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2 years ago

A treasure map directs you to start at palm tree

Physics

1 answer:

masya89 [10]2 years ago

6 0

The direction of the 90° turns are the possible directions used for the calculations

The four distances and directions are;

24.2 m, 65.6° West of North

29.7 m, 42.7° East of North

24.2 m, 65.6° East of North

29.7m, 47.7° West of North

Reason:

Let point A represent the motion of the treasure hunter, we have:

Turning West then West;

Walking due north to location (0, 15)

Turn 90° West and walk 22.0 m to the location (-22, 15)

Turn 90° West again and walk 5.00 m. to the location (-22, 10)

The location of point A = (-22, 10)

$Direction \ of \ point \ A = arctan \left(\dfrac{10}{-22} \right) \approx -24.4^{\circ}$

Direction to North = 90° - 24.4° ≈ 65.6°

Distance = √((-22)² + 10²) ≈ 24.2

Therefore, we have;

24.2 m, 65.6° West of North

Turning East then West:

Turn 90° East and walk 22.0 m to the location (22, 15)

Turn 90° West again and walk 5.00 m. to the location (22, 20)

The location of point A = (22, 20)

$Direction \ of \ point \ A = arctan \left(\dfrac{20}{22} \right) \approx 42.7^{\circ}$

Direction to North = 90° - 42.3° ≈ 47.7° East

Distance = √((-22)² + 20²) ≈ 29.7

Therefore, we have;

29.7 m, 42.7° East of North

Turning East then East:

Turn 90° East and walk 22.0 m to the location (22, 15)

Turn 90° East again and walk 5.00 m. to the location (22, 10)

The location of point A = (22, 10)

$Direction \ of \ point \ A = arctan \left(\dfrac{10}{22} \right) \approx 24.4^{\circ}$

Direction to North = 90° - 24.4° ≈ 65.6° East

Distance = √((-22)² + 10²) ≈ 24.2

Therefore, we have;

24.2 m, 65.6° East of North

Turning West then East:

Turn 90° West and walk 22.0 m to the location (-22, 15)

Turn 90° East and walk 5.00 m. to the location (-22, 20)

The location of point A = (-22, 20)

$Direction \ of \ point \ A = arctan \left(\dfrac{20}{-22} \right) \approx -42.7^{\circ}$

Direction to North = 90° - 42.7° ≈ 47.7° West

Distance = √((-22)² + 20²) ≈ 29.7

Therefore, we have;

29.7m, 47.7° West of North

Learn more here:

brainly.com/question/11489059

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A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration

Oksi-84 [34.3K]

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

Therefore, the acceleration of the box is 3 m/s²

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3 years ago

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3 years ago

¿Qué distancia recorrió un avión que viajaba a 750 km/h después de 2 h y media de vuelo?

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3 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

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3 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

Part A: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

Part B: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$