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ExtremeBDS [4]
2 years ago
13

What is the amount of matter in a substance?​

Physics
2 answers:
kifflom [539]2 years ago
8 0

Answer:

Mass is the amount of matter in a substance

professor190 [17]2 years ago
6 0

Answer:

mass

Explanation:

Mass (M) is the measure of the amount of matter in an object. Mass is measured in grams (g). Mass is measured on a balance by comparing the object against other objects with known masses.

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A 650-kg roller coaster starts from rest at the top of a 80 m hill.
Flura [38]

Answer:

or a roller coaster loop, if it were perfectly circular, we would have a minimum speed of vmin=√gR at the top of the loop where g=9.8m/s2 and R is the radius of the 'circle'. However, most roller coaster loops are actually not circular but more elliptical.

Explanation:

7 0
2 years ago
Guys help me with this question.​
sleet_krkn [62]

Answer:

Explanation:

Brownian motion is a random (irregular) motion of particles e.g smoke particle. The set up in the diagram can be used to observe the motion of smoke.

1. The apparatus used are:

A is a source of light

B is a converging lens

C is a glass smoke cell

D is a microscope

2. The uses of the apparatus are:

A - produces the light required to so as to see clearly the movement of the particles.

B - converges the rays of light from the source to the smoke cell.

C - is made of glass and used for encamping the smoke particles so as not to mix with air.

D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.

4 0
3 years ago
There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
PLZ HELP ME FAST A relationship between two variables is called:
Irina18 [472]

Answer:

B- Correlation

Explanation:

6 0
2 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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