Answer:
a)
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
the required distance is 40.98 m
Explanation:
Given that;
velocity of the river u = 1.70 m/s
velocity of boat v = 14.0 m/s
Now to get the velocity of the boat relative to shore;
( north of east), we say
a² + b² = c²
(1.70)² + (14.0)² = c²
2.89 + 196 = c²
198.89 = c²
c = √198.89
c = 14.1028 m/s
tan∅ = v/u = 14 / 1.7 = 8.23529
∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east
Therefore, the velocity of the boat relative to shore is;
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
width of river = 340 m,
ow far downstream has the boat moved by the time it reaches the north shore in meters = ?
we say;
340sin( 90° - 83.0765°)
⇒ 340sin( 6.9235°)
= 40.98 m
Therefore, the required distance is 40.98 m
Answer:
276.135 J
Explanation:
Given that:
mass of Fe = 30.0 g
initial temperature = 24.5°C
final temperature = 45.0°C
specific heat of Fe = 0.449 J/g°C
We can determine the thermal energy added by using the formula;
Q = mcΔT
Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C
Q = 276.135 J
Answer;
The temperature change for the second pan will be lower compared to the temperature change of the first pan
Explanation;
-The quantity of heat is given by multiplying mass by specific heat and by temperature change.
That is; Q = mcΔT
This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.
-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.
Answer:
Neutrally charged!!!!!!!!!!!!!!!!!!!!!
Explanation:
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.
v is the speed of cat, 
So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
