Answer:
magnitude = 7.446 km, direction = 75.22° north of east
Explanation:
From the questions,
To get the the magnitude of the resultant vector we use Pythagoras theorem
a² = b²+c²
From the diagram,
y² = 1.9²+7.2²
y² = 55.45
y = √(55.45)
y = 7.446 km.
The direction of the dolphin is given as,
θ = tan⁻¹(7.2/1.9)
θ = tan⁻¹(3.7895)
θ = 75.22° north of east
Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Answer:
72.98 km
Explanation:
Her displacement is simply the distance from her final position to her initial position.
Now, I've drawn and attached a triangle diagram to depict this her movement.
Point O is her initial starting point.
Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.
From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.
Thus;
x² = 62² + 26² - 2(62 × 26)cos 120
x² = 4520 + 806
x² = 5326
x = √5326
x = 72.98 km
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.
Explanation:
By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:
Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.
When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:
By replacing the known values it is possible to find the plate's weight:
When the plate kept to 30° from the vertical the moment equation balance is written as:
The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:
