Answer:
NFC Near Field Communication
Explanation:
The Near Field Communication is a communication protocol, for extra short distance with a maximum of 10 centimeters, but usually used in 4 to 5 cm. Its intended to be used in contactless pay systems and in transportation card. Actually has been used to transfer multimedia from cell phones and other devices. The maximum data rate is around 424 kbit/s, with mean in 250 Kbp
Answer:
<em>765,000Joules or 765kJ</em>
Explanation:
The Quantity of heat required is expressed as;
Q = (mcΔt)al + (mcΔt)water
m is the mass
c is specific heat capacity
Δt is the change in temperature
Q = (3(900)(90-5)) + (1.5(4200)(90-5))
Q = 2700*85 + 6300*85
Q = (2700+6300)85
Q = 9000*85
<em>Q = 765,000</em>
<em>Hence the amount of energy needed is 765,000Joules or 765kJ</em>
Frequency is given by the formula cycles÷seconds
So if the mosquito can complete 600 cycles in one second, our formula will be
f=600÷1
f=600!
The frequency is 600 Hz
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know
where:
a= acceleration
we have:
initial velocity,
Using the eq. of motion:
where:
v= final velocity after the separation of spring with the bullet.
(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,
∵At maximum height the final velocity will be zero
Using the equation of motion:
where:
h= height
g= acceleration due to gravity
is the height from the release position of the spring.
So, the height from the latched position be:
Answer:
If by 1.5 MJ you mean 1.5E6 Joules then
W = P t = power X time
W / t = P power
P = 1.5E6 J / 600 sec = 2500 J / s
P = I V
a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps
b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs
c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules
