All categories

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!

You might be interested in

A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2

madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0

3 years ago

Which substance might lead to potentially addiction if abuse

uranmaximum [27]

Heroin is the number 1 most addictive substance.

4 0

3 years ago

Read 2 more answers

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f

sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

$f=\frac{v_s}{2L}$ (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

$f=\frac{343m/s}{2(0.47228m)}=362.36Hz$

Closed tube:

$f'=\frac{v_s}{4L'}$

L': length of the closed tube = 0.702821m

$f'=\frac{343m/s}{4(0.702821m)}=122.00Hz$

Next, you use the following formula for the beat frequency:

$f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz$

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0

3 years ago

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
$V=IR$
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length $L=2.90 m$ , while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

7 0

4 years ago

How much work is done in accelerating a 2000 kg car from rest to a speed of 30 m/s?

Otrada [13]

The work done is the same as the amount of energy increase. The formula for kinetic energy is 122
1
2
m
v
2
.

The initial KE of the car is 12(1000)×202=200,000
1
2
(
1000
)
×
20
2
=
200
,
000
joules.

The final KE of the car is 12(1000)×302=450,000
1
2
(
1000
)
×
30
2
=
450
,
000
joules.

The difference between these is the amount of work done: 450,000−200,000=250,000
450
,
000
−
200
,
000
=
250
,
000
joules.
14.2K viewsView upvotes

20

3

8 0

3 years ago