When liquid silver nitrate and liquid sodium chloride are combined, solid silver chloride forms along with a new liquid, sodium
1 answer:
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The tangential velocity of the car is 124.5 m/s
Explanation:
The centripetal acceleration of an object in uniform circular motion is given by:
where
a is the acceleration
v is the tangential velocity
r is the radius of the circle
For the car in this problem, we have:
is the centripetal acceleration
r = 155 m is the radius of the turn
Solving for v, we find the tangential velocity of the car:
Learn more about centripetal acceleration:
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Answer:
The displacement of the net from player 2 in component form = (-47.498î - 26.812j)
The displacement of the net from player 2 in statement form is 54.54 m and 29.44° (S of W) or 60.56° (W of S)
Explanation:
The sketch of the bearings described in the question is presented in the attached image to this solution.
Method 1
Using component method
Taking the player 1's position as the origin,
The displacement of the player 2 from the origin is (25î) m
The displacement of the net from the origin is 35[(sin θ)î + (cos θ)j]
But θ is the angle of the net's displacement reading from the positive x-axis in the anticlockwise direction. θ = 230°
Displacement of the net from the origin = 35[(cos 230°) + (cos 230°)]
= 35[-0.6428î - 0.7660j]
= (-22.498î - 26.812j) m
In component form, taking note of the directions of the respective displacements calculated (check the attached image)
(The displacement of the net from player 1) = (The displacement of player 2 from player 1) + (The displacement of the net from player 2)
Since we have agreed that player 1 is the origin
(The displacement of the net from origin) = (The displacement of player 2 from origin) + (The displacement of the net from player 2)
(-22.498î - 26.812j) = (25î) + (The displacement of the net from player 2)
The displacement of the net from player 2 = (-22.498î - 26.812j) - (25î) = (-47.498î - 26.812j)
The magnitude of this displacement = √[(-47.498)² + (-26.812)²]
= √(2,256.060004 + 718.883344) = 54.54 m
Direction = tan⁻¹ (-26.812/-47.498) = 209.44° (the signs on the components show that the direction is the third quadrant from the positive x-axis in the anti-clockwise direction)
Hence, the displacement of the net from player 2 is 54.54 m and 29.44° (S of W)
Method 2
Using trignometry,
We will use cosine and sine rule to obtain the required magnitude and direction of the displacement of the net from player 2
Cosine rule
Magnitude = √[35² + 25² - (2×25×35×cos 130°)] = √2,974.8783169514 = 54.54 m
Sine rule
(Sin θ)/35 = (Sin 130°)/54.54
Sin θ = (35 × Sin 130°)/54.54 = 0.4916
θ = Sin⁻¹ (0.4916) = 29.44°
This answer matches the answers from method 1.
Hope this Helps!!!
