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sukhopar [10]
1 year ago
12

(YOU WILL GET BRAINLIEST)Matter may be classified as a pure substance or a mixture. Where on the Venn diagram would you insert t

he term "pure substance"?
A) A
B) B
C) C
D) D

Physics
2 answers:
inessss [21]1 year ago
3 0
I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms
Klio2033 [76]1 year ago
3 0

Answer: Option (B) is the correct answer.

Explanation:

Anything that has space and occupies volume is known as matter.

A pure substance can be molecule of an element or molecule of a compound.

A molecule can have atoms of either same or different element. Whereas a compound has atoms of different elements.

For example, H_{2} is a molecule with similar atoms, that is, hydrogen element.

H_{2}O is a compound that contains hydrogen and oxygen atom.

Thus, we can conclude that pure substance contains molecule of an element or molecule of a compound.

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Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp
maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = \dfrac{N}{C}m

c = Speed of light = m/s

Momentum is given by

\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s

The unit of MeV/c in SI fundamental units is kg m/s

5 0
1 year ago
A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*
4 0
2 years ago
Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)
Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force,(FS_p)s=-k\Delta s-q(\Delta s)^3

k=350N/m

q=750 N/m^3

We have to find the work must you do to compress this spring 15 cm.

\Delta s=15 cm=0.15 m

Using 1 m=100 cm

Work done=\int_{0}^{0.15}-Fd(\Delta s)

W=-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)

W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}

W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750

W=4.033 J

Ideal spring work=0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J

Percentage increase in work=\frac{4.033-3.938}{3.928}\times 100=2.4%

6 0
2 years ago
To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?
koban [17]

<em>To determine the y component of velocity of a projectile </em><u><em>sine </em></u><em>operation is performed on the angle of launch.</em>

<u>Answer:</u> <em>sine</em>

<u>Explanation:</u>

Thus a_x=0,a_y=g

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion v= u+at

Thus velocity along x direction v_x=u cos θ

Velocity along y direction v_y= u sinθ -gt

Sign of g is negative.

3 0
1 year ago
Read 2 more answers
A 0.15-m-diameter pulley turns a belt rotating the driveshaft of a power plant pump. The torque applied by the belt on the pulle
Sveta_85 [38]

Answer:

F= 2666.66 N

N=334.22 RPM          

Explanation:

Given that

Diameter of pulley ,d= 0.15 m

Radius ,r= 0.075 m

Torque ,T= 200 N.m

Power ,P = 7 kW

Lets take force = F

T = F .r

F=\dfrac{T}{r}

Now by putting the values

F=\dfrac{200}{0.075}\ N

F= 2666.66 N

Lets take rotational speed = N RPM

We know that  

P=\dfrac{2\pi N\ T}{60}

N=\dfrac{60\times P}{2\pi \ T}

Now by putting the values in the above equation

N=\dfrac{60\times 7000}{2\pi \times 200}\ RPM

N=334.22 RPM

6 0
1 year ago
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