What is the mass of a bullet moving at 970m/s if the bullet’s KE is 3.9x 10^3J

Answer:

6.22 N/m

Explanation:

From Hooke's law we deduce that F=kx where F is the applied force and k is the spring constant while x is the extension or compression of the spring. Making k the subject of the above formula then

[tex]k=\frac {F}{x}[/tex]

We also know that the force F is equal to mg where m is the mass of an object and g is acceleration due to gravity hence substituting F with mg we get that

[tex]k=\frac {mg}{x}[/tex]

Substituting m with 425 g which is equivalent to 0.425 kg and g with 9.81 then 0.67 for x we get that

[tex]k=\frac {mg}{x}=\frac {0.425\times 9.81}{0.67}=6.222761194 N/m\approx 6.22\ N/m[/tex]

Therefore, the spring constant is approximately 6.22 N/m

Answer:

Explanation:

A law and theory are distinct levels in the scientific method. They do not lead to one another.

A law is a description of an observed phenomenon in the natural world. Laws are always true and do not provide explanations as to why they hold true.

A theory is an explanation of an observed phenomenon. It is usually based on experimental evidence and bounded most time in the limits of available data.

Answer:

Explanation:

A fire alarm notification appliance is an active fire protection component of a fire alarm system. The primary function of the notification appliance is to alert persons at risk.

If want the audible public mode signal to be hear clearly then, we need to have a sound level that is at least 15dB above the average ambient sound level or 5dB above the maximum sound level of at least 1minute

In this case the,

The average ambient sound level is 62dB,

And the maximum sound level is 68dB

Then, the public mode signal should be at least

1. 62dB+ 15dB=77dB

Or

2. 68dB +5dB =73dB.

Then the public mode signal hearing must be at least 77dB.

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Given data :-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that Kinetic energy (KE) = ½ mv²

Where m= mass of the body

           v=velocity of the body

Substituting the values of respective mass and velocity from the above given data-

KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

Answer:

The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

Explanation:

1 lb of body fat = 4.1 k cal

1 k cal = 4.184 Kilo joule

1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule

Athlete expends 3480 Kilo joule in one hour

⇒ Time required to expand 3480 Kilo joule  = 60 minute

⇒ Time required to expand 1 Kilo joule = [tex]\frac{60}{3480}[/tex] [tex]\frac{min}{KJ}[/tex]

⇒ Time required to expand 17.1544 Kilo joule = [tex]\frac{60}{3480}[/tex] × 17.1544 = 0.296 min

Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

Explanation:

Below is an attachment containing the solution

Answer:

The charge on the particle = -0.00075 C = -0.75 mC = -750 μC

Explanation:

The solution to this question is presented in the attached image to this answer.

The Biot Savart's formula for calculating magnetic field due to moving point charge is used in this calculation.

Hope this Helps!!!

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

[tex]\Delta U = 0[/tex]

so from 1st law of thermodynamics we have

[tex]dQ = W + \Delta U[/tex]

[tex]dQ = W[/tex]

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

Answer:

Explanation:

The distance of the student from the wall is L

After hearing the reflection she claps her hand again

A complete cycle is the distance to the wall and back to the boy, so the total distance travel then is 2L, then the wave length is 2L

λ=2L

So the frequent is f

Then using wave equation

v=f λ

Since our λ=2L

Then, v=f×2L

v=2fL

The speed of sound in air is 2fL

Final answer:

To find the speed of sound in air based on the frequency of clapping and its reflection off a wall, we can use the rearranged equation V = 2fL, where V is the speed of sound, f is the frequency of clapping, and L is the distance to the wall.

Explanation:

The question involves calculating the speed of sound in air based on the frequency at which a person claps their hands and the time it takes for the sound to reflect off a wall and travel back to them. To solve this, we can use the formula f = ½(V / L), where f is the frequency of clapping (in Hz), V is the speed of sound in air (in m/s), and L is the distance from the person to the wall (in meters). However, the provided formulas from various attempts don't directly address the question but hint at the relationship between the speed of sound, frequency, and distance. Thus, by understanding that the sound needs to travel twice the distance (L) to reach the person again and taking into account the time interval represented by the frequency of clapping, we can rearrange the formula to solve for the speed of sound as V = 2fL. It's understood that the time taken for the sound to travel to the wall and back is inversely related to the frequency of clapping.

Answer:

A. Electric flux

Explanation:

Electric flux is the rate of flow of the electric field through a given area (see ). Electric flux is proportional to the number of electric field lines going through a virtual surface.

Electric flux has SI units of volt metres (V m), or, equivalently, newton metres squared per coulomb (N m2 C−1). Thus, the SI base units of electric flux are kg·m3·s−3·A−1.

Final answer:

The average force that the net exerts on the man is approximately -5.02 x 10^7 N. The negative sign indicates that the force is in the opposite direction to the motion.

