State the conservation of momentum theorem

(1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing nort

Veronika [31]

Answer:28.8 knots

Explanation:

The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.

First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.

The distance between them is described with Pythogorean theorem, which is $D=\sqrt{A^{2} +B^{2} }$ and when we replace the values A and D, we find Distance (D) to be 182 miles.

Now, let's make the notations clear. The velocity of A and B is notated as $\frac{dA}{dt}$ and $\frac{dB}{dt}$ . The rate of change of distance is also notated as $\frac{dD}{dt}$ . Now, we have to find $\frac{dD}{dt}$ from the Pythogorean theorem. If we derive the Pythogorean expression $D=\sqrt{A^{2} +B^{2} }$ , we would have:

$\frac{dD}{dt} =\frac{1}{2} *(A^{2} +B^{2} )^{-1/2} *(2*A*\frac{dA}{dt} + 2*B*\frac{dB}{dt} )$

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:

$\frac{dC}{dt} =\frac{1}{2}*(142^{2} +114^{2})^{-\frac{1}{2} }*(2*142*22 + 2*114*19)$ and the answer is 28.6 knots.

6 0

3 years ago

Which conditions are usually the effect of a low air pressure system? clear, dry weather cloudy, wet weather cold, dry weather h

qaws [65]

Your answer is hot dry waether

5 0

3 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago

If the current in the circuit is 10.5 A,

sergiy2304 [10]

Answer:

E) 147 V

Explanation:

Ra+Rb+Rc = 14.0

V = IR

V = (10.5)(14.0)

V = 147 V

7 0

3 years ago

A 4 kg billiard ball moving on a horizontal surface has a speed of 16 m/s when it strikes a horizontal coiled spring is brought

ruslelena [56]

Answer:

spring constant of the spring is 1820.44 N/m

Explanation:

given data

ball mass = 4 kg

speed = 16 m/s

distance = 0.75 m

to find out

spring constant of the spring

solution

we know that kinetic energy of ball = energy store in spring as compression

so we can express it as

0.5 × m × v² = 0.5 × k × x² ....................1

so put here value we get spring constant k

m × v² = k × x²

4 × 16² = k × 0.75²

solve it we get

k = 1820.44 N/m

so spring constant of the spring is 1820.44 N/m

6 0

4 years ago

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