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3 years ago

The law of conservation of mass states that mass can never be created or destroyed, only changed. true or faulse

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

true

Explanation:

The law of conservation of mass states that in a chemical reaction mass is neither created nor destroyed. For example, the carbon atom in coal becomes carbon dioxide when it is burned. The carbon atom changes from a solid structure to a gas but its mass does not change.

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Which statements about kinetic energy are true?

Kruka [31]

Answer: All 3 but the last one

Explanation:

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3 years ago

A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti

Zigmanuir [339]

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: $Vf^{2} = Vo^{2} - 2gh$ and $h=Vo*t - \frac{g*t^{2} }{2}$ , First we need to find the height to time iqual to 3.04 s using the formula: $h=Vo*t - \frac{g*t^{2} }{2}$ and remember that golf ball was released from the rest (Vo= 0 (m/s)) so $h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}$ , we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using $Vf^{2} = Vo^{2} - 2gh$ solving for Vf, we get: $Vf = \sqrt{Vo^{2}-2*g*h }$ replacing the values given $Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }$ , so we get: Vf = 29.79 (m/s).

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3 years ago

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An unknown element shiny in color, malleable, conducts electricity, and is in group 2 on the periodic table is a non-metal.

kaheart [24]

Ummm I believe true..

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3 years ago

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

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3 years ago

A particle with a charge of −1.24×10−8c is moving with instantaneous velocity v⃗ = (4.19×104m/s)i^ + (−3.85×104m/s)j^ . part a w

Bezzdna [24]

Answer:

F = 0i (in the x-direction), 0j (in the y-direction),-8.59*10^-4 N k (In the z-direction)

Explanation:

The force given by charged particles moving in a magnetic field is given below (cross is cross product, they don't have that format in the equation tool):

$F=qv (cross) B\\$

Now we can perform the cross product between v and B

$v(cross)B = \left[\begin{array}{cc}4.19*10^{4} &-3.85*10^{4}\\1.8&0&\en[tex]v(cross)B = 69300 (kg*m/(s^2*C))\\$ d{array}\right][/tex]

Now multiply by Q (charge) to get the force

$F = -1.24*10^-8 * 69300\\F = -8.59*10^-4N$

F = -8.59*10^-4 N k

F = 0i, 0j, (-8.59*10^-4)k

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3 years ago

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