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3 years ago

describe how a physical property, such as mass or texture, can change without causing a change in the substance

1 answer:

6 0

On the mass subject you could simply change the weight, on the texture subject you could smooth it out like polishing fresh wood, or you could change the color of water with a drink-mix packet. Hope it helped!

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If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

aksik [14]

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

8 0

3 years ago

The diagram below shows different weights on the see-saw. Will the see-saw move?

oee [108]

Answer:

yes it will

Explanation:

8 0

2 years ago

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

6 0

3 years ago

What is the energy difference between the second excited state and first excited state of an electron in the "box" of size L=1nm

jolli1 [7]

I thinks it’s A, tell me if you get it right

4 0

3 years ago

An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

Leya [2.2K]

Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
$F=qvB \sin \theta$
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
$\theta$ the angle between the directions of v and B.

Re-arranging the formula, we find:
$v= \frac{F}{qB \sin \theta}$
and by substituting the data of the problem (the charge of the electron is $q=1.6 \cdot 10^{-19} C$ ), we find the velocity of the electron:
$v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s$

4 0

3 years ago