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il63 [147K]
2 years ago
14

An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.

Physics
1 answer:
Aleksandr-060686 [28]2 years ago
6 0

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

  • T(u) is tension in upper rope
  • T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

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3 0
3 years ago
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A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

6 0
3 years ago
What is the weight of a spring balance when the point is 30​
andrezito [222]

Answer:

A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)

Explanation:

8 0
3 years ago
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0
3 years ago
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