Answer: Hey
B an encyclopedia entry about an experiment
A primary resource is information that was documented during an event that was written about.
Please wait tell you get another answer because im not sure about this
Answer:
Grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules = 54.3 g
Explanation:
Heat required to increase temperature
H = mcΔT
m = mass of material
C = specific heat of material
ΔT = Change in temperature.
Here we need to find how many grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules
That is
H = 12500 J
Specific heat of water, C = 4186 J/kg°C
ΔT = 75 - 20 = 55
Substituting
12500 = m x 4186 x 55
m = 0.0543 kg
m = 54.3 g
Grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules = 54.3 g
Because the ones that have better adaptions survive and spread on the genes for that adaptation
Answer:
1.97 m/s.
Explanation:
From the question,
Using the law of conservation of energy,
The energy stored in the charged plate = Kinetic energy of the mass
1/2(qV) = 1/2mv².......................... Equation 1
Where q = charge, V = voltage, m = mass, v = velocity.
make v the subject of the equation
v = √(qV/m)......................... Equation 2
Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg
Substitute these values into equation 2
v = √(6.5×10⁻⁶×12000 /0.02)
v = √3.9
v = 1.97 m/s.
Answer:
Explanation:
Given
mass of bus along with travelers travelling in North direction is ![m_1=1.6\times 10^4 kg](https://tex.z-dn.net/?f=m_1%3D1.6%5Ctimes%2010%5E4%20kg)
speed of bus towards North ![v_1=15.2 km/h\approx 4.22\ m/s](https://tex.z-dn.net/?f=v_1%3D15.2%20km%2Fh%5Capprox%204.22%5C%20m%2Fs)
mass of bus travelling in South direction is ![m_2=1.578\times 10^4 kg](https://tex.z-dn.net/?f=m_2%3D1.578%5Ctimes%2010%5E4%20kg)
speed of bus ![v_2=12.2 km/h\approx 3.38\ m/s](https://tex.z-dn.net/?f=v_2%3D12.2%20km%2Fh%5Capprox%203.38%5C%20m%2Fs)
mass of each Passenger in south moving bus ![m_0=64.8 kg](https://tex.z-dn.net/?f=m_0%3D64.8%20kg)
Momentum of North moving bus
![P_1=m_1\times v_1](https://tex.z-dn.net/?f=P_1%3Dm_1%5Ctimes%20v_1)
![P_1=1.6\times 10^4\times 4.22](https://tex.z-dn.net/?f=P_1%3D1.6%5Ctimes%2010%5E4%5Ctimes%204.22)
![P_1=6.768\times 10^4 kg-m/s](https://tex.z-dn.net/?f=P_1%3D6.768%5Ctimes%2010%5E4%20kg-m%2Fs)
Momentum with south moving bus
![P_2=m_2\times v_2+n\cdot m_0\times v_2](https://tex.z-dn.net/?f=P_2%3Dm_2%5Ctimes%20v_2%2Bn%5Ccdot%20m_0%5Ctimes%20v_2)
![P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38](https://tex.z-dn.net/?f=P_2%3D%281.578%5Ctimes%2010%5E4%2Bn%5Ccdot%2064.8%20%29%5Ccdot%203.38)
For total momentum to be towards south
should be greater than 0
thus for least value of n
![P_2=P_1](https://tex.z-dn.net/?f=P_2%3DP_1)
![(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4](https://tex.z-dn.net/?f=%281.578%5Ctimes%2010%5E4%2Bn%5Ccdot%2064.8%20%29%5Ccdot%203.38%3D6.768%5Ctimes%2010%5E4)
![1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4](https://tex.z-dn.net/?f=1.578%5Ctimes%2010%5E4%2Bn%5Ccdot%2064.8%3D2.0023%5Ctimes%2010%5E4)