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Studentka2010 [4]
2 years ago
7

Suppose a particle is accelerated through space by a constant 10-N force. Suddenly the particle encounters a second force of 10-

N in a direction opposite to that of the first force. The particle
Physics
1 answer:
Harlamova29_29 [7]2 years ago
3 0

Answer:

The particle will continue moving at constant velocity

Explanation:

When the particle encounters the second force of 10 N, the net force acting on the particle becomes zero, because the two forces are equal in magnitude but opposite in direction:

F_{net}=10 N-10 N=0

For Newton's second Law, the acceleration of the particle is proportional to the net force acting on it:

F_{net}=ma

Therefore, since F_{net}=0, the acceleration is zero (a=0) and so the particle will keep a constant velocity.

You might be interested in
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α   Formula  (1)

where:

τ : It is the moment applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :  

ωf = ω₀+ α*t  Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (rad)

Data  

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel  

We replace data in the formula (2):  

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F  = tangential force (N)

R  = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0
1 year ago
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
Greeley [361]

Answer:

I = 215.76 A  

Explanation:

The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.  

Fm = u_o*I^2*L/2*π*d

where I is the current in each rod, d = 0.0082 m is the distance 27rId  

between each, L = 0.85 m is the length of each rod.

Fm = 4π*10^-7*I^2*1.1/2*π*0.0083

The mass of each rod is m = 0.0276 kg  

F_m = mg

4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8

I = 215.76 A  

note:

mathematical calculation maybe wrong or having little bit error but the method is perfectly fine

5 0
2 years ago
Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una
EleoNora [17]

Answer:

a)  f = 615.2 Hz      b)  f = 307.6 Hz

Explanation:

The speed in a wave on a string is

         v = √ T / μ

also the speed a wave must meet the relationship

          v = λ f

           

Let's use these expressions in our problem, for the initial conditions

            v = √ T₀ /μ

             √ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

         T = 2T₀

          √ (T /μ) = λ f

          √( 2To /μ) = λ f

         √2  √ T₀ /μ = λ f

we substitute

         √2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

           λ₀ = λ

           f = f₀ √2

           f = 435 √ 2

           f = 615.2 Hz

b) The tension is cut in half

         T = T₀ / 2

         √ (T₀ / 2muy) =  f = λ f

          √ (T₀ / μ)  1 /√2 = λ f

           fo / √2 = f

           f = 435 / √2

           f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

         v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

          v= λ f  

           

Usemos estas expresión en nuestro problema, para las condiciones iniciales

            v= √ To/μ

             √ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

         T = 2T₀

          √ ( T/μ) = λf

          √ ) 2T₀/μ = λ f

         √ 2 √ T₀/μ = λ f

         

substituimos  

         √2    ( λ₀ f₀)  =  λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

                 λ₀ =  λ

           f = f₀ √2

           f = 435 √2

           f = 615,2 Hz

b)  La tension se reduce a la mitad

         T = T₀/2    

         RA ( T₀/2μ)  =  λ  f

          Ra(T₀/μ) 1/ra 2  =  λ f

           fo /√ 2 = f

           f = 435/√2

           f = 307,6 Hz

5 0
1 year ago
An alloy is made of a material of specific gravity 7.87 and another material of specific gravity 4.50. The alloy of mass 750g ha
julsineya [31]

Answer:

13.9

Explanation:

Apparent weight is the normal force.  Sum of the forces on the alloy when it is submerged:

∑F = ma

N + B − W = 0

N + ρVg − mg = 0

6.6 + (0.78 × 1000) V (9.8) − (0.750) (9.8) = 0

V = 9.81×10⁻⁵

If x is the volume of the first material, and y is the volume of the second material, then:

x + y = 9.81×10⁻⁵

(7.87×1000) x + (4.50×1000) y = 0.750

Two equations, two variables.  Solve with substitution:

7870 (9.81×10⁻⁵ − y) + 4500 y = 0.750

0.772 − 7870 y + 4500 y = 0.750

0.0222 = 3370 y

y = 6.58×10⁻⁶

x = 9.15×10⁻⁵

The ratio of the volumes is:

x/y = 13.9

8 0
2 years ago
Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t
Zepler [3.9K]

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    p_{f} = m₁ v₁ + m₂ v₂

   p₀ =  p_{f}

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

6 0
1 year ago
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