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3 years ago

11

A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They al

l roll without slipping. In what order do they reach the bottom of the ramp

1 answer:

pav-90 [236]3 years ago

8 0

Answer:

sphere, disk, hoop

Explanation:

See attached file

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Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par

Murrr4er [49]

Answer:

X₃₁ = 0.58 m and X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

F₁₃ - F₂₃ = 0

F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

F₁₃ = k q₁ q₃ / r₁₃²

F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

r₁₃ = x₃- 0

r₂₃ = 2 –x₃

We substitute

k q q / x₃² = k 4q q / (2-x₃)²

q² (2 - x₃)² = 4 q² x₃²

4- 4x₃ + x₃² = 4 x₃²

5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5

x₃ = [-4 ± 9.80] 10

X₃₁ = 0.58 m

X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0

3 years ago

A new planet is discovered orbiting a distant star. Observations have confirmed that the planet has a circular orbit with a radi

Mariulka [41]

Answer:

Mass of star is $1.31\times10^{35}$ kg.

Explanation:

The cube of orbital radius is equal to the square of its orbital time period is known as Kepler's law.

$T^{2} = (\frac{4\pi^{2} }{GM})r^{3}$ .....(1)

Here T is time period, r is orbital radius, G is universal gravitational constant and M is the mass of the star.

According to the problem,

Time period, T = 109 days = 109 x 24 x 60 x 60 s = 9.41 x 10⁶ s

Orbital radius, r = 18 AU = 18 x 1.496 x 10¹¹ m = 2.70 x 10¹² m

Gravitational constant, G = 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻²

Substitute these values in equation (1).

$(9.41\times10^{6}) ^{2} = (\frac{4\pi^{2} }{6.67\times10^{-11}\times M})(2.70\times10^{12}) ^{3}$

M = $1.31\times10^{35}$ kg

3 0

3 years ago

Pierre is a 375 kg great white with an average speed of 3 m/s. When Pierre spots a seal, he increases his velocity to 7 m/s. Aft

tangare [24]

Answer:

1890J

Explanation:

375+45 = 420kg (total mass)

kinetic energy = 1/2 × mass × velocity²

1/2 × 420 × 3² = 1890J

5 0

3 years ago

if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet

klio [65]

Answer: $3765.66 \frac{m^{3}}{s}$

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

$Q=\frac{\pi r^{4}\Delta P}{8\eta L}$

Where:

$Q$ is the Volume flow rate

$r=23 cm \frac{1 m}{100 cm}=0.23 m$ is the effective radius

$L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m$ is the length

$\Delta P=1.88(10)^{4} Pa$ is the difference in pressure

$\eta=3(10)^{-3} Pa.s$ is the viscosity of blood

Solving:

$Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}$

$Q=3765.66 \frac{m^{3}}{s}$

4 0

4 years ago

An adult (mass 98.36kg) is standing on top of a 4.45m cliff right next to a river. He grabs a vine of length 30.1m, which the po

Tom [10]

Answer:

1. 2.98m/s

2. 0.28m

Explanation:

The energy equation would work great in this scenario:

E=K+U. Since all of our energy comes from gravitational potential energy, and we are interested in finding the kinetic energy, all our mechanical energy must be in kinetic form, therefore:

$mgy=1/2mv^2+0\\98.36*4.45=1/2*98.36*v^2\\437.7=49.18v^2\\8.9=v^2\\2.98=v$

We can use energy to find max height.

For energy, set the equation E=K+U as 437.7=(mass adult+mass senior)gh:

$437.7=155.89gh\\\\2.8=gh\\0.28=h$

8 0

2 years ago