Orb spiders make silk with a typical diameter of 0.15 mm. a. A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch?b. What is the maximum weight that a single thread of this silk could support?

Answer:

i got u its a

Explanation:

Answer:

X = 69.1 x 10⁻⁶ m = 69.1 μm

Explanation:

The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:

d = mλ/2

where,

d = distance moved by the mirror = X = ?

m = No. of Fringes counted = 246

λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m

Therefore,

X = (246)(5.62 x 10⁻⁷ m)/2

Therefore,

X = 69.1 x 10⁻⁶ m = 69.1 μm

Answer:

c. (H-h/2) above the ground

Explanation:

The mirror must be at least half as tall as the alien, and its base must be located at half of the distance between the alien's eyes and the ground (assuming that the alien doesn't float or levitate).

This question is about the Law of Reflection which states that the angle of reflection = angle of incidence.

I attached an image that can help you understand the concept, although the alien is not included.

Answer:

None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave

Explanation:

Answer:it’s A

Explanation:

because i took the quiz

Answer:

D is the correct answer, not A

Explanation:

Answer:

A) refraction experiment   n = n₁ sin θ₁ / sin θ₂

B)  n_A = 1.19 ,    n_B = 1.53

Explanation:

A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index

            n₁ sin θ₁ = n₂ sinθ₂

            n₂ = n₁ sin θ₁₁ /sin θ₂

If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is

            n = n₁ sin θ₁ / sin θ₂

B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be

material A

             n_A = sin 50 / sin 40

             n_A = 1.19

material B

              n_B = sin 50 / sin30

              n_B = 1.53

Answer:

F = 1.63 x 10⁻⁹ N

Explanation:

Complete question is as follows:

The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers. Find the magnitude of Gravitational Force?

Answer:

The gravitational force is given by Newton's Gravitational Law as follows:

F = Gm₁m₂/r²

where,

F = Gravitational Force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of each ball = 7 kg

r = distance between balls = 2 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(7 kg)(7 kg)/(2 m)²

F = 1.63 x 10⁻⁹ N

Answer:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. ... If that velocity is zero, then the object remains at rest.

Explanation:

Answer:

Newton's First Law is about inertia; objects at rest stay at rest unless acted upon and objects in motion continue that motion in a straight line unless acted upon. The amount of inertia an object has is simply related to the mass of the object.

Answer:

163.35

__________________________________________________________

We are given:

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

Kinetic Energy of the object:

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)²            [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

Answer:

56 J

Explanation:

The following data were obtained from the question:

Energy 1 (E₁) = 7 J

Extention 1 (e₁) = 1.8 cm

Extention 2 (e₂) = 1.8 + 3.6 = 5.4 cm

Energy 2 (E₂) =?

Energy stored in a spring is given by the following equation:

E = ke²

Where E is the energy.

K is the spring constant.

e is the extension.

E = ke²

Divide both side by e²

K = E/e²

Thus,

E₁/e₁² = E₂/e₂²

7/ 1.8² = E₂/ 5.4²

7 / 3.24 = E₂/ 29.16

Cross multiply

3.24 × E₂ = 7 × 29.16

3.24 × E₂ = 204.12

Divide both side by 3.24

E₂ = 204.12 / 3.24

E₂ = 63 J

Thus, the additional energy required can be obtained as follow:

Energy 1 (E₁) = 7 J

Energy 2 (E₂) = 63 J

Additional energy = 63 – 7

Additional energy = 56 J

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

Answer:

2.56 nC

Explanation:

       [tex]C =\frac{Q}{V} (1)[/tex]

       [tex]Q = \sigma * A (2)[/tex]

        [tex]V = E*d (3)[/tex]

        where d is the distance between plates.

       [tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]

       [tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]

        [tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]

       [tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]

Answer:

C. 4.3 seconds

Explanation:

B 0.56 s is the time period of a twirlers baton.

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

a = 4π²R/T²

Given values are:

a = 47.8 m/s²

D = 0.76 m  so , R = 0.76/2 = 0.38m

Using this formula,

47.8*T² = 4π² x0.38

T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have  time period of 0.56 s.

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.

