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Schach [20]
1 year ago
7

20 points please help!!

Physics
1 answer:
IgorC [24]1 year ago
4 0

Answer:

Sample Response: If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.

Explanation:

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If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.
olya-2409 [2.1K]

Answer:

F=mkv

Explanation:

Given that

v = v_i - kx

We know that acceleration a given as

a=\dfrac{dv}{dt}

v = v_i - kx

\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}

\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}

We know that

F=m\dfrac{dv}{dt}

F=-mk\dfrac{dx}{dt}

F=-mkv

So the magnitude of force F

F=mkv

5 0
1 year ago
When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
gayaneshka [121]

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

5 0
1 year ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr
mash [69]

Answer:

the inductive reactance of the coil is 1335.35 Ω

Explanation:

Given;

inductance of the coil, L = 250 mH = 0.25 H

effective current through the coil, I = 5 mA

frequency of the coil, f = 850 Hz

The inductive reactance of the coil is calculated as;

X_l = \omega L = 2\pi f L\\\\X_l = 2\pi \times 850 \times 0.25\\\\X_l = 1335.35 \ ohms

Therefore, the inductive reactance of the coil is 1335.35 Ω

6 0
1 year ago
A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d
AlladinOne [14]
The current is defined as the amount of charge transferred through a certain point in a certain time interval:
I= \frac{Q}{\Delta t}
where
I is the current
Q is the charge
\Delta t is the time interval

For the lightning bolt in our problem, Q=6.0 C and \Delta t= 2.0 \cdot 10^{-3}s, so the average current during the event is
I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A
4 0
2 years ago
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