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Lapatulllka [165]
3 years ago
13

In pairs figure-skating competition, a 65-kg man and his 45-kg female partner stand facing each other both at rest on the ice. I

f they push apart and the woman has a velocity of 1.5 m/s eastward, what is the velocity of her partner? (Neglect friction.)
Physics
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

The final velocity of her partner is approximately -1.04 m/s or 1.04 m/s in the opposite direction to her direction of motion

Explanation:

The given parameters are;

The mass of the man, m₁ = 65 kg

The mass of the woman, m₂ = 45 kg

Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

The initial velocity of the man, v₁₁ = 0 m/s

The initial velocity of the man, v₁₂ = 0 m/s

The final velocity of the woman, v₂₂ = 1.5 m/s

The final velocity of the man = v₂₁

Therefore, we have, by the conservation of momentum principle;

The total initial momentum = The total final momentum

Which gives;

m₁ × v₁₁ + m₂ × v₁₂ = m₁ × v₂₁ + m₂ × v₂₂

Substituting the known values;

65 × 0 + 45 × 0 = 65 × v₂₁ + 45 × 1.5

∴ 65 × v₂₁ + 45 × 1.5 = 0

45 × 1.5 = - 65 × v₂₁

v₂₁ = 45 × 1.5/(-65) ≈ -1.04 m/s

The final velocity of the man, her partner = v₂₁ ≈ -1.04 m/s.

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A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli
fredd [130]

Answer:

Part a)

A = 1.78 m

Part b)

f = 2 rev/s

Explanation:

Part A)

As we know that time period of the motion is given as

T = 2.68 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.68}

\omega = 2.34 rad/s

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

mg = m\omega^2 A

so we have

A = \frac{g}{\omega^2}

A = \frac{9.81}{2.34^2}

A = 1.78 m

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

mg = m\omega^2 A

g = \omega^2 (0.0623)

\omega = 12.5 rad/s

so now we have

2\pi f = 12.5

f = 2 rev/s

3 0
3 years ago
A rock falls off a cliff. How fast will it be going after falling for 4.33 seconds?
bixtya [17]

Answer:42.43m/s

Explanation:According to vf=vi+at, we  can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s

4 0
11 months ago
A. Obtain the following: microwave, ruler, something meltable (e.g. candy bar, marshmallows) B. A microwave works by setting up
Makovka662 [10]

Answer:

E = 124.7 N / C

Explanation:

Let's analyze the exercise: the microwave creates an electromagnetic wave of frequency F = 2.45 GHz, this wave is introduced into the microwave cavity and is reflected on the metal walls, which is why one or more standing waves are formed.

The electric field of the standing wave is

            I = E²

            E =√I

where I is the intensity of the radiation.

What is it

             I = P / A

where P is the effective emission power, almost all the power of the microwave and A is the area of ​​the cavity, in the most used microwaves

P = 700 W and the area is A = 25 x 18 cm² = 0.045 m²

             I = 700 / 0.045

             I = 15555.56 W/m²

let's calculate the electric field

            E = √15555.56

            E = 124.7 N / C

7 0
3 years ago
Calculate the power of convex lens of focal length 20cm​
maxonik [38]

D = 1/f, where D is the power in diopters and f is the focal length in meters.

D=1/20

<u>D=0.05</u>

6 0
3 years ago
What is a charge-coupled device (CCD), and how is it used in astronomy?
Sedaia [141]

Answer:

Charge-coupled device  (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device  (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

6 0
3 years ago
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