Answer:
128.21 m
Explanation:
The following data were obtained from the question:
Initial temperature (θ₁) = 4 °C
Final temperature (θ₂) = 43 °C
Change in length (ΔL) = 8.5 cm
Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)
Original length (L₁) =.?
The original length can be obtained as follow:
α = ΔL / L₁(θ₂ – θ₁)
17×10¯⁶ = 8.5 / L₁(43 – 4)
17×10¯⁶ = 8.5 / L₁(39)
17×10¯⁶ = 8.5 / 39L₁
Cross multiply
17×10¯⁶ × 39L₁ = 8.5
6.63×10¯⁴ L₁ = 8.5
Divide both side by 6.63×10¯⁴
L₁ = 8.5 / 6.63×10¯⁴
L₁ = 12820.51 cm
Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
12820.51 cm = 12820.51 cm × 1 m / 100 cm
12820.51 cm = 128.21 m
Thus, the original length of the wire is 128.21 m
Answer:
speed = 7.9 m/s
Explanation:
speed = total distance / time taken
speed = 300 / 38
speed = 7.89473684 m/s
to the nearest tenth
speed = 7.9 m/s
Answer:
Gravitational
Tension
Normal
Friction.
Explanation:
The forces acting on the sled are:
Tension: the tension from the rope, this is the force that "moves" the sled.
Friction: kinetic friction between the sled and the ground as the sled moves.
There are another two forces that also act on the sled, but that "has no effect"
Gravitational force: This force pulls the sled down, against the floor.
Normal force: This force "opposes" to the gravitational one, so they cancel each other.
These two forces cancel each other, so they have no direct impact on the movement of the sled. BUT, the friction force depends on the weight of the moving object, and the weight of the moving object depends on the gravitational force, so we need gravitational force in order to have friction force.
Then we can conclude that the forces acting on the sled are:
Gravitational
Tension
Normal
Friction.
Answer:
Explanation:
At the beginning, we have:
V = 4.0 V potential difference across the capacitor
charge stored on the capacitor
Therefore, we can calculate the capacitance of the capacitor:
Later, the battery is replaced with another battery whose voltage is
V = 5.0 V
Since the capacitance of the capacitor does not change, we can calculate the new charge stored:
Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from 
to 
. Therefore, the additional charge that moved to the positive plate is
