All categories

3 years ago

~choices below~

2 answers:

6 0

Answer : The correct option is, (A) Physical properties can be observed without changing the identity of the substance, while chemical properties can not.

Explanation :

Physical properties : It can be observed or measured without changing the chemical property of the substance.

In the physical property, no new compound or substance is formed only changes occur in the phase.

Chemical properties : It can be observed or measured by changing the chemical property of the substance.

In the chemical property, a new compound or substance is formed and no changes occurs in the phase.

Hence, the correct option is, (A)

mart [117]3 years ago

5 0

A is the correct answer

You might be interested in

Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.

Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is $t=0.00129 s$

suppose u is the incoming velocity and v is the final velocity after collision

$m=mass\ of\ ball$

Impulse exerted is given by change in momentum of the particle.

Initial momentum $P_i=m\times u$

Final momentum $P_f=m\times v$

Change in momentum $\Delta P=P_i-P_f$

Impulse $J=F_{avg}\cdot t=\Delta P$

$J=F_{avg}\cdot t=m(u-v)$

$F_{avg}=\frac{m(v-u)}{t}$

$F_{avg}=775.2\times m(v-u)\ N$

5 0

4 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$