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pickupchik [31]
3 years ago
9

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). if a particular disk is spun at 79

2.7 rad / s while it is being read, and then is allowed to come to rest over 0.234 seconds, what is the magnitude of the average angular acceleration of the disk?
Physics
1 answer:
NeX [460]3 years ago
4 0
(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²
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An archer shoots an arrow at a 74.0 m distant target, the bull's-eye of which is at same height as the release height of the arr
GuDViN [60]

A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °

B. The arrow will go over the branch.

<h3>A. How to determine the angle</h3>
  • Range (R) = 74 m
  • Initial velocity (u) = 33 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Angle (θ) = ?

R = u²Sine(2θ) / g

74 = 33² × Sine (2θ) / 9.8

Cross multiply

74 × 9.8 = 33² × Sine (2θ)

725.2 = 1098 × Sine (2θ)

Divide both sides by 1098

Sine (2θ) = 725.2 / 1098

Sine (2θ) = 0.6605

Take the inverse of sine

2θ = Sine⁻¹ 0.6605

2θ = 41.3

Divide both sides by 2

θ = 41.3 / 2

θ = 20.7 °

<h3>B. How to determine if the arrow will go over or under the branch</h3>

To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:

  • Initial velocity (u) = 33 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Angle (θ) = 20.7 °
  • Maximum height (H) = ?

H = u²Sine²θ / 2g

H = [33² × (Sine 20.7)²] / (2 ×9.8)

H = 6.94 m

Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).

Therefore, we can conclude that the arrow will go over the branch

Learn more about projectile motion:

brainly.com/question/20326485

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3 0
2 years ago
The law of reflection states that the angle of reflection is equal to the angle of
Natali [406]

The law of reflection states that the angle of reflection is equal to the angle of Incidence .

3 0
4 years ago
A) Millions of years of rain and wind
barxatty [35]

Answer:

B. Millions of years of high pressure and temperature.

  • Millions of years of high pressure and temperature the ancient forests turn into fossil fuels.

Explanation:

Fossil fuels were formed from the dead remains of living organisms millions of years ago. They were formed under high pressure and high temperature.

  • Coal : about 300 million years ago the earth had this forest in low lying wetland areas. Due to natural processes like flooding, these forest god buried under the soil. As more soil deposited over them, they were compressed. Under high pressure and high temperature, dead plants got slowly converted into coal.
  • Petroleum : Petroleum was formed from organisms living in the sea. As these organisms died, their bodies settled at the bottom of the sea and got covered with layers of soil and clay. Over millions of years, absence of air, high pressure and high temperature transformed the dead organisms into Petroleum.

Coal, Petroleum and natural gas are fossil fuels.

6 0
2 years ago
A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o
IrinaK [193]

Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as \sigma

\sigma=\frac{Q}{4\pi r_{out}^2}, and the strength of the electric field outside the sphere E=\frac{\sigma _{new}}{\epsilon _o}

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

5 0
3 years ago
A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is
GarryVolchara [31]

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)

so we have

H = 1.47 m

Part b)

Similarly when spring is compressed by 4 cm

then we have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)

so we have

H = 5.94 m

8 0
4 years ago
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