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3 years ago

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A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

40 m. If he completes the 200 m dash in 24.0 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track?

1 answer:

34kurt3 years ago

5 0

Answer:

The centripetal acceleration of the runner is $1.73\ m/s^2$ .

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

v is the velocity of runner

$v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s$

Centripetal acceleration,

$a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2$

So, the centripetal acceleration of the runner is $1.73\ m/s^2$ . Hence, this is the required solution.

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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

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3 years ago

Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de

-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

$s(t)^2={h^2+x^2}$ ...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

$2s(t)s'(t)=2x(t)x'(t)$

$x'(t)=\dfrac{s(t)s'(t)}{x(t)}$

When x = 40, $s(t)=\sqrt{40^2+4^2}=40.19\ m$

$x'(t)=\dfrac{-240s(t)}{x(t)}$

$x'(t)=\dfrac{-240\times 40.19}{40}$

$x'(t)=-241.14\ m/s$

So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.

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Bernoulli's principle can be used to explain the lift force on an airplane wing. How must an airplane's wing be designed to ensu

mihalych1998 [28]

Answer:

Airplane wings must be designed to ensure that air molecules move more rapidly over the top surface of the wing, creating a region of lower pressure.

Explanation:

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3 years ago

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Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

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2 years ago

A uniform metal tube of length 5m and mass 9kg is suspended by two vertical wires attached at 50cm and 150cm respectively from t

Elena-2011 [213]

Answer:

force (tension) of 29.4 N (upward) in 100 cm

force (tension) of 58.4 N (upward) in 200 cm

Explanation:

Given:

Length of tube = 5 m (500 cm)

Mass of tube = 9

Suspended vertically from 150 cm and 50 cm.

Computation:

Force = Mass × gravity acceleration.

Force = 9.8 x 9

Force = 88.2 N

So,

Upward forces = Downward forces

D1 = 150 - 50 = 100 cm

D2 = 150 + 50 = 200 cm

And F1 = F2

F1 x D1 = F2 x D2

F1 x 100 = F2 x 200

F = 2F

Total force = Upward forces + Downward forces

3F = 88.2

F = 29.4 and 2F = 58.8 N

force (tension) of 29.4 N (upward) in 100 cm

force (tension) of 58.4 N (upward) in 200 cm

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3 years ago