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Elanso
1 month ago
6

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2m/s2
for 6.5 s, after which the gazelle continues continues at a steady speed.
a. What is the gazelle's top speed?
b. A human would win a very short race with a gazelle. The best time for a 30 m sprint for a human runner is 3.6 s. How much time would the gazelle take for a 30 m race?
c. A gazelle would win a longer race. The best time for a 200 m sprint for a human runner is 19.3s.
How much time would the gazelle take for a 200 m race?
Physics
1 answer:
Sav [3.1K]1 month ago
4 0

Answer:

a) 27.3 m/s

b) 3.78 seconds

c) 10.576 seconds

Explanation:

Definitions: t = time elapsed, u = initial speed, v = final speed, s = distance covered, a = acceleration

a)

v=u+at\\\Rightarrow v=0+4.2\times 6.5\\\Rightarrow v=27.3\ m/s

The gazelle's maximum velocity is calculated to be 27.3 m/s.

b)

s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0t+\frac{1}{2}\times 4.2\times t^2\\\Rightarrow t=\sqrt{\frac{30\times 2}{4.2}}\\\Rightarrow t=3.78\ s

It would take the gazelle 3.78 seconds to complete a 30-meter dash.

c)

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 6.5+\frac{1}{2}\times 4.2\times 6.5^2\\\Rightarrow s=88.725\ m

The gazelle accelerates over a distance of 88.725 meters before maintaining constant speed.

Time = Distance divided by Speed

\text{Time}=\frac{200-88.725}{27.3}=4.076\ s

Total time for the gazelle to cover 200 meters equals 6.5 plus 4.076, resulting in 10.576 seconds.

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A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximate
inna [3103]

Answer:

0.0031 m

Explanation:

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

\lambda = Wavelength = 550 nm

d = Pupil diameter

\theta = Angle

We have the expression

tan\theta=\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{281\times 10^{-6}}{1.3}

We have the expression

sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow d=\dfrac{1.22\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1.22\times 550\times 10^{-9}}{sin\left(tan^{-1}\frac{281\times 10^{-6}}{1.3}\right)}\\\Rightarrow d=0.0031\ m

The pupil diameter calculates to 0.0031 m

4 0
10 days ago
Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
ValentinkaMS [3465]

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

V=\frac {1}{3}\times \pi\times r^2\times h

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Additionally, we differentiate the volume expression in relation to time:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given that \frac {dV}{dt} = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

The rate at which the root beer level is decreasing is 0.08603 cm/s.

3 0
1 month ago
9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J
Maru [3345]

Answer:

The energy expected to be released is calculated to be 4182 Joules.

Explanation:

The total mass of coke is 2 kg, which is equivalent to 2000 g

1 calorie per gram corresponds to 4.184 Joules of energy

4.184 J/gC * 2000g results in 8368 J

1 food calorie approximates to 4186 J

By subtracting, we find 8368 - 4186

Hence, the total energy that will be released amounts to 4182 Joules.

3 0
23 days ago
A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0
kicyunya [3294]

Answer:

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

U_1=k\dfrac{q_1q_2}{x_1}

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

W=-\Delta U

W=-(U_2-U_1)

W=(U_1-U_2)

W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Consequently, the work done by the electrostatic force on the moving charge is kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}}). Therefore, this concludes the solution.

3 0
27 days ago
If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius
ValentinkaMS [3465]

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

8 0
1 month ago
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