+46xVf2

Vf2≅ 12.87mph

7 0

3 years ago

A 1 kg object can be accelerated at 10 m/s^2. If you apply this same force to a 4kg object, what will its acceleration be?

AleksAgata [21]

Answer:

$a2 = 2.5~m/s^2$

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force F exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

Assume we apply some given force F to an object of m1=1 Kg that produces an acceleration $a1=10 m/s^2$ , then:

F = m1.a1

The same force F is now applied to a second object m2=4 Kg that produces an acceleration a2, then:

F = m2.a2

Dividing both equations:

$\displaystyle 1=\frac{m1.a1}{m2.a2}$

Solving for a2:

$\displaystyle a2=\frac{m1.a1}{m2}$

Substituting values:

$\displaystyle a2=\frac{1*10}{4}$

$a2 = 2.5~m/s^2$

7 0

2 years ago

Emilio tries to jump to a nearby dock from a canoe that is floating in the water.Instead of a canoe that is floating in the wate

elixir [45]

Oooo that ones hard. ummm... idk i think we should just leave it to the experts ya know.

7 0

3 years ago

Sports managers have the potential to earn more than a million dollars per year.<br> True<br> false

andriy [413]

Answer:

true

Explanation:

hope this helped!

4 0

2 years ago