Explanation:

To determine the average force that the net exerts on the man, you can use the equation:

Force = mass x acceleration

First, calculate the acceleration of the man using the formula:

acceleration = change in velocity / time taken

In this case, the final velocity is 0 m/s, the initial velocity is 28 m/s, and the time taken is the distance the net stops the man divided by his initial velocity. Since the net stops the man within a distance of 1 mm (0.001 m), the time taken is:

time taken = distance / initial velocity = 0.001 m / 28 m/s = 3.571 x 10-5 s

Now, you can calculate the acceleration:

acceleration = (0 - 28) m/s / 3.571 x 10-5 s = -7.838 x 105 m/s2

Next, calculate the force:

force = mass x acceleration = 64 kg x (-7.838 x 105 m/s2)

force = -5.02 x 107 N

The average force that the net exerts on the man is approximately -5.02 x 107 N. The negative sign indicates that the force is in the opposite direction to the motion.

Answer:

Explanation:

Give that

I=16A

The current is flowing to the right

Equal amount of positive and negative charges.

i.e, total charge is q= Q+Q=2Q. i.e magnitude

Then, the rate of charge that pass though is dq/dt

Give that,

q=it

2Q=it

Let differentiate with respect to t

2dQ/dt=i

Then, dQ/dt=i/2

Since i=16A

Then, dQ/dt=16/2

dQ/dt= 8 A/s. Toward the left

Answer:

Therefore the required solution is

[tex]U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t[/tex]

Explanation:

Given vibrating system is

[tex]u''+\frac{1}{4}u'+2u= 2cos \omega t[/tex]

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

[tex]-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t[/tex]

[tex]\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t[/tex]

Equating the coefficient of sinωt and cos ωt

[tex]\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2[/tex]

[tex]\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0[/tex].........(1)

and

[tex]\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0[/tex]

[tex]\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0[/tex]........(2)

Solving equation (1) and (2) by cross multiplication method

[tex]\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}[/tex]

[tex]\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]

[tex]\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]   and        [tex]B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]

Therefore the required solution is

[tex]U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t[/tex]

Answer:

yes it is important.

Explanation:

It is important to study the climate of other planet to understand the climate of the earth because it is useful to know which things are harmful for our earth and which are important for our earth climate comparing with the other planets.

Final answer:

The work done by the spring force to ready the pen for writing is 0.404 J.

Explanation:

To find the work done by the spring force, we can use the formula for work: work = (1/2)k(x²), where k is the spring constant and x is the displacement of the spring from its unstressed length.

In this case, the spring is initially compressed 5.1 mm and then compressed an additional 6.1 mm. So the total displacement is 5.1 mm + 6.1 mm = 11.2 mm.

Substituting the values into the formula, we have work = (1/2)(257 N/m)((11.2 mm / 1000)²) = 0.404 J. The calculated work of 0.404 J represents the energy transferred to or from the spring as it undergoes the specified compressions, providing insight into the mechanical behavior of the system under the influence of the spring force.

Answer:

Supply-side bonding jumper.

Explanation:

A supply-side bonding jumper is a conductor installed on the supply side of a service or separately derived system to ensure the required electrical conductivity between metal parts required to be electrically connected.

The supply side bonding jumper was referred to as the equipment bonding jumper prior to 2011 NEC when its name changed.

The wire runs from the source of the separately derived system to the first disconnecting means.

Final answer:

The emission spectrum corresponds to energy emitted by electrons as they fall back from an excited state to a lower energy state. Energy levels are the quantized orbits around the nucleus, and quantum refers to the specific amount of energy in these processes. The excited state is a temporary, higher energy level for an electron.

Explanation:

Lets match the vocabulary with their definitions:

Now, to give a more in-depth understanding:

Explanation:

Porosity is the percentage of spaces,(hollow) within a rock or a material that can contain air or fluid. This can be used in geology( study of rocks), soil mechanics, engineering and pharmaceutics. Industrial CT scanning can be used to test for porosity of a substance. While permeability is the ability of water or other fluids to flow through a rock or material that has spaces ( that is, that are porous). This is also applicable in the fields of chemical engineering and geology.

Porosity refers to the amount of void space in a material that can hold water, while permeability measures how well those spaces are connected, affecting water movement. Materials with high permeability have larger, well-connected pores, allowing water to flow easily; materials with low permeability do not. The hydraulic conductivity is a measure of permeability that accounts for both material and fluid properties.

Difference Between Porosity and Permeability

The difference between porosity and permeability concerns the storage and movement of water in subsurface materials like rock or sediment. Porosity refers to the percentage of open space within a material that can potentially hold water, expressed as a ratio of the volume of voids to the total volume of the material. Permeability, on the other hand, is about how well those pores are interconnected, determining the ease with which water can move through the material. Higher permeability indicates the presence of larger, well-connected pores, enabling water to flow with less friction. Conversely, materials with low permeability have fewer, smaller, and poorly connected pores, which restricts water movement.

