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3 years ago

Jogger runs 400 m East in 120 s, then turns around and runs 250 m West in 90 seconds. Calculate the following values.

1 answer:

8 0

First, we must recall that distance is the total length traveled by an object in which the direction of the motion does not matter. Meanwhile, displacement is the distance of the object from its starting point which means the directions matters for displacement.

Now, speed is distance over time while velocity is displacement over time. Since we're talking about the same object's motion, the total time traveled is (120 + 90) = 210 seconds.

Now, the total distance traveled by the object is (400 + 250) = 650 m. Meanwhile the total displacement traveled by the object is 400 m, East + 250 m, West = 150 m, East.

Now, to find the speed and velocity, we just divide the values of distance and displacement, respectively, over time. Thus, we have

speed = 650 m / 210s ≈ 3.095 m/s
velocity = 150 m, East /210 s ≈ 0.714 m/s, East

Now, rounding the values up to 2 significant digits, we have

a) speed = 3.1 m/s
b) velocity = 0.71 m/s, East

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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

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3 years ago

What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?

mote1985 [20]

The last choice. Two arrows and the arrow up is shorter than the arrow down. Since the guy is falling and he’s opened his chute, he’s slowing down but he’s still falling meaning the force of gravity is stronger than the air resistance.

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4 years ago

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Select the correct answer.

Murrr4er [49]

Answer:

I will choose the answer A

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3 years ago

What is the frequency heard by a person driving at 15 m/s toward a factory whistle emitting a

zmey [24]

Answer:

835.29 Hz

Explanation:

When moving towards the source of sound, frequency will be given by

f*=f(vd+v)/v

Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.

Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then

f*=800(15+340)/340=835.29411764704 Hz

Rounded off, the frequency is approximately 835.29 Hz

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4 years ago

A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tal

masya89 [10]

Answer:

0.918 sec

Explanation:

time of flight:

t=u²/g

=3²/9.8=0.918sec

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3 years ago

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