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Kruka [31]
3 years ago
13

A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope

, and the box moves upward with a velocity magnitude that varies in time according to v(t)=(2.00m/s2)t+(0.600m/s3)t2. Part A What is the tension in the rope when the velocity of the box is 15.0 m/s ?
Physics
1 answer:
Alex3 years ago
4 0

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

      T - W = m a

The weight

      W = mg

The acceleration can be found by derivatives

     a = dv / dt

     v = 2 t + 0.6 t²

     a = 2 + 0.6 t

We replace

      T - mg = m (2 + 0.6t)

      T = m (g + 2 + 0.6 t)               (1)

Let's look for the time for the speed of 15 m / s

       15 = 2 t + 0.6 t²

       0.6 t² + 2 t - 15 = 0

We solve the second degree equation

        t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6

        t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

       t = 3.6 s

Let's calculate from equation 1

       T = 2.00 (9.8 + 2 + 0. 6  3.6)

        T = 27.92 N

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Answer:

Explanation:

A )

speed of swimming in still water is given by the expression

distance / time

= 50 / 25

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In lane 1 , 1.2 cm/s current is flowing in the direction that the swimmers are going so swimmers will cover distance at the rate of  2 + 1.2 = 3.2 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / 3.2 = 15.625 s

In lane 8 , 1.2 cm/s current is flowing against  the direction that the swimmers are going so swimmers will cover distance at the rate of  2 - 1.2 = .8 m /s.

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3 years ago
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2 years ago
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What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?
iren [92.7K]

Answer:

\lambda=1090nm

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where R_H is the Rydberg constant for hydrogen and n_1, n_2 are the lower energy state and the higher energy state, respectively.

\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm

4 0
3 years ago
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
4 years ago
The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli
Elenna [48]

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

4 0
3 years ago
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