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4 years ago

What can you infer about copper and silver based on their position relative to each other?

Physics

2 answers:

horrorfan [7]4 years ago

6 0

Both are metals and are good conductors of electricity and heat.

Alenkasestr [34]4 years ago

6 0

Answer:

Explained

Explanation:

Both silver and copper are placed in the same group in the periodic table. In the D block. The elements in the same group have almost similar physical and chemical properties.

Both element have only one electron in their outer shell. Both are very good conductor of electricity and heat. Both are ductile in nature as both can be drawn into wire. The atomic number of Cu is 29 and that of the Ag is 47. Both are highly malleable in nature.

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A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.

Blizzard [7]

Answer:

Approximately $1.6\times 10^{3}\; \rm N$ .

Explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum will be equal to the impulse that is applied to her.

The momentum $p$ of an object is equal to the product of its mass $m$ and velocity $v$ . That is: $p = m \cdot v$ .

Let $v(\text{before})$ and $v(\text{after})$ represent the velocity of the woman before and after the landing. Let $m$ represent the woman's mass.

The woman's momentum before the landing would be $m \cdot v(\text{before})$ .
The woman's momentum after the landing would be $m \cdot v(\text{after})$ .

Therefore, the change in this woman's momentum would be:

$\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}$ .

On the other hand, impulse is equal to force multiplied by the duration of the force. Let $F$ represent the average force on the woman. The impulse on her during the landing would be $F \cdot t$ .

Apply the Impulse-Momentum Theorem.

Impulse: $F\cdot t$ .
Change in momentum: $m \cdot (v(\text{after})- v(\text{before}))$ .

Impulse is equal to the change in momentum:

$F \cdot t = m \cdot (v(\text{after})- v(\text{before}))$ .

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, $v(\text{after}) = 0\; \rm m \cdot s^{-1}$ .

$\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}$ .

3 0

4 years ago

The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con

faust18 [17]

When light travels from a medium with greater refractive index $n_1$ to a medium with smaller refractive index $n_2$ , there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
In this problem, we know that the critical angle is $\theta_c = 53.7^{\circ}$ , so we can find the ratio between the refractive indices of the two mediums:
$\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81$
and since the second medium is air (n=1.00), the refractive index of the first medium is
$n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23$

In the second part of the problem, we have light entering from air ( $n_i = 1.00$ ) at angle of incidence of $\theta_i = 45.0 ^{\circ}$ , into the second medium with $n_r = 1.23$ . By using Snell's law, we can find the angle of refraction of the light inside the medium:
$n_i \sin \theta_i = n_r \sin \theta_r$
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574$
$\theta_r = \arcsin(0.574)=35.1^{\circ}$

3 0

3 years ago

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripe

jonny [76]

To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:

$a_c= \frac{v^2}{r}=r\omega^2$