Answer:
a. 
Explanation:
The equation of the forces along the directions parallel and perpendicular to the slope are:
- Along the parallel direction:
where
:
m = 6.0 kg is the mass of the box
g = 9.8 m/s^2 the acceleration of gravity
is the angle of the slope
is the coefficient of friction
R is the normal reaction
a is the acceleration
- Along the perpendicular direction:
From the 2nd equation, we get an expression for the reaction force:
And substituting into the 1st equation, we can find the acceleration:
Solving for a,
Answer:
4.123 * 10∧4 kg
Explanation:
mass of car M' = 3.34 * 10∧4 kg
energy loss E = 19/100 K
from law of conservation of momentum ,
M' V' = ( M' + M'' ) v
V = M' V' / ( M' + M'' )
Initial kinetic energy is K' = (M' V')² / 2
final kinetic energy K" = 81 K' /100
= (M' V')² 81 / (200) = ( M' + M'' ) v²/ 2
therefore , M' / ( M' + M" ) = 0.81
mass of caboose is , M" = 1.234 M' - M'
M" = .234 M'
= 0,234 ( 3.34 * 10∧4 kg)
= 4.123 * 10∧4 kg
It acquires a positive electric charge.
The answer is 35 minutes
The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ
T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant
step 1. Calculate constant k:
We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?
T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ*¹⁰
200 = 68 + 142 * e⁻¹⁰ⁿ
200 - 68 = 142 * e⁻¹⁰ⁿ
132 = 142 * e⁻¹⁰ⁿ
e⁻¹⁰ⁿ = 132/142
e⁻¹⁰ⁿ = 0.93
Logarithm both sides with natural logarithm:
ln(e⁻¹⁰ⁿ) = ln(0.93)
-10n * ln(e) = -0.07
-10n * 1 = - 0.07
-10n = -0.07
n = -0.07 / - 10
n = 0.007
Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
Ta = 68°F
To = 210°F
n = 0.007
T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79
Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = ln(0.79)
-0.007x * ln(e) = -0.24
-0.007x * 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35