Answer:

it's an insulator

Explanation:

Insulators provides resistance

Answer:

C. insulator

Explanation:

Answer:

20m/s

Explanation:

u = 0m/s

t = 2s

a = +g = 10m/s²

t = 2d

v = ?

v = ut + 1/2at²

v = 0(2) + 1/2(10)(2)²

v = 0 + 5(4)

v = 20m/s

Answer:

1 W = 1 V x 1 A

Explanation:

Other dude is wrong this is right

Answer:

Explanation:

Force = mass * acceleration

Given

Mass = 1500kg

Get the acceleration using the equation of motion;

v² = u²+2aS

20² = 0+2s(200)

400 = 400a

a = 400/400

a = 1m/s²

Get the upward force required

F = 1500 * 1

F = 1500N

Hence the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s is 1500N

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then   m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

Answer:

 x = 10.75 m

Explanation:

For this problem we will solve it in two parts, the first using energy and the second with kinematics

Let's use the energy work relationship to find the velocity of the block as it exits the ramp

       W = [tex]Em_{f}[/tex] - Em₀

Starting point. Higher

       Em₀ = U = m g h

the height from the edge of the ramp of the graph has a value

        h = 9-3 = 6 m

Final point. At the bottom of the ramp

       Em_{f} = K = ½ m v²

Friction force work

      W = - fr  d

The friction force has the formula

      fr = μ N

On the ramp, we can use Newton's second law

         N - W cos θ = 0

         N = W cos θ

where the angle is obtained from the graph

         tan θ = (9-3) / (0.5-4) = -6 / 3.5

         θ = tan⁻¹ (-1,714)

         θ = -59.7º

the distance d is

         d = √ (Δx² + Δy²)

         d = √ [(0.5-4)² + (9-3)²]

         d = 6.95 m

for which the work is

       W = - μ mg cos 59.7 d

we substitute

        W = Em_{f} -Em₀

        - μ mg cos 59.7 d = ½ m v² - m g h

In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2

        - μ g cos 59.7 d = ½ v² - g h

         v² = 2g (h - very d coss 59.7)

let's calculate

         v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)

         v = √ 103.8546

         v = 10.19 m / s

in the same direction as the ramp

in the second part we use projectile launch kinematics

let's look for the components of velocity

         v₀ₓ = vo cos -59.7

         [tex]v_{oy}[/tex] = vo sin (-59,7)

         v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s

         v_{oy} = 10.19 if (-59.7) = -8.798 m / s

Let's find the time to get to the floor (y = o)

          y = y₀ + v_{oy} t - ½ g t²

to de groph y₀=3 m

          0 = 3 - 8.798 t - ½ 9.8 t²

          t² - 1.796 t - 0.612 = 0

we solve the quadratic equation

          t = [1.796 ±√(1.796² + 4 0.612)] / 2

          t = [1,795 ± 2,382] / 2

          t₁ = 2.09 s

          t₂ = -0.29 s

since time must be a positive quantity the correct value is t = 2.09 s

we calculate the horizontal displacement

          x = v₀ₓ t

          x = 5.14 2.09

          x = 10.75 m

The motion of the box, after it exits the incline is the motion and trajectory

of a projectile.

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor is approximately 1.24613 m.

Reasons:

Mass of the block,  m = 3.5 kg

Coefficient of kinetic friction, μ = 1.2

Location of the = 0.5 m from the origin

Required:

Horizontal distance between the block's point of contact with the floor and

the bottom right-hand edge of the incline.

Solution:

Let θ represent the angle the incline make with the horizontal.

The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)

Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L

Rise of the incline = 10 - 3 = 7

Run of the incline = 4

L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]

Let ΔP.E.₁  represent the potential energy transferred to kinetic energy

and work along the incline, we have;

Energy of the block at the bottom of the incline, M.E.₂, is found as follows;

K.E.₂ = mgh - m·g·μ·cos(θ)·L

[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]

v ≈ 6.1456 m/s

The vertical component of the velocity is therefore;

[tex]v_y = v \cdot sin(\theta)[/tex]

[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]

From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;

ΔP.E.₁ = 3.5×9.81×7

3 = 5.33588·t + 0.5×9.81·t²

Factorizing, the above quadratic equation, we get;

The time it takes the block to reach the floor, t ≈ 0.40869 seconds

Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]

The horizontal distance, x = vₓ × t

∴ x = 3.04908 × 0.40869 ≈ 1.08194

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor, x ≈ 1.24613 m.

