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liraira [26]
2 years ago
13

19. Assume you have been chewing a piece of bread for 5 or 6 minutes. How would you expect its taste to change during this inter

val? Why?​
Physics
1 answer:
I am Lyosha [343]2 years ago
7 0
Well 1 minute in it would taste like normal bread but 2-3 would the bread will break down and then get really soggy and 4-6 it would be into really tiny peices or almost gone and really wet from your saliva
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A 2.50 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. What is the common velocity of th
yarga [219]

Answer:

V = 2.5*7.0 / ( 2.5 + 70 )

8 0
2 years ago
The bulk modulus of water is 2.2 x 109 Pa, and its density is 1.0 x 103 kg/m3. What is the speed of sound in water?
yarga [219]

hi guys! Lets be friends!

7 0
2 years ago
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so
lubasha [3.4K]

Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

4 0
3 years ago
Use your periodic table to answer the following question.
Pavel [41]
Helium has only 2 electrons in it's valence shell

So, option D is your answer.

Hope this helps!
6 0
3 years ago
Ml(d^2θ/dt^2) =-mgθ
Nata [24]

The equation of motion of a pendulum is:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,

where \ell it its length and g is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (\theta \ll 1), we can use:

\sin\theta \simeq \theta.

Additionally, let us define:

\omega^2\equiv\dfrac{g}{\ell}.

We can now write:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.

The solution to this differential equation is:

\theta(t) = A\sin(\omega t + \phi),

where A and \phi are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.

This justifies that the period depends only on the pendulum's length.

4 0
3 years ago
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