PHYSICS COLLEGE

What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?A. 1.6J
B. 7.9J
C. 15J
D. 20J

Answers

Answer 1

Answer:

Answer:

D. 20J

Explanation:

Answer 2

Answer:

Answer:

20 J

Explanation:

yes

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Consider an astronomical telescope with a 48 centimeter focal-length objective lens and a 10 centimeter focal-length eyepiece. Approximately how many centimeters apart should the lenses be placed

Answers

Answer:

D = 58 cm

Explanation:

Given that,

Focal length of the objective lens, $f_o=48\ cm$

Focal length of the eye piece, $f_e=10\ cm$

We need to find how many cm apart should the lenses be placed. Let d be the distance between lenses. It is equal to the sum of focal lengths of objective lens and eye-piece

D = 48 cm + 10 cm

= 58 cm

Hence, the object is placed at a distance of 58 cm.

Final answer:

In an astronomical telescope, the lenses should be placed at a distance equal to the sum of their focal lengths. In this case, this distance would be 58 cm.

Explanation:

In an astronomical telescope, the distance between the objective lens and the eyepiece should be equal to the sum of their focal lengths for the telescope to produce clear and sharp images. Here, the focal length of the objective lens is 48 cm and the focal length of the eyepiece is 10 cm.

Therefore, calculating: Objective lens focal length + Eyepiece focal length = 48 cm (objective) + 10 cm (eyepiece) = 58 cm

This means that the objective lens and the eyepiece should be approximately 58 centimeters apart.

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The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.

Answers

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

$v = \sqrt(T)/(\mu) {}$

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

$v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(2.5)/(6.25*10^(-3)) } \n\nv = 20 \ m/s$

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?

Answers

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = (1)/(12)ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= (1)/(12)ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$((1)/(12)ML^2 +2mr_1^2) \omega_i = ((1)/(12)ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s$

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

The landing gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described determine the values of k and c that limit the amplitude of vibration of the airplane (x) to 0.1 m. assume

Answers

The land of airplane gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described

5. The order is to give 600 mg of Ampicillin IM q8h. The directions for dilution on the 2 gm vial reads: Reconstitute with 4.8 mL of sterile water to obtain a concentration of 400 mg per mL. How many mL will you administer per dose?

Answers

The volume of Ampicillin IM q8h to be administered per dose is 1.5 mL when the order is to give 600 mg of it from a concentration of 400 mg per mL prepared by the dilution of 2 g in 4.8 mL of sterile water.

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