Answer:
38,437.5
Explanation:
Density(d)= 102.5g/ml
Volume (v)=375ml
Mass(m) = ?
D =m/v
102.5= m/375
102.5*375=m
38,437.5=m
therefore Mass = 38,437.5g/ml.
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds
Answer:
x = 76.5 m
Explanation:
Let's use Newton's second law at the point of contact between the wheel and the floor.
fr = m a
fr = miy N
N-W = 0
N = W
μ mg = m a
a = miu g
a = 0.600 9.8
a = 5.88 m / s²
Having the acceleration we can use the kinematic relationships to find the distance
² = v₀² + 2 a x
= 0
x = -v₀² / 2 a
Acceleration opposes the movement by which negative
x = - 30²/2 (-5.88)
x = 76.5 m
Answer:
C. 85%
Explanation:
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?
A. 15%
B. 30%
C. 85%
D. 100%
work done by the system will be
W=PdV
p=pressure
dV=change in volume
3tam will be changed to N/m^2
3*1.01*10^5
W=3.03*10^5*(1.5-1)
convert 0.5L to m^3
5*10^-4
W=3.03*10^5*5*10^-4
W=152J
therefore
to find the percentage used
152/1000*100
15%
100%-15%
85% uf the fuel's energy was lost to friction and heat