What would happen if a solid temperature increases

A car speedometer has a 6% uncertainty. What is the range of possible speeds when it reads 100 km/h?

MAVERICK [17]

<span>Range = 88.5 Km/h - 94.5 Km/h</span><span>
</span>

3 0

3 years ago

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

cluponka [151]

Answer:

The thickness is $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 480 \ nm = 480*10^{-9} \ m$

The first order of the dark fringe is $m_1 = 16$

The second order of dark fringe considered is $m_2 = 6$

Generally the condition for destructive interference is mathematically represented as

$y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=> $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=> $\Delta y = 2.4 *10^{-6} \ m$

8 0

3 years ago

If you were trying to build a soundproof room, which of the following materials would you choose in order to absorb the most sou

leonid [27]

Explanation :

Absorption coefficient of a material determines how much sound is absorbed by the material.

To build a soundproof room, Heavy curtains and carpet can be used. They reduce reverberation.

Reverberation means an echoing sound which persists for some time. For example, when we bang on a huge piece of metal, we hear the reverberation even after we stop banging.

Hence, option (A) and (D) are correct.

5 0

4 years ago

Read 2 more answers

An object has a mass of 60 kg. It decelerates from 50 m/s to 20 m/s when a resultant force of 300 N acts on it. For how long doe

Pavlova-9 [17]

Force=300N
Mass=60kg

Find Acceleration

$\\ \rm\longmapsto F=ma$

$\\ \rm\longmapsto a=\dfrac{F}{m}$

$\\ \rm\longmapsto a=\dfrac{-300}{60}$

$\\ \rm\longmapsto a=-5m/s^2$

Now

u=50m/s
v=20m/s

Using 1st equation of kinematics

$\\ \rm\longmapsto v=u+at$

$\\ \rm\longmapsto t=\dfrac{v-u}{a}$

$\\ \rm\longmapsto t=\dfrac{20-50}{-5}$

$\\ \rm\longmapsto t=\dfrac{-30}{-5}$

$\\ \rm\longmapsto t=6$

4 0

2 years ago

Time left 0:43:35

ss7ja [257]

Answer:

Bro where is image Hope you understand me

7 0

3 years ago