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Answer:

It isn't moving

Explanation:

Answer:

the speed of the mass at the bottom of its swing is 6.61m/s

Explanation:

Applying energy conservation

[tex]\frac{1}{2}m(Vlowest)^2 = mg(2R) + \frac{1}{2}m(Vtop)^2[/tex]

There is no potential energy at the bottom as the body will have a kinetic energy there.

h= 2R = 1.6m as the diameter of the circle will represent the height in the circle.

g = 9.8m/s^2

m will cancel out, so the net equation becomes.

[tex]\frac{(Vbottom)^2}{2} = 2gR + \frac{(Vtop)^2}{2}[/tex]

                 = [tex]2*9.8*0.8 + \frac{(3.5)^2}{2}[/tex]

                  =  15.68+ 6.125

         [tex]\frac{(Vbottom)^2 }{2}[/tex]     =  21.805

(Vb)^2 = 2*21.805

     = 43.64

Vb = 6.61m/s

Answer:

C. The net force applied to the ball is zero.

Explanation:

From Newton's second law of motion;

F = ma

Where F is the force on an object, m is its mass and a is its acceleration.

Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.

Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.

So that;

F = m x 0

  = 0

No force is applied on the object.

Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.

Answer:

Explanation:

Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.

the expression is Ft=mv

where F= force

 m= mass

t= time

v= velocity

Step one:

given data

mass m= 110g= 0.11kg

velocity v= 6.5m/s

time t= 0.19seconds

Step two:

we also know that the force on impulse is given as

Ft=mv

F=0.11*6.5/0.19

F=0.715/0.19

F=3.76N

The magnitude of the average force on the wall if the duration of the collision is 0.19 is 14N

Answer:

V₀ = 45.81 m/s

H = 70.45 m

T = 5.36 s

Explanation:

The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 201.24 m

V₀ = Initial Speed = ?

θ = Launch Angle = 35°

g = 9.8 m/s²

Therefore,

201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²

V₀ = √[(201.24 m)/(0.095 m/s²)

V₀ = 45.81 m/s

Now, for maximum height:

H = V₀² Sin² θ/g

H = (45.81 m/s)² Sin² 35°/9.8 m/s²

H = 70.45 m

For the total time of flight:

T = 2 V₀ Sin θ/g

T = 2(45.81 m/s) Sin 35°/9.8 m/s²

T = 5.36 s

Answer:

Explanation:

Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.

the expression is Ft=mv

where F= force

           m= mass

             t= time

            v= velocity

Step one:

given data

mass m=3kg

velocity v= 25m/s

time t= 0.10seconds

Step two:

we also know that the force on impulse is given as

Ft=mv

F=3*25/0.10

F=75/0.10

F=750N

The magnitude of the average force on the heavy bag if the duration of the collision is 0.1 is 750N

We know that Period of a wave is the inverse of its Frequency

So,  Period = 1 / Frequency

From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period

Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times

Answer:

The value  is  [tex]\theta =407.3 \ radian[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]

    The first time is  [tex]t_1 = 1.00 \ s[/tex]

    The second time [tex]t_2 = 6.10 \ s[/tex]

Generally the angular velocity is mathematically represented as

     [tex]w = \int\limits {\alpha } \, dt[/tex]

=>  [tex]w = \int\limits {7t + 8 } \, dt[/tex]

=>  [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]

Generally the angular displacement  is mathematically represented as

[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]

=>  [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]

=>  [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]

=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]

=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]

=> [tex]\theta =407.3 \ radian[/tex]

Answer:

The two RF signals are in phase.

Explanation:

A wave is a disturbance that travels through a medium which transfers energy from one point to another in the medium without causing any permanent displacement of the particles of the medium.

Characteristics of waves include frequency, wavelength, velocity, etc.

Two types of waves are longitudinal and transverse wave. Radio Frequency (RF) signals travel in the form of transverse waves which have regions of maximum and minimum displacements called crests and troughs.

Travelling waves with  the same frequency may be said to be in phase or out of phase depending on whether their crests/peaks or troughs/valleys are reached at the same instant of time.

When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment or in step, they are said to be in phase.

The phase of a wave involves the relationship between the position of the amplitude peaks and valleys of two waveforms.