1. A 5.00 kg box is being pulled at constant speed up a 30o incline by a force of 45.0 N parallel to the incline. Determine the coefficient of kinetic friction between the box and the plane.

Answer:

A mixture is a combination of two or more substances in any proportion. This is different from a compound, which consists of substances in fixed proportions. ... The lemonade pictured above is a mixture because it doesn't have fixed proportions of ingredients.

Explanation:

Answer:

6000 joules

Explanation:

I jus learned dis

Answer:6000j

Explanation:

Hope that helps

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

[tex]\Sigma F = F_{E}-W = 0[/tex] (Eq. 1)

Where:

[tex]F_{E}[/tex] - Electrostatic force exerted on human, measured in Newton.

[tex]W[/tex] - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

[tex]q\cdot E-m\cdot g = 0[/tex]

[tex]q = \frac{m\cdot g}{E}[/tex] (Eq. 2)

[tex]E[/tex] - Electric field, measured in Newtons per Coloumb.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravity acceleration, measured in meters per square second.

[tex]q[/tex] - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that [tex]m = 60\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]E = -150\,\frac{N}{C}[/tex], the charge that a 60-kg human must have to overcome weight is:

[tex]q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }[/tex]

[tex]q = -3.923\,C[/tex]

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

[tex]F = \kappa \cdot \frac{q^{2}}{r^{2}}[/tex] (Eq. 3)

Where:

[tex]\kappa[/tex] - Electrostatic constant, measured in Newton-square meter per square Coulomb.

[tex]q[/tex] - Electric charge, measured in Coulomb.

[tex]r[/tex] - Distance between two people, measured in meters.

If we know that [tex]\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]q = -3.923\,C[/tex] and [tex]r = 100\,m[/tex], then the force of repulsion between two people is:

[tex]F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right][/tex]

[tex]F = 13.851\times 10^{6}\,N[/tex]

The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

Answer:

4 bobux

Explanation:

one bobux

two bobux

three bobux

four bobux

Answer:

luster

Explanation:

Complete question:

use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is 2.2 x 10⁻⁵ km/s/Lyr

Answer:

The distance to the quasar is 1.02 x 10¹⁰ Lyr

Explanation:

Given;

speed of light, v = 300, 000 km/sec

Hubble's constant, H₀ = 2.2 x 10⁻⁵ km/s/Lyr

percentage of the quasar recession = 75% of speed of light

Hubble's Law is given by;

[tex]v =H_od\\\\d = \frac{v}{H_o}\\\\d= \frac{(0.75*300,000)}{2.2*10^{-5}}.Lyr\\\\d = 1.02*10^{10} \ Lyr[/tex]

Therefore, the distance to the quasar is 1.02 x 10¹⁰ Lyr

Answer:

The observed wavelength is  [tex] \lambda = 700nm[/tex] (color -  Red)

Explanation:

From the question we are told that

   The wavelength of the emitter is  [tex]\lambda_ e  = 500 nm  =  500 *10^{-9} \  m[/tex]

  The redshift is  R  =  0.4  

     Generally red shift is mathematically represented as

    [tex]R =  \frac{ \lambda -  \lambda_e }{\lambda_e}[/tex]

=>  [tex]0.4  =  \frac{ \lambda -   500 *10^{-9} }{500 *10^{-9} }[/tex]

=>  [tex] \lambda - 500*10^{-9} = 200*10^{-9}  [/tex]

=>   [tex] \lambda = 700 *10^{-9}[/tex]

=>   [tex] \lambda = 700nm[/tex]

Complete question is;

Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at

(x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)

Compute the length of the bungee cord at t = 7 seconds. (Round your answer to three decimal places.)

Answer:

Length of bungee cord = 85.734 ft

Explanation:

We are told that Adrian moves 9ft/sec. Thus, at 5.5 seconds, distance he moved is; 9 ft/sec × 5.5sec = 49.5 ft in the negative x (-x) direction. Therefore, the coordinate is (-49.5, 0).

Now, Allyson has moved 10ft/sec. Thus, at 7 seconds, distance he moved would be; 10 ft/sec x 7sec = 70 feet in the positive (+y) direction. Therefore, the coordinate is (0, 70).

Now, since they started from the origin, it means (0, 0) is a coordinate. Thus, we now have 3 coordinates which are; (0, 0), (0, 70) & (-49.5,0). These 3 coordinates would therefore combine to form a right triangle.

The hypotenuse is the distance between Allyson and Adrian.

Thus, from pythagoras theorem, we can find the distance between them which is same as the length of the cord.

Thus;

(-49.5)² + 70² = D².

