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maxonik [38]
3 years ago
14

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

displacement). Round your answer to the appropriate significant figures. Round your angle to the nearest degree.
Physics
1 answer:
levacccp [35]3 years ago
5 0

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

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A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its
Nostrana [21]

Answer:

Its speed is maximum and its acceleration is zero

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- Concerning the speed:

In a simple harmonic motion, the speed can be determined through the law of conservation of energy. In fact, we have that the total mechanical energy of the system, which is sum of elastic potential energy U and kinetic energy K, is constant:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2

where k is the spring constant, x is the displacement, m is the mass and v is the speed.

We see that when the displacement is 0, x = 0: this means that the elastic potential energy U is also 0, therefore the kinetic energy K is maximum, and so the speed v is also maximum.

- Concerning the acceleration:

According to Newton's second law, the acceleration of the system is proportional to the net force:

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the net force is just the restoring force of the spring, given by Hooke's law:

F=-kx

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4 0
4 years ago
This is the last question that I need help with, and I'll try and do the rest on my own. Thank you to all of the tutors that hel
ivann1987 [24]

Question 7.

Given:

Mass, m = 2200 kg

Time, t = 8 s

Velocity = 12 m/s

Let's find the average net force required to stop the car.

To find the average net force required to stop the car, let's first find the acceleration of the car.

v=v_o+at

Where:

Initial velocity, vo = 12 m/s

Final velocity, v = 0 m/s (velocity of car when it stops)

t is the time = 8 seconds

a is the acceleration.

Rewrite the formula for a:

a=\frac{v-v_o}{t}

Hence, we have:

\begin{gathered} a=\frac{0-12}{8} \\  \\ a=\frac{-12}{8} \\  \\ a=-1.5m/s^2 \end{gathered}

The acceleration is -1.5 m/s² (negative because the car is deccelerating).

Now, to find the net force, we have:

\sum ^{}_{}F=ma

Where:

m is the mass = 2200 kg

a is the acceleration = -1.5 m/s²

Hence, we have:

\sum ^{}_{}F=2200\times(-1.5)=-3300\text{ N}

The negative sign indicates that the force acts in the opposite direction the car is travelling.

Therefore, average net force required to stop the car is 3300 Newtons in the opposite direction.

ANSWER:

3300 N in the opposite direction

8 0
1 year ago
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