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3 years ago

A girl runs once around a circular track with a radius of 100m at speed of 10m/s.

Physics

1 answer:

goblinko [34]3 years ago

5 0

Answer:

0 m

Explanation:

Displacement is the shortest distance from one ppont to another. Here, the girl returns to the same spot after running. Hence, displacement is 0 m.

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Which formula can be used to find the magnitude of the resultant vector? R2 = Rx2 + Ry2 R = Rx + Ry R = Rx(cosθ) R = Rx(sinθ)

12345 [234]

R^2 = rx^2 + ry^2 !!!!!!!!!!

5 0

3 years ago

Read 2 more answers

water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49

vesna_86 [32]

When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
$\theta = \arcsin ( \frac{n_2}{n_1} )$
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
$\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}$

6 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago

Given that renewable sources provide only a small percentage of our energy and that nuclear power is so expensive, in your opini

Katena32 [7]

Answer:

No getting rid of fossil fuel is not at all possible.

Explanation:

Dream of 100% renewable energy is possible through electrification and use of energy from sun, water and wind. Since, the high production cost of power plant for extracting energy from these sources in itself is a big challenge and hence decarbonized energy in current scenario is not realistic.

3 0

3 years ago

Four velcro-lined air-hockey disks collide with each other in a perfectly

Reil [10]

Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, $\Sigma P_{initial}$ is given as follows;

$\Sigma P_{initial}$ = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ $\Sigma P_{initial}$ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ $\Sigma P_{initial}$ = -20·i + 130·j

By the law of conservation of linear momentum, we have;

$\Sigma P_{initial} = \Sigma P _{final}$ = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

$\Sigma P _{final}$ = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, $\left | v \right |$ = √((-0.08)² + (0.52)²) ≈ 0.526

$\left | v \right |$ ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0

3 years ago