22) The sensory cells associated with the deep layers of the epidermis are

Simple curiosity is not a legitimate reason to complete a scientific investigation

guapka [62]

Why not ? ? ?
If you can get the money, the laboratory to use, the experimental
equipment, and the people you need to work on it with you, then
curiosity is a fine, honorable, and perfectly good reason to go ahead,
do what you gotta do, follow your heart, satisfy your curiosity, and learn
new stuff.
Neither the Large Hadron Collider, nor the Mars Rover, nor the Hubble
Space Telescope, nor the New Horizons probe, was built to cure cancer.

5 0

4 years ago

What are five of the most important things you learned in physics? ASAP

scoundrel [369]

Answer:

The definitions of mass, energy, time and space as used in Physics are circular

The Conservation of Mass and Energy

The definition of zero is not sufficiently defined in Physics

Newton’s Laws of Motion are related by Calculus

Knowing the terms of Physics is most important.

Please mark as brainliest

7 0

3 years ago

A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it

Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

$\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}$

$\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}$

$\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81$

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

8 0

3 years ago

Select all the expressions below which represent a 20% discount off the price of an item that originally costs d dollars.

dmitriy555 [2]

A - 0.8d

multiply the item (d) times 80% (which shows a 20% discount since 100%-20%=80%). Convert 80% to a decimal = .8

therefore, 0.8d

7 0

3 years ago

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has dr

Wittaler [7]

Answer:

A. ) K =126. 7 J

B. ) h= 91.1 m.

Explanation:

A)

Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:

$K = \frac{1}{2}*m*v^{2} = \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)$

B)

Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:

$U_{o} = K_{f} = 126. 7 J (2)$

⇒ m*g*h = 126. 7 J

Solving for h, we get:

$h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)$

4 0

3 years ago