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Trava [24]
3 years ago
10

You volunteer for chalkboard cleaning duty and you get to work with the erasers. As you rub the dust off of the chalkboard, whit

e chalk dust enters the air and makes you cough. As you clean the chalkboard with the eraser, are you causing a physical change or a chemical change? Explain your answer.
Physics
1 answer:
Anna71 [15]3 years ago
3 0

Answer:

A physical change

Explanation:

You are wiping chalk particles from the board to the eraser which is just changing the location of the chalk particles, a physical change.

you're not changing the chalk into a new substance or causing a chemical reaction, which would be a chemical change.

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Suppose, the same angular momentum is transferred to two rotating bodies of different moment of inertia , how will you compare t
Sonbull [250]

Answer:

As per the law of conservation of angular momentum, the angular velocity will be higher for the body with a lower moment of inertia and vice versa.

Explanation:

Angular momentum L of a body is given by:

L=I\times \omega

Now when the same angular momentum is transferred to two different bodies with different moment of inertia, the body with a higher moment of inertia will have lower angular velocity and vice versa.

4 0
3 years ago
AYUDA!!!!!
ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 \frac{g}{cm^{3} } o 790 \frac{g}{cm^{3} }

b) El peso especifico es 7749.9\frac{N}{m^{3} }

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

d=\frac{m}{V}

En este caso:

  • masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
  • volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

d=\frac{237 g}{300cm^{3} } →  d=0.79 \frac{g}{cm^{3} }

d=\frac{0,237 kg}{0,0003m^{3} }→  d=790 \frac{g}{cm^{3} }

<u><em>El valor de la densidad es 0.79 </em></u>\frac{g}{cm^{3} }<u><em> o 790 </em></u>\frac{g}{cm^{3} }<u><em></em></u>

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 \frac{m}{s^{2} }= 2,32497 N

el peso especifico es calculado como:

Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }

Pe= 7749.9\frac{N}{m^{3} }

<u><em>El peso especifico es 7749.9</em></u>\frac{N}{m^{3} }<u><em></em></u>

8 0
3 years ago
A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose
dmitriy555 [2]

Answer:

x=0.478\ m is the compression in the spring

Explanation:

Given:

  • mass of the bullet, m=10\ g=0.01\ kg
  • mass of block, M=2\ kg
  • stiffness constant of the spring, k=19.6\ N.m^{-1}
  • initial velocity of the spring just before it hits the block, u=300\ m.s^{-1}

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

m.u=(M+m).v

0.01\times 300=(2+0.01)\times v

v=1.4925\ m.s^{-1}

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

\rm Kinetic\ energy=Spring\ potential\ energy

\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2

\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2

x=0.478\ m is the compression in the spring

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Answer:

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remove contamination; pollution ; test ground water and dirt .

4 0
3 years ago
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