Ask question

Ask question

All categories

3 years ago

13

What is another name for the dependent variable in an experiment? A. A manipulated variable B. A responding variable C. A contro

lled variable D. An independent variable

2 answers:

anzhelika [568]3 years ago

8 0

A. a. manipulated

abcdefg

SIZIF [17.4K]3 years ago

7 0

The answer to your question is,

A. A manipulated variable.

-Mabel <3

You might be interested in

Blending three primary colors of light

Tju [1.3M]

Answer: white light

Explanation:

if you mix red and blue light together, you get magenta and when you mix blue and green light together, you get cyan and when you mix red and green light together, you get yellow but when you mix all three primary colors of light together, you end up with white light or as we see it.

5 0

3 years ago

A client with a history of chronic alcoholism is brought to the emergency department after falling. He is disoriented, with peri

Flauer [41]

Answer:

Magnesium, potassium, calcium and sodium

Explanation:

Chronic alcoholism induces renal tubular dysfunction leading to loss of renal Magnesium called Hypomagnesemia. This state often induce peripheral resistance to parathyroid hormone leading to hypocalcemia (reduction of calcium). There is equally loss of or reduction in gastrointestinal absorption causing reduction in phosphate. Magnesium reduction also contribute to reduction in potassium. The reduction in magnesium has been attributed to dysrhythmias

7 0

3 years ago

Read 2 more answers

Radio waves coming from a radio station are tuned into by a student in class. What form of energy transfer does the radio statio

Bezzdna [24]

Radio waves are a type of electromagnetic (EM) radiation with wavelengths in the electromagnetic spectrum longer than infrared light.

8 0

3 years ago

Help me pleaseeeeee?!!!!!!!

Gnom [1K]

There are 8 neutrons and 8 protons in the nucleus of an oxygen atom.

4 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago