The famous astronomer, Kepler, determined that the planetary orbits were: 1. ellipses 2. circles 3. epicentric 4. geocentric

If you are 5.00ft and 10.0in tall, what is your height in meters

Stolb23 [73]

The answer that I got was 5'8333..., which is the total for height and in inches, all of this would be 70 inches, or 1.78 meters, or 178 centimeters, or 1780 millimeters, but just go for the height that person is going to be at, 5'833... (this is repeated so just round it up to 5'8 if you want to.)

Hope this helped!

Nate

8 0

3 years ago

. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder

andrezito [222]

Answer:

As we can see, a string is attached with block A, and three string is folded with ply which is attached with B

x

B

=3x

A

Now differentiate with respect to x

V

B

=3V

A

Given,

V

A

=0.6m/s(totheright)

So,

V

B

=0.6×3

=1.8m/s(downward)

Explanation:

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5 0

3 years ago

Read 2 more answers

A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

4 0

4 years ago

Read 2 more answers

Person 1 moves an object 10 meters and exerts a force of 50N in 5 seconds. Person 2 moves an object 10 meters and exerts a force

mel-nik [20]

Answer:

Explanation:

Work is a force time the distance moved in the direction of that force, time is not a variable. Provided that the 50 N forces were applied in the same direction, the work done is identical. Assuming both applied force and direction of motion are horizontal W = Fd = 50(10) = 500J.

If the reason that one was slower is because the second person applied his force at an angle, let's say 60° below the horizontal, then the work done by the second person is 50cos60(10) = 250 J

Time IS a consideration for Power, the RATE of doing work. Provided the force and motion are horizontal, the first person applied twice as much Power as the second person doing an identical amount of work in half the time.

7 0

3 years ago

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column

JulsSmile [24]

Answer:

1) f = 214 Hz , 2) answer is c , 3) f = 428 Hz , 4) f₂ = 428 Hz , f₃ = 643Hz

Explanation:

1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore

L = λ / 2

λ = 2L

λ = 2 0.8

λ = 1.6 m

wavelength and frequency are related to the speed of sound (v = 343 m / s)

v =λ f

f = v / λ

f = 343 / 1.6

f = 214 Hz

2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced

λ' = 2 L ’

as L ’<L₀

λ' <λ₀

f = v / λ'

f' > fo

the correct answer is c

3) in this case the length is L = 0.40 m

λ = 2 0.4 = 0.8 m

f = 343 / 0.8

f = 428 Hz

4) the different harmonics are described by the expression

λ = 2L / n n = 1, 2, 3

λ₂ = L

f₂ = 343 / 0.8

f₂ = 428 Hz

λ₃ = 2 0.8 / 3

λ₃ = 0.533 m

f₃ = 343 / 0.533

f₃ = 643 Hz

4,1) as we have two maximums at the ends, all integer multiples are present

the answer is C

E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested

in this pipe there is a maximum in the open part and a node in the closed part, so the expression

L = λ / 4

L = 1.6 / 4

L = 0.4 m

the answer is C

F) in this type of pipe the general expression is

λ = 4L / n n = 1, 3, 5 (2n + 1)

therefore only odd values can produce standing waves

λ₃ = 4L / 3

λ₃ = 4 0.4 / 3

λ₃ = 0.533

f₃ = 343 / 0.533

f₃ = 643 Hz

5 0

4 years ago