Answer:
A) It takes the truck 8 s to catch the motorcycle.
B) The motorcycle has traveled 160 m in that time.
C) The velocity of the truck is 40 m/s at that time.
Explanation:
The equations of the position and velocity of an object moving in a straight line are as follows:
x = x0 +v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
(A) When the the truck catches the motorcycle, both have the same position. Notice that the motorcycle moves at constant speed so that a = 0:
x truck = x motorcycle
x0 +v0 · t + 1/2 · a · t² = x0 + v · t
Placing the origin of the frame of reference at the point where the truck starts, both have an initial position of 0. The initial velocity of the truck is 0. Then:
1/2 · a · t² = v · t
solving for t:
t = 2 v/a
t = 2 · 20 m/s/ 5 m/s²
t = 8 s
It takes the truck 8 s to catch the motorcycle.
(B) Using the equation of the position of the motorcycle, we can calculate the traveled distance in 8 s.
x = v · t
x = 20 m/s · 8 s
x = 160 m
(C) Now, we use the velocity equation at time 8 s.
v = v0 + a · t
v = 0 m/s + 5 m/s² · 8 s
v = 40 m/s
Answer:
= 17º C
Explanation:
This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)
Let's reduce the magnitudes to the SI system
Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg
L = 3.33 105 J / kg
Water M = 860 g = 0.860 kg
= 4186 J / kg ºC
Q₁ = m L
Q₁ = 0.080 3.33 10⁵
Q₁ = 2,664 10⁴ J
Now let's see what this liquid water temperature is when this heat is released
Q = M ΔT = M (T₀₁ -
)
Q₁ = Q
= T₀₁ - Q / M ce
= 26.0 - 2,664 10⁴ / (0.860 4186)
= 26.0 - 7.40
= 18.6 ° C
The initial temperature of water that has just melted is T₀₂ = 0ª
The initial temperature of the liquid water is T₀₁= 18.6
m + M 
= M T₀₁ - m T₀₂o2
= (M To1 - m To2) / (m + M)
= (0.860 18.6 - 0.080 0) / (0.080 + 0.860)
= 17º C
gg
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.
Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²
We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m
Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
= </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
