<span>D. A statement that explains an observation
</span>
Refer to the figure shown below.
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
Answer:
The time for the cake to cool off to room temperature is
approximately 30 minutes.
Let
=
F be the temperature and T that of the body
Explanation:
Our Tm = 70, the initial-value problem is
= <em>k</em>(T − 70), T(0) = 300
Solving the equation, we get
= <em>kdt</em>
In [T-70]= <em>kt </em>+
T = 70 +

Finding he value for
using the initial value of T (0)= 300, therefore we get:
300=70+
= 230 therefore
T= 70+ 230 
Finding the value for <em>k </em>using T (3) = 200, therefore we get
T (3) = 200
= 
<em>K </em>=
in 
= -0.19018
Therefore
T(t) = 70+230
Answer:
A. 
B. 
C. ΔK
Explanation:
From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

A. Since the two vehicles become entangled the final mass is:

From linear momentum we got that:




B. The change in velocity of both vehicles are:
For the car

For the truck

C. The change in kinetic energy is:
ΔK=
ΔK=
ΔK
Explanation:
radio waves, which include visible light waves.