A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate

Answer:

[tex]\phi=628.3[/tex]

Explanation:

From the question we are told that

Radius [tex]r=10.0 cm[/tex]

Magnetic field[tex]B=2T[/tex]

Generally the equation for area of circular path is mathematically given by

 [tex]Area=\pi r^2[/tex]

 [tex]A=\pi 10^2[/tex]

 [tex]A=314.15m^2[/tex]

Generally the equation for Magnetic flux is mathematically given by

 [tex]\phi=BA[/tex]

 [tex]\phi=2*314.15[/tex]

 [tex]\phi=628.3[/tex]

Given :

A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.

The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.

To Find :

At what location are the kinetic energy and the potential energy the same.

Solution :

Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.

So,

[tex]P.E = K.E\\\\\dfrac{kx^2}{2} = \dfrac{mv^2}{2}\\\\x = v\sqrt{\dfrac{m}{k}}\\\\x = 1.5 \times \sqrt{\dfrac{0.5}{20}}\ m\\\\x = 0.238 \ m[/tex]

Hence, this is the required solution.

Answer:

To increase the conductivity of the material.

Explanation:

Generally , the group 4 elements are non conductor but in certain conditions, such as doping or the increase in temperature, they becomes conductor.

The doping is the process of mixing of pentavalent or the trivalent material into tetra valent material in the very small amount, so that the material becomes conductor.

In making a diode we need two types of the materials, n type semiconductor and p type semi conductor.

When the trivalent impurity is added in the tetra valent element, the semiconductor becomes n type because an electron is left for the conduction.

When the pentavalent impurity is added in the tetra valent element, the semiconductor becomes p type because a hole is left for the conduction.

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.

Given-

velocity of the ice skater is 12 m/ sec.

Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.

[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]

The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,

[tex]W_{f}=\bigtriangleup KE +0[/tex]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,

[tex]W_{f} =u_{f} mgx[/tex]

Here,  [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .

Equate the value of kinetic energy and work done of friction for further result, we get,

[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]

[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]

[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]

[tex]u_{f} =0.077[/tex]

Hence, the coefficient of kinetic friction between the ice and skates is 0.077.

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Answer:

3 pls give me brainliest

Explanation:

a particle of violet light has exactly the same energy as a particle of red light

Answer:

Golgi apparatus (bodies).

Explanation:

A cell can be defined as the fundamental or basic functional, structural and smallest unit of life for all living organisms. Some living organisms are unicellular while others are multicellular in nature.

Generally, cells have the ability to independently replicate themselves. In a cell, the "workers" that perform various functions or tasks for the survival of the living organism are referred to as organelles. Some examples of cell organelles found in all living organisms such as trees, birds, and bacteria include; nucleus, cytoplasm, cell membrane, golgi apparatus, mitochondria, lysosomes, ribosomes, chromosomes, endoplasmic reticulum, vesicles, etc.

Golgi apparatus is also referred to as Golgi bodies and it functions as a packaging unit in living organisms, especially eukaryotic cells because it prepares protein and lipid molecules for export by chemically tagging them.

Hence, the Golgi apparatus (bodies) is an organelle involved in cell secretion through the transportation (export) of proteins and lipids out of a cell.

Answer:

Horizontal Component of Fish's Velocity = 2.6 m/s

Explanation:

In this scenario, we will neglect the effects of the air resistance on the small fish. Since there is no resisting force available in the horizontal direction. Therefore, the horizontal component of the velocity of the fish will remain equal to the horizontal component of the velocity of the seagull and it will remain the same throughout the whole motion.

Horizontal Component of Fish's Velocity = Constant Horizontal Speed of Seagull

Horizontal Component of Fish's Velocity = 2.6 m/s

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= [tex]480+720[/tex]

= [tex]1200 \ units[/tex]

The total number of particles will be:

= [tex]50+100[/tex]

= [tex]150[/tex]

hence,

Energy of each A particle or each B particle will be:

= [tex]\frac{1200}{150}[/tex]

= [tex]8 \ units[/tex]

Answer:

Explanation:

For time period of revolution , the expression is as follows .

T² =  4π² R³ /GM , M is mass of the earth.

Putting the values

T² =  4π² (7880 x 10³)³ /(6.67 x 10⁻¹¹ )( 5.97 x 10²⁴ )

T² = 4.846 x 10⁷ s

T = 6.961 x 10³ s

= 6961 s

= 116 minutes .

Answer:

a)   p = 4.167 cm, b)   m = + 6

Explanation:

a) For this exercise we must use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm

Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.

       [tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]

      [tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]

      [tex]\frac{1}{ p}[/tex] = 024

      p = 4.167 cm

b) angular magnification

     m = h ’/ h = - q / p

     m = - (-25) /4.167

     m = + 6

the positive sign indicates that the image is straight and enlarged

Answer:

Explanation:

the spheres cant be electrons as they will repel each other.

so the ans is D. Atoms

Answer:

D. Atoms

Explanation:

The spheres in this model represents the atoms. So, option (D) is correct answer.

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

Answer: hello your question is incomplete attached below is the complete question

answer :  1/2 KD^2  ( option A )

Explanation:

P.E ( potential energy ) = mgd

In case 1 P.E = 0   i.e. mgd = 0  

Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring

The potential energy of the M in case 2

= P.E of M at rest + P.E of the spring

= 0 + 1/2 KD^2

Answer:

We cannot tell from the information given

Explanation:

Given;

mass of the box, m = 5 kg

first force, F₁ = 10 N

second force, F₂ = 5 N

(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;

∑Fx = 10 N - 5 N

      = 5 N

Apply Newton's second law of motion;

∑Fx = ma

a = ∑Fx/m

a = 5 / 5

a = 1 m/s² in the direction of the 10 N force.