D² = 2450.25 + 4900

D = √7350.25

D = 85.734 ft

Answer:

Answer and steps in the pic

Answer:

C. Some social patterns are helpful, while others are harmful.

Explanation:

Hope this was helpful, Have an amazing,spooky Halloween!!

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

Answer: The correct answer is A) The track was not level and was tilted slightly downward.

Explanation: This is because of the two values: 0.4 kg and 0.5 kg. I won't go into much detail but due to this difference of mass, we know that the track was not level.

"The track was not level and was tilted slightly downward" could explain the difference in the mass.

Thus Option A is appropriate.

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Answer:

1.152 miles

Explanation:

Given: central angle = 1 minute = [tex](\frac{1}{60}) ^{o}[/tex]

           radius of the earth = 3960 miles

The length of an arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r

where: [tex]\alpha[/tex] is the central angle, and r is the radius.

Thus,

Distance along the arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r

Distance along the arc = [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 2 x [tex]\frac{22}{7}[/tex] x 3960

                                      = [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 24891.4286

                                      = 1.1524

The required distance along an arc is 1.152 miles.

Complete Question

Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:

τs = -κ θ.

Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.

Required:

a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.

b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device? Assume the full scale deflection is 60° from the spring's equilibrium position

Answer:

a

[tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

b

[tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

Explanation:

From the question we are told that

   The number of turns is  N  =  50  

   The radius is  r =  1.55 cm  =  0.0155 m

   The  magnetic field is  B  =  0.35 T

   The induced current is  [tex]I =  45 \mu A =  45 *10^{-6} \  A[/tex]

Generally the area of  loop is mathematically represented as

      [tex]A =  \pi r^2[/tex]

=>   [tex]A =3.142 *  0.0155^2[/tex]

=>   [tex]A =0.000755\ m^2[/tex]

Generally the maximum torque is mathematically represented as

     [tex]\tau_{m} =  N  *  B  * I  *  A[/tex]

=>  [tex]\tau_{m} =  50  *  0.35  * 45 *10^{-6} *  0.000755[/tex]

=>  [tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

Generally the  torque 60° from the spring's equilibrium position is mathematically represented as

      [tex]\tau  =  N  *  B  *  I  *  A  *  sin (60)[/tex]

=>    [tex]\tau  =  50  *  0.35  *  45 *10^{-6} *  0.000755  *  sin (60)[/tex]

=>   [tex]\tau  = 2.973 *10^{-7} \  N  \cdot m [/tex]

Generally the toque constant of the spring is mathematically represented as

     [tex]\beta =  \frac{\tau}{60}[/tex]

=>   [tex]\beta =  \frac{\tau}{\frac{\pi}{3}}[/tex]

=>   [tex]\beta =  \frac{2.973 *10^{-7}}{\frac{\pi}{3}}[/tex]

=>    [tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

Answer:

The angle is 3.95 rad.

Explanation:

The angle can be calculated as follows:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]

Where:

[tex]\omega_{f}[/tex]: is the final angular speed

ω₀: is the initial angular speed = 0 (it starts from rest)

α: is the angular acceleration

θ: is the angle=?

The centripetal acceleration is:

[tex]a_{c} = \omega_{f}^{2}*r[/tex]

And the tangential acceleration is:

[tex] a_{T} = \alpha*r [/tex]

Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:

[tex]a_{c} = 7.9a_{T}[/tex]

[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]

Now, the angle is:

[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]

[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]

Therefore, the angle is 3.95 rad.

I hope it helps you!          

The angular distance traveled by the electric drill is 3.95 radians.

The given parameters;

The angular distance traveled by the electric drill is calculated as follows;

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

The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;

[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]

Substitute the value of angular acceleration into the first equation;

[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]

Thus, the angular distance traveled by the electric drill is 3.95 radians.

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Answer:

Hyperbole

Explanation:

this is an extreme exaggeration or overstatement/ magnification

Answer:

d. kT ln 3

Explanation:

Given;

k as Boltzmann constant

Let initial initial microstate, = Фi

Let the final microstate, = Фf = 3Фi

at constant temperature, T

The energy input to gas as heat is given by;

Q = TΔS = Tk(lnФf - lnФi)

[tex]Q= kT*\frac{ln \phi _f}{ln \phi_i} \\\\Q= kT*\frac{3ln \phi _i}{ln \phi_i}\\\\Q= kT ln3[/tex]

Therefore, the energy input to gas as heat is kT ln 3

Answer:

μk = 0.58

Explanation:

       [tex]F_{apph} = F_{app} * sin\theta = 126 N * sin 25 = 53.3 N[/tex]

       ⇒  [tex]F_{n} = - F_{apph} = -53.3 N[/tex]

