Answer:
- The total distance traveled is 28 inches.
- The displacement is 2 inches to the east.
Explanation:
Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector 
pointing in the west direction, the ant start at position
Then, moves to
so, the distance traveled here is
after this, the ant travels to
so, the distance traveled here is
The total distance traveled will be:
The displacement is the final position vector minus the initial position vector:
This is 2 inches to the east.
<h3><u>Answer;</u></h3>
33.9 pounds
<h3><u>Explanation</u>;</h3>
In order for the ladder to be in equilibrium, the net torque should be equal to zero. Therefore, the torque in the opposite directions should equal each other:
Clockwise torque = Counter clockwise torque
Torque is the product of the applied force and the distance between that force and the axis of rotation.
Wι (7.5 ft) cos 53° + Wb (6 ft) cos 53° = F (15 ft) sin 53 °
Substitute the values for the weights of the ladder and the boy, respectively.
(20 lb) (7.5 ft)cos 53° + (75 lb) (6 ft) cos 53° = F(15 ft) sin 53°
Solving for F;
F = ((30 ×7.5 × cos 53°) + (75 × 6 × cos 53°))/ (15 × sin 53°)
= 33.9 lb
<u>= 33.9 Pounds</u>
Answer:
<h2>
242.5kPa</h2>
Explanation:
According to one of the gas laws, 
Given P1 = 240.0kPa, T1, 15.0°C, T2 = 18.0°C, P2 = ?
Substituting this values into the equation, we have;
Cross multiplying we have;
288P2 = 240*291
The new pressure is 242.5kPa
Answer:
$ 18.75
Explanation:
given,
capacity of the Dave’s car = 12 gallons
Assuming that the tank is 3/8 full before
Cost of gas per gallon = $2.50 per gallon of gas
Volume Dave have to fill
=
=
of full tank
=
volume Dave have to fill = 7.5 gallons
total money Dave spent
= 7.5 gallons x $2.50
= $ 18.75
Dave have to spend $ 18.75 to fill tank to it capacity.
Answer:
The second knife-edge must be placed 46.2 cm from the zero mark of the rod.
Explanation:
From the law of equilibrium, ΣF = 0 and ΣM = 0.
Let R be the reaction at the knife edge. Since the weight of the rod and zinc load act downward, and we take downward position as negative
-32 N - 2 N + R = 0
-34 N = -R
R = 34 N
Also, let us assume the knife-edge is x cm from the zero mark. Taking moments about the weight and assuming the knife-edge is right of the weight of the rod. Taking clockwise moments as positive and anti-clockwise moments as negative,
-(45 - 25)2 + (x - 45)R = 0
-(20)2 + (x - 45)34 = 0
-40 = -(x - 45)34
x - 45 = 40/34
x - 45 = 1.18
x = 45 + 1.18
x = 46.18 cm
x ≅ 46.2 cm
The second knife-edge must be placed 46.2 cm from the zero mark of the rod.
