Which is described as the flow of charge particles?

A seashell is suspended in the air at rest by two strands of spider silk. Each strand of silk makes an angle of 75 degrees above

Ksenya-84 [330]

The mass of the seashell suspended by the two strands of spider silk is 0.28 kg.

The given parameters:

<em>Tension on each strand of silk, T = 1.42 N</em>
<em>Angle of inclination of each strand, θ = 75⁰</em>

The mass of the seashell at equilibrium is calculated by applying Newton's second law of motion;

$\Sigma F = 0\\\\T_1 sin(\theta) + T_2 sin(\theta) - W = 0$

where;

W is the weight of the seashell

The weight of the seashell is calculated as;

$1.42 \times sin(75) \ + \ 1.42 \times sin(75) \ - W = 0\\\\2.743 - W = 0\\\\W = 2.743 \ N$

The mass of the seashell is calculated as follows;

$W = mg\\\\m = \frac{W}{g} \\\\m = \frac{2.743}{9.8} \\\\m = 0.28 \ kg$

Thus, the mass of the seashell suspended by the two strands of spider silk is 0.28 kg.

Learn more about equilibrium forces here: brainly.com/question/8045102

5 0

3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is <u>6.4m</u>

5 0

3 years ago

Sally is concerned about the X-ray images her dentist plans to take of her teeth. She has read that X-rays can cause tissue effe

taurus [48]

Answer:

The dentist uses X-ray radiography, which takes too short a time to cause those problems.

Explanation:

I took the test on Edg

6 0

4 years ago

What is the equivalent capacitance of the three capacitors in the figure (figure 1)?

vekshin1

The equivalent capacitance of the combination is $\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}$ where C1 and C2 are the capacitance of both capacitors in series.

<h3>What is equivalent capacitor?</h3>

Let the capacitance of both capacitors be C1 and C2. For a series connected capacitors, same charge flows through the capacitors but different voltage flows through them.

Let Q be the amount of charge in each capacitors,

V be the voltage across each capacitors

C be the capacitance of the capacitor.

Using the formula Q = CV where V = Q/C... (1)

For the large capacitor with capacitance of the capacitor C1,

$Q = C_1V_1;$

$V_1 = \dfrac{Q}{C_1}...(2)$

where $V_1$ is the voltage across $C_1,$

For the small capacitor with capacitance of the capacitor $C_2$ ,

$Q = C_2V_2$ ;

$V_2 = \dfrac{Q}{C_2} ... (3)$

where $V_2$ is the voltage across $C_2$ ,

Total voltage V in the circuit will be;

$V = V_1+V_2... (4)$

Substituting equation 1, 2 and 3 in equation 4, we have;

$\dfrac{Q}{C} = \dfrac{Q}{C_1} +\dfrac{ Q}{C_2}$

$\dfrac{Q}{C} = Q({\dfrac{1}{C_1}+\dfrac{1}{C_2})$

Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;

$\dfrac{1}{C} = ({\dfrac{1}{C_1}+\dfrac{1}{C_2})$

This gives the equivalent capacitance of the combination.

To know more about equivalence capacitance follow

brainly.com/question/5626146

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7 0

2 years ago

A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is

igor_vitrenko [27]

Answer: M is equal to m.

Explanation:

The question gives us two important informations:

M is initially at rest
m finishes at rest after collision.

In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

p₁ = mv₁ + M.0 = p₂ = m.0 + Mv₂ ⇒ mv₁ = Mv₂

From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

k₁ = 1/2 m(v₁)² + 1/2 M.0 = k₂ = 1/2m.0 + 1/2M(v₂)² ⇒ m(v₁)² = M(v₂)²

Mathematically, the only way in which both equations be true, should be with v₁ = v₂, which is only possible if m=M too.

In this type of collision, it is said that the energy transfers from one mass to the other.

7 0

3 years ago