Ask question

Ask question

All categories

2 years ago

11

Ammonia and _________ are waste products of animal cells. A. oxygen B. urine C. hydrogen D. carbon dioxide

1 answer:

pickupchik [31]2 years ago

7 0

B or D, one of these are the correct answer

You might be interested in

ILL ADD BRAINLIEST

Marina86 [1]

This is kinda confusing. I wish u just to a screenshot of the problem but here goes...

Forest at highest latitudes- Hardwood trees/deer, squirrel, foxes

Praries/temperate climate- Mostly small mammals/scrubs/steppes

High humidity/rainfall near equator- Abundant thick vegatation/manny species

No trees/ polar bears/ mosses- 25cm rain/few animals

6 0

3 years ago

Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c

amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

$F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F$

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0

3 years ago

What Is one of many factors that can determine the rate of alcohol absorption

Alla [95]

The percentage of the drink that finds the target and lands in your mouth.

5 0

2 years ago

The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li

melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate $Q$ of a hypodermic needle:

$Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}$ (1)

Where the dimensions of each one is:

Volume flow rate $Q=\frac{L^{3}}{T}$

Radius of the needle $R=L$

Length of the needle $L=L$

Pressures at opposite ends of the needle $P_{2}$ and $P_{1}=\frac{M}{LT^{2}}$

Viscosity of the liquid $\eta=\frac{M}{LT}$

We need to find the value of $n$ whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

$\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}$ (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}$ (3)

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}$ (4)

As we can see $n$ must be 4 if we want the exponent to be 3:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}$ (5)

Finally:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}$ (6)

8 0

3 years ago

A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

y (x, t) = A*cos(k*x -w*t)

Where k = stiffness and w = angular frequency

Hence,

k = 172 and w = 2730

- Calculate wave speed V:

V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

T = W = Y*V^2 = (w/L*g)*V^2

W = (0.0126 / 1.8*9.81)*(13.78)^2

W = 0.135 N

4 0

3 years ago