![HEY THERE !!](https://tex.z-dn.net/?f=%3Cb%3EHEY%20THERE%20%21%21)
![\huge{COMPOUND:-}](https://tex.z-dn.net/?f=%5Chuge%7BCOMPOUND%3A-%7D)
Compounds contain two or more different elements.
Water is a molecule because it contains molecular bonds . Water is also a compound because it is made from more than one kind of element. (oxygen and hydrogen).
So If you like, you can say that water is a molecular compound.
HOPE IT HELPED YOU.
Answer:
Explanation:
Check the attachment for solution
Answer:
The SI units of the “A” is m (meters)
The SI units of the “B” is m/s^2
Explanation:
Given the distance = d meters.
Time taken to travel = t (seconds)
Function of the distance, d = A + Bt^2
Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.
Thus, the SI units of the “A” is = m (meters)
The SI units of the “B” is = m/s^2
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
![m_1u_1+m_2u_2=(m_1+m_2)V](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3D%28m_1%2Bm_2%29V)
V is the common velocity after the collisions
![V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bm_1u_1%7D%7B%28m_1%2Bm_2%29%7D%5C%5C%5C%5CV%3D%5Cdfrac%7B11000%5Ctimes%2021%7D%7B%2811000%2B23000%29%7D%5C%5C%5C%5CV%3D6.79%5C%20m%2Fs)
So, the two car train will move with a common velocity of 6.79 m/s.
Answer:
Yes
Explanation:
Yes, bluetooth devices work in a frequency range between 2.4 - 2.485GHz. Outside this frequency the devices will not communicate with each other correctly. This frequency equals a wavelength of around 1cm. Therefore, any change in the amplitude or wavelength would need to be in relation to each other in order to maintain the frequency in the required range for the bluetooth device to work accordingly. If one increases while the other remains the same it can easily change the frequency to outside the range.