(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;

∑Fx = 10 N + 5 N

∑Fx = 15 N

a = 15 / 5

a = 3 m/s²

Therefore, the information given is not enough to determine the acceleration of the box.

The distance covered by the car for the speed of 2.40 m/s² for 15 seconds is 36 meters

To find the distance the given datas are:

Speed = 2.40 m/s²

Time = 15 seconds.

               Distance = Speed × Time.

Substituting the given datas in the formula,

Distance = 2.40 × 15

               = 36 m

The Car went at the distance of about 36 meters.

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Answer:

 [tex]Q=2C[/tex]

Explanation:

From the question we are told that:

Force [tex]F=10000000N[/tex]

Distance [tex]d=60m[/tex]

Where

 [tex]Q_1=Q_2[/tex]

Generally the equation for Force is mathematically given by

 [tex]F=\frac{KQ^2}{r^2}[/tex]

Where

 [tex]K=9*10^9[/tex]

Therefore

 [tex]Q^2=\frac{Fr^2}{K}[/tex]

 [tex]Q^2=\frac{10000000*60^2}{98*10^9}[/tex]

 [tex]Q=\sqrt{\frac{10000000*60^2}{9*10^9}}[/tex]

 [tex]Q=2C[/tex]

What is the primary evidence used to determine how the Moon formed?

[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]

A. Moon craters and Earth craters were caused by the same asteroid strike. ✅

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]

Answer:

θ = 33°

Explanation:

Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:

[tex]T = \frac{2\ u\ Sin\theta}{g} \\\\Sin\theta = \frac{Tg}{2u}[/tex]

where,

θ = launch angle = ?

T = Total time of flight = 0.6 s

g = acceleration due to gravity = 9.81 m/s²

u = launch speed = 5.4 m/s

Therefore,

[tex]Sin\theta = \frac{(0.6\ s)(9.81\ m/s^2)}{(2)(5.4\ m/s)}\\\\\theta = Sin^{-1}(0.545)[/tex]

θ = 33°

Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point =  mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

Answer:

 d = 178.57 10⁻⁹ m

Explanation:

For this exercise we must find the thickness to minimize the reflection, so the interference for the reflection must be destructive.

To find the expression we must take into account, two things:

* When the light goes from an index mordant medium to one with a higher refractive incoe, it undergoes a phase change of 180 (pi radians)

* within the film the wavelength of light is modulated by the index of refraction

           λₙ = λ₀/ n

In this case the light passes from the air to the reflective layer and undergoes a phase change of ∫π rad,  then it is reflected in the film-glass layer where it undergoes another phase change of π rad, therefore the total change of phase is 2π radians, this change is the or changes its value

period of the trigonometric functions, therefore its value does not change

  the expression for destructive interference is

           d sin θ = (me + ½) λₙ

           d sin θ = (m + ½) λ₀ / n

the minimum thickness occurs for m = 0 and if we take perpendicular incidence the sine = 1

          d = λ₀ /2 n

l

et's calculate

          d = 500 10⁻⁹ /( 2  1.4)

         d = 178.57 10⁻⁹ m

Given :

On the way home from school, Mr. X drives the first 10 miles at 55 mi/hr, the next 20 miles at 70 mi/hr, and the last 5 miles at 35 mi/hr.

To Find :

How far does Mr. X live from school.

Solution :

To find the distance between Mr. X residence from school is simply given by summing all the distance he travelled .

So, distance = 10 + 20 + 5 miles

distance = 35 miles.

Hence, this is the required solution.

Answer:

T = 6.43 x 10⁻⁵ N.m

Explanation:

First, we will calculate the deceleration of the apparatus by using the third equation of motion:

[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]

where,

α = angular decelration = ?

θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad

ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s

ωf = final angular speed = 0 rad/s

Therefore,

[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]

negative sign shows deceleration

Now, for torque:

T = Iα

where,

T = Torque = ?

I = moment of inertia = 0.076 kg.m²

Therefore,

T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)

T = 6.43 x 10⁻⁵ N.m

Answer: 28

Explanation:

Given

Circuit breaker current is [tex]I=20\ A[/tex]

Power of the light bulb is [tex]P=100\ W[/tex]

Voltage of the DC-circuit is [tex]V=120\ V[/tex]

If the resistance are connected in parallel, they must have same voltage i.e. 120 V

So, Resistance is given by

[tex]\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega[/tex]

For the 20 A current and 120 V battery, net resistance is

[tex]\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega[/tex]

Suppose there are n resistance in the circuit connected in parallel.

[tex]\Rightarrow \dfrac{144}{n}=R_{net}\\\\\Rightarrow n=\dfrac{144}{6}\\\\\Rightarrow n=28.8\approx 28\ \text{for current to be less than 20A}[/tex]

Thus, there can maximum of 28 bulbs.

Answer:

RIGHT,  SAME SIZE,   VIRTUAL

Explanation:

Plane mirrors comply with the law of reflection where the angle of incidence is equal to the angle of reflection

therefore to complete the sentences:

A plane mirror produces images of objects that have an orientation that is RIGHT __, a size that is _SAME  SIZE____ (compared to that of the object) and a type that is VIRTUAL_____

Answer:

6.24 x 1018 electrons.

Explanation:

So I think C

Answer:

The speed will be "0.144 rad/s".

Explanation:

Given that,

Diameter,

d = 7.50 km

Radius,

R = [tex]\frac{7.5}{2} \ Km[/tex]

Acceleration on inner curve,

= 8 times

Now,

As we know,

⇒ [tex]\omega^2R=8g[/tex]

or,

⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]

⇒     [tex]=\sqrt{\frac{78.4}{3750} }[/tex]

⇒     [tex]=\sqrt{0.0209}[/tex]

⇒     [tex]=0.144 \ rad/s[/tex]