Understanding both porosity and permeability is crucial when discussing groundwater storage and extraction, as they define an aquifer's ability to store and transmit water. Soil texture plays a significant role in determining these properties. Coarse-grained soils with larger pores tend to have both higher porosity and permeability, whereas fine-grained soils like clay can have high porosity but low permeability due to poorly connected pores.

Answer:

(a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

Explanation:

Given that,

Spring stiffness = 205 N/m

Mass = 0.6 kg

Compression of spring = 13 cm

Initial speed = 3 m/s

(a). We need to calculate the maximum stretch during the motion

Using conservation of energy

[tex]E_{initial}=E_{final} [/tex]

[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2[/tex]

[tex]x_{m}=\sqrt{\dfrac{4.43\times2}{205}}[/tex]

[tex]x_{m}=20.7\ cm[/tex]

(b). Maximum speed comes when stretch is zero.

We need to calculate the maximum speed during the motion

Using conservation of energy

[tex]E_{initial}=E_{final} [/tex]

[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2[/tex]

[tex]v'=\sqrt{\dfrac{4.43\times2}{0.6}}[/tex]

[tex]v'=3.84\ m/s[/tex]

(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system

We need to calculate the time period

Using formula of time period

[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]

Put the value into the formula

[tex]T=2\pi\sqrt{\dfrac{0.6}{205}}[/tex]

[tex]T=0.33\ sec[/tex]

We need to calculate the energy

Using formula of energy

[tex]E=\dfrac{P}{t}[/tex]

Put the value into the formula

[tex]E=\dfrac{0.02}{0.33}[/tex]

[tex]E=0.060\ Watt[/tex]

Hence, (a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

Answer:

v' = 0.714 m/s

Explanation:

Solution:

- Assuming no external torque is acting on the system then the angular momentum is conserved for the system.

- The initial momentum angular Mi and final angular momentum Mf are as follows:

                                  Mi = Mf

                                  m*L*v = m*x*L*v'

Where,

             m : mass of the telescope

             L : Length of teether line

             v: Initial speed

             v' : Changed speed.

- Then we have:

                                  L*v = x*L*v'

                                  v' = v / x

                                  v' = 2 / 2.8

                                  v' = 0.714 m/s

Answer:

The answer to the question is

The linear speed of the telescope will be 5.6 m/s if the length of the line is changed to x*L where  x = 2.8; and initial velocity v = 2 m/s

Explanation:

Speed = v₁ = ωL = 2 m/s

When the line is changed to x*L where x = 2.8 the linear speed will be

v₂ = 2.8 × L× ω = 2.8× 2 = 5.6 m/s

The linear speed varies with the angular speed following the relation v/r =ω where

ω = angular speed

v = linear speed and

r = radius of the path of travel of the object at the vertex

Answer:

Explanation:

- The volume of water displaced by immersing the object is equal the amount of water spilled and caught by Dr. Hewitt.

- The amount of water is proportional to the volume of object of fraction of object immersed in water will lead to the same fraction of water displaced and caught by Dr. Hewitt.  

- When the object is immersed the force of Buoyancy acts against the weight and reducing the scale weight.

- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.

Answer:

[tex]3.34\Omega[/tex]

Explanation:

The resistance of a metal rod is given by

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity

L is the length of the rod

A is the cross-sectional area

The resistivity changes with the temperature as:

[tex]\rho(T)=\rho_0 (1+\alpha (T-T_0))[/tex]

where in this case:

[tex]\rho_0[/tex] is the resistivity of silver at [tex]T_0=21.0^{\circ}C[/tex]

[tex]\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}[/tex] is the temperature coefficient for silver

[tex]T=180.0^{\circ}C[/tex] is the current temperature

Substituting,

[tex]\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0[/tex]

The length of the rod changes as

[tex]L(T)=L_0 (1+\alpha_L(T-T_0))[/tex]

where:

[tex]L_0[/tex] is the initial length at [tex]21.0^{\circ}C[/tex]

[tex]\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}[/tex] is the coefficient of linear expansion

Substituting,

[tex]L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0[/tex]

The cross-sectional area of the rod changes as

[tex]A(T)=A_0(1+2\alpha_L(T-T_0))[/tex]

So, substituting,

[tex]A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0[/tex]

Therefore, if the initial resistance at 21.0°C is

[tex]R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega[/tex]

Then the resistance at 180.0°C is:

[tex]R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega[/tex]

Answer:

Ptolemy made geocentric model of the solar system using epicycles

Explanation:

Ptolemy made geocentric model of the solar system using epicycles.

