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Kobotan [32]
2 years ago
13

if two masses 5.2kg and 4.8kg are attached to ends of inextensible string passed over a friction less pulley then acceleration o

f system
Physics
1 answer:
Alexus [3.1K]2 years ago
4 0

Answer:

If two masses 5.2 kg and 4.8 kg are attached to the ends of inextensible strings passed over a frictionless pulley then the acceleration of system is 0.4 ms-2.

Explanation:

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What is the metric unit for time?
spin [16.1K]
The base unit of time in the metric and SI system is the second.
5 0
3 years ago
In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
The quadriceps muscles pull on the patella simultaneously. Below are the forces from each
Nostrana [21]

Based on the calculation of the resultant of vector forces:

  1. the resultant force due to the quadriceps is 1795 N
  2. the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.

<h3>What is the resultant force due to the quadriceps?</h3>

The resultant of more than two vector forces is given by:

  • F = √Fₓ² + Fₙ²

where:

  • Fₓ is the sum of the horizontal components of the forces
  • Fₙ is the sum of the vertical components of the forces
  • Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
  • Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 480 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55

Fx = -280.6 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55

Fₙ = 1773.1 N

then:

F = √(-280.6)² + ( 1773.1)²

F = 1795.16 N

F ≈ 1795 N

Therefore, the resultant force due to the quadriceps is 1795 N

<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>

From the new information provided:

  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 720 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55

Fx = -142.95 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55

Fₙ = 1969.72 N

then:

F = √(-142.95)² + ( 1969.72)²

F = 1974.9 N

F ≈ 1975 N

Therefore, the resultant force due to the quadriceps is 1975 N.

Training and strengthening the vastus medialis results in a greater force of muscle contraction.

Learn more about resultant of forces at: brainly.com/question/25239010

3 0
2 years ago
What kind of employment is regarded as national level employment?
yanalaym [24]

Answer:

In most cases, the sources of information for both levels are same as follows:

Verbal or personal contacts : This information is transferred through verbal method. ...

Different media: Media is one of the best sources of information in these days.

Explanation:

7 0
2 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
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