Answer:
16 m/s.
Explanation:
The following data were obtained from the question:
Mass of truck = 5000 Kg
Velocity of truck = 8 m/s
Mass of car = 2500 kg
Velocity of car =..?
Next, we shall determine the momentum of the truck. This can be obtained as follow:
Mass of truck = 5000 Kg
Velocity of truck = 8 m/s
Momentum of truck =.?
Momentum = mass × velocity
Momentum = 5000 × 8
Momentum of the truck = 40000 Kg.m/s
Finally, we shall determine the velocity of the car as follow:
From the question given above, we were told that the car and truck has the same momentum.
This implies that:
Momentum of the truck = momentum of car = 40000 Kg.m/s
Thus, the velocity of the car can be obtained as shown below:
Mass of car = 2500 kg
Momentum of the car = 40000 Kg.m/s
Velocity of car =..?
Momentum = mass × velocity
40000 = 2500 × velocity
Divide both side by 2500
Velocity = 40000/2500
Velocity = 16 m/s
Therefore, the velocity of the car is 16 m/s.
Answer:
a) 3.37 x 
b) 6.42kg/
Explanation:
a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .
Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x 
. So density of metal = mass of metal / volume of metal = 5 / 14 x = 3.37 x
b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/
Answer:
it would help if we knew the question and other answers
Explanation:
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.
