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
Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
<em />
They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
From the information given and if the question is complete then;
Absolute temperature is the temperature in Kelvin
To convert degree Celsius to kelvin we normally add 273
that is Kelvin = deg Celsius + 273
Thus since we have been given that the air was at -70 degrees celcius;
then; - 70° C + 273 = 203 K
Therefore; the absolute temperature is 203 K
kinematic equation
v squared = u squared + 2 a x s
v= sq root (0 + 2 10 x 65)
i thimk
Consider 20 deg.C. as room temperature.
From tables,
Silver has a resistivity of 1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m
At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.
The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx
Answer: 20 + 3421 = 3441 °C
