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Zigmanuir [339]
3 years ago
11

1.What is a non example of constant? 2.What is an example of constant?

Physics
2 answers:
ser-zykov [4K]3 years ago
6 0

Answer:

a science experiment, the controlled or constant variable is a variable that does not change. For example, in an experiment to test the effect of different lights on plants, other factors that affect plant growth and health, such as soil quality and watering, would need to remain constant. a dependant variable(opposite)would be the properties of the plants themselves, which the scientist is observing in relation to the changing light. These properties might include the plants' color, height and general health of the plant

hope this helped

Dafna1 [17]3 years ago
3 0

Answer:

1. A non example of a constant in physics is a value that changes. For example, the temperature of Hong Kong, let's say, is not a constant, since it changes and does not stay at the same value. It gets higher in the summer and lower in the winter.

2. An example of a constant in physics would be the gravitational acceleration of the Earth. Since the Earth's mass' changes is negligible when compared with the overall mass of the Earth (unless something big happens, of course, like a meteor or asteroid), the gravitational acceleration on Earth is largely unchanged, even constant from our viewpoint.

Explanation:

Hope this helped!

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Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
How do the chemical properties of the halogens compare to those of the noble gases?
jeyben [28]
The inert gases do not ordinarily for compound. the halogens are extremely reactive whereas the inert gases tends to be unreactive....hope helps
5 0
3 years ago
A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?
Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, \theta=21^{\circ}

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0
3 years ago
Object 1 has a mass of 3m and is moving to the right at a velocity of
andrew11 [14]

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

6 0
3 years ago
Nabil wants to work out the density of a pebble. he has a displacement can and a beaker. What other piece of equipment will he n
Svetllana [295]

Answer:

1) Measuring cylinder

2) Top Pan Balance

Explanation:

To find the density of the pebble Nabil should first accurately measure the mass of the pebble using the top pan balance

He should then fill the displacement can till the water reaches the outlet pipe at the to of the cylinder

He should then place the pebble into the cylinder carefully ans allow the water that would start coming out to stop

The volume of water in the displaced cylinder is then measured and the density is found by dividing the mass of the pebble by the volume of water in the measuring cylinder

Density \ of \ pebble =  \dfrac{Pebble \ mass}{Volume \ of \ water \ in\ the \ measuring \ cylinder} =  \dfrac{Pebble \ mass}{Water \ volume \ displaced}

3 0
3 years ago
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