       [tex]F_{g} = m_{b} * g = 8.5 Kg * (-9.8 m/s2) = -83.3 N[/tex]

       [tex]F_{appv} = F_{app}* cos\theta = 126 N * cos 25 = 114.2 N[/tex]

      [tex]F_{ff} = \mu_{k}* F_{n} =\mu_{k} * (-53.3 N)[/tex]

        μk = 0.58

Answer:

12.5 m

Explanation:

The first thing we would do is to calculate the wavelength. To do this, we use the formula

v = fλ, where

v = wave speed

f = frequency

λ = wavelength

If we make wavelength the formula, we have

wavelength = speed / frequency

Now, we substitute the values we had been given and we have

wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m

half of this said wavelength will be

= 3.50 / 2

= 1.75 m

As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,

(10 / 1.75) = 5.7

So the separation will have to be 7 half wavelengths

= (7 * 1.75) = 12.5 m

Answer:

Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".

Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.  

What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.  

The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.  

It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this,  it is still nonsense on screen in Hollywood and video.  

Answer:

can exchange energy with its surroundings through heat and work transfer. In other words, work and heat are the forms that energy can be transferred across the system boundary.

Answer:

100 N

Explanation:

The following data were obtained from the question:

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 50 cm

Force (F) =.?

Next, we shall convert 50 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

50 cm = 50 cm × 1 m / 100 cm

50 cm = 0.5 m

Therefore, 50 cm is equivalent to 0.5 m.

Finally, we shall determine the magnitude of the net force on the ball by using the following formula:

F = mv²/r

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 0.5 m

Force (F) =.?

F = mv²/r

F = 2 × 5²/ 0.5

F = 2 × 25/ 0.5

F = 50 / 0.5

F = 100 N

Therefore, the magnitude of the net force on the ball is 100 N.

The magnitude of the net force on the ball will be "100 N".

According to the question,

Mass, m = 2 kg

Velocity, v = 5 m/s

Radius, r = 50 cm or,

               = 50 × [tex]\frac{1}{100}[/tex]

               = 0.5 m

We know the relation,

Force, F = [tex]\frac{mv^2}{r}[/tex]

By substituting the values, we get

              = [tex]\frac{2\times (25)^2}{0.5}[/tex]

              = [tex]\frac{2\times 25}{0.5}[/tex]

              = [tex]\frac{50}{0.5}[/tex]

              = 100 N

Thus the above response is appropriate.

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Answer:

b. The bicyclist is exerting 49 N of force

Explanation:

Given;

mass of bicyclist start, m = 50 kg

acceleration, a = 1 m/s²

density of air, ρ = 1.2 kg/m³

velocity, v = 2 m/s

Area of the bicyclist body, A = 0.5 m²

The drag force on the bicyclist is given by ;

Fd = 0.5CρAv²

where;

C is drag coefficient = 0.9 for bicycle

Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²

Fd = 1.1 N

The force of the bicyclist is given by;

F = ma

F = 50 x 1

F = 50 N

The effective force exerted by the bicyclist is given by;

Fe = F - Fd = 50 N - 1.1 N

Fe = 49 N

Therefore, the force exerted by the bicyclist is 49 N

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G[tex]\frac{m1m2}{r}[/tex]

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

Answer:

The displacement of the canoe is 1.46 m

Explanation:

Given that,

Mass of canoe = 61 kg

Mass of man = 86.5 kg

Length = 4 m

Let the the displacement of the canoe is x'

We need to calculate the displacement of the man

Using formula of displacement

[tex]x=x_{2}-x_{1}[/tex]

Put the value into the formula

[tex]x=4-(0.75+0.75)[/tex]

[tex]x=2.5\ m[/tex]

We need to calculate the displacement of the canoe

Using conservation of momentum

[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]

[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]

[tex]86.5\times2.5=(61+86.5)\times x'[/tex]

[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]

[tex]x'=1.46\ m[/tex]

Hence, The displacement of the canoe is 1.46 m

Answer:

(a)  24.56 N

(b) 142.28 N

Explanation:

(a)

The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.

The centripetal function is generated from either scenario by Equation:

⇒  [tex]Fc = \frac{mv^2}{r}[/tex]

On putting the values, we get

⇒       [tex]=\frac{12\times 1.33^2}{0.864}[/tex]

⇒       [tex]=24.56 \ N[/tex]

(b)

Use T to denote whatever arm stress we can get at the bottom including its circle:

⇒  [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]

⇒  [tex]T = mg + Fc[/tex]

⇒      [tex]=12\times 9.81+24.56[/tex]

⇒      [tex]=142.28 \ N[/tex]