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scZoUnD
28 days ago
7

The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov

es faster than this but he'd just woken up. Determine his final position. 2. A hedgehog starts at a position of 0 m and moves with an average speed of 0.75 m/s for 4.0 s. Determine the final position of the hedgehog. 3. A tortoise starts at a position of 1.0 m and moves with an average speed of 1.25 m/s for 3.0 s, what is the final position of the tortoise? Cintia, in a 100 m race, has an average speed of 1.25 m/s. What was her race time? 5. Angel runs a marathon with an average speed of 5 miles/hour. How long did it take to finish?​
Physics
1 answer:
Yuliya22 [3.3K]28 days ago
5 0

Answer:

(1). The position he ends up at is 1.5 m.

(2). The hedgehog's final location is 3 m.

(3). The tortoise's final location

(4). She completed the race in 80 sec.

(5). The time taken to complete is 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to determine the final position

Using the average speed formula

v_{av}=\dfrac{x_{f}+x_{i}}{t}

Where, x_{f} = final position

x_{i} = Initial position

t = total time

Insert the values into the formula

0.50=\dfrac{x_{f}+0}{3.0}

x_{f}=0.50\times3.0

x_{f}=1.5\ m

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using the average speed formula

v_{av}=\dfrac{x_{f}+x_{i}}{t}

Insert the values into the formula

0.75=\dfrac{x_{f}+0}{4.0}

x_{f}=0.75\times4.0

x_{f}=3\ m

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to find the final position

Using the average speed formula

v=\dfrac{x_{f}+x_{i}}{t}

Insert the values into the formula

1.25=\dfrac{x_{f}+1.0}{3.0}

x_{f}=1.25\times3.0-1.0

x_{f}=2.75\ m

(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We are to determine the time

Using the time formula

t=\dfrac{d}{v}

Insert the values into the formula

t=\dfrac{100}{1.25}

t=80 sec

(5). Given that,

Average speed = 5 miles/hr

Assuming the distance = 25 miles

We need to find the time

Using the time formula

t=\dfrac{d}{v}

Insert the values into the formula

t=\dfrac{25}{5}

t=5\ hr

Thus, (1). The position he ends up at is 1.5 m.

(2). The hedgehog's final location is 3 m.

(3). The tortoise's final location

(4). She completed the race in 80 sec.

(5). The time taken to complete is 5 hr.

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solution:

the spring force applied by a spring with spring constant k can be expressed as

F(x)=-kx

where k acts as the spring constant

and x indicates the spring's deformation

to determine the work completed by the spring

W=\int\limits^L_0 {} \, dW

the amount of work done by the spring when moving from x=0 to x=L

W=-kx^2/2

substituting the limits x=0 and x=L

we derive the work done in terms of k and L

ANSWER

W=-kL^2/2

4 0
1 month ago
 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer
serg [3582]

Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

a)

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}

We have y_{f}=0 and v_{y0}=0, allowing us to simplify the equation to:

0=y_{0}+\frac{1}{2}at^{2}

Now, we can calculate for t:

-y_{0}=\frac{1}{2}at^{2}

-2y_{0}=at^{2}

\frac{-2y_{0}}{a}=t^{2}

t=\sqrt{\frac{-2y_{0}}{a}}

We know y_{0}=1.20m and a=g=-9.8m/s^{2}

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

V_{x}=\frac{x}{t}

Solving for x, we have:

x=V_{x}t

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

a=\frac{v_{f}-v_{0}}{t}

The initial vertical velocity being zero simplifies our calculations:

a=\frac{v_{f}}{t}

Thus, we can determine the final velocity:

V_{yf}=at

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

V_{yf}=(-9.8m/s^{2})(0.495s)

This leads to:

V_{yf}=-4.851m/s

Now, we ascertain the velocity components:

V_{xf}=1.5m/s and V_{yf]=-4.851m/s

Next, we find the speed by calculating the vector's magnitude:

|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}

<pThus:

|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

\theta = tan^{-1} (\frac{V_{y}}{V_{x}})

<pFulfilling this provides:

\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})

<pLeading to:

\theta = -72.82^{o}

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2 months ago
An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac
ValentinkaMS [3465]

Answer:

arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

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By applying the Law of Sines,

sin \theta = \frac{sin \phi R}{ l + R}

Based on Newton's Law,

mg = N\sqrt{\mu^2+1}

And the final equation also derived from Newton's Law,

\mu N = mgsin\phi

Then by consolidating all the equations together,

\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\

sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}

Thus,

\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

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You are designing a spacecraft intended to monitor a human expedition to Mars (mass 6.42×1023kg, radius 3.39×106m). This spacecr
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3 0
1 month ago
A horizontal jet of water is made to hit a vertical wall with a negligible rebound. If the speed of water from the jet is 'v', t
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Mass can be calculated as density multiplied by volume:

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Flow rate is determined by velocity multiplied by the cross-sectional area:

F = ρ (v A) v

This simplifies to F = ρ A v²

The area of a circle is calculated as pi times the square of the radius, or as pi/4 times the diameter squared:

F = ρ (π/4 d²) v²

Hence, F = π/4 ρ d² v²

3 0
1 month ago
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