This model accounted for the apparent motions of the planets in a very direct way, by assuming that each planet moved on a small circle, called an epicycle, which moved on a larger circle, called a deferent.

Therefore, Ptolemy is the answer.

a) 6.9 m/s

b) 4.9 m/s

c) After 4.50 cm

d) 5.2 m/s

Explanation:

a)

In this case, there is no resistive force. Therefore, according to the law of conservation of energy, all the initial elastic potential energy stored in the spring when it is compressed is converted into kinetic energy of the ball as it leaves the barrel; so we can write:

[tex]\frac{1}{2}kx^2=\frac{1}{2}mv^2[/tex]

where:

k = 400 N/m is the spring constant

x = 6.00 cm = 0.06 m is the compression of the spring

m = 0.03 kg is the mass of the ball

v is the velocity of the ball as it leaves the barrel

Solving the equation, we can find the speed of the ball:

[tex]v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(400)(0.06)^2}{0.03}}=6.9 m/s[/tex]

b)

In this case, there is a constant resistive force of

[tex]F_r = 6.00 N[/tex]

acting on the ball as it moves along the barrel.

The work done by this force on the ball is:

[tex]W=-F_rd[/tex]

where

d = 6.00 cm = 0.06 m is the length of the barrel

And where the negative sign is due to the fact that the force is opposite to the motion of the ball

Substituting,

[tex]W=-(6.00)(0.06)=-0.36 J[/tex]

Therefore, part of the initial elastic potential energy stored in the spring has been converted into thermal energy due to the resistive force. So, the final kinetic energy of the ball will be less than before:

[tex]\frac{1}{2}kx^2+W=\frac{1}{2}mv^2[/tex]

And solving for v, we find the new speed:

[tex]v=\sqrt{\frac{kx^2}{m}+\frac{2W}{m}}=4.9 m/s[/tex]

c)

The (forward) force exerted by the spring on the ball is

[tex]F=k(x_0-x)[/tex]

where

k = 400 N/m

x is the distance covered by the ball

[tex]x_0=0.06 m[/tex] is the maximum displacement of the spring

While the (backward) resistive force is instead

[tex]F_r=6.0 N[/tex]

So the net force on the ball is

[tex]F=k(x-x_0)-F_r[/tex]

The term [tex]k(x-x_0)[/tex] prevails at the beginning, so the ball continues accelerating forward, until this term becomes as small as [tex]F_r[/tex]: after that point, the negative resistive force will prevail, so the ball will start delecerating. Therefore, the greatest speed is reached when the net force is zero; so:

[tex]k(x_0-x)-F_r=0\\x=x_0-\frac{F_r}{k}=0.06-\frac{6.00}{400}=0.045 m[/tex]

d)

After the ball has covered a distance of

x = 0.045 m

The work done by the resistive force so far is:

[tex]W=-F_r x =-(6.00)(0.045)=-0.27 J[/tex]

The total elastic potential energy of the spring at the beginning was

[tex]E_e=\frac{1}{2}kx_0^2 = \frac{1}{2}(400)(0.06)^2=0.72 J[/tex]

While the elastic potential energy left now is

[tex]E_e'=\frac{1}{2}k(x_0-x)^2=\frac{1}{2}(400)(0.015)^2=0.045 J[/tex]

So, the kinetic energy now is:

[tex]K=E_e-E_e'+W=0.72-0.045-0.27=0.405 J[/tex]

And by using the equation

[tex]K=\frac{1}{2}mv^2[/tex]

We can find the greatest speed:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.405)}{0.03}}=5.2 m/s[/tex]

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

[tex]N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1[/tex]

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

[tex]P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}[/tex]

= 1250W

d) Final peak=

P= Ik/e

= [tex]= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W[/tex]

P = 2.5 × 10⁷W

Explanation:

Below is an attachment containing the solution.

Answer:

Explanation:

Check attachment for solution

The analytical method of the components allows to find the results for the questions about the sum vector are:

     a) In the attached we have a scheme of the vectors

     b) the modulus is C = 96.3 m

     c) The angle is θ = 27.0º

Given parameters

To find

    a) Draw the vectors and their sum.

    b) The module.

    c) the adirection.

The sum of vectors has several methods of resolution:

The analytical method consists:

In the attached we have a diagram of each vector and its sum. Let's use trigonometry to find the component of each vectors.

Vector A

            cos θ₁ = [tex]\frac{A_x}{A}[/tex]  

            sin θ₁ = [tex]\frac{A_y}{A}[/tex]  

            Aₓ = A cos θ₁

            [tex]A_y[/tex] = A sin θ₁

            Aₓ = 40 cos (-20) = 37.59 m

            [tex]A_y[/tex]= 40 sin (-20) = -13.68 m

Vector B

           cos θ₂ = [tex]\frac{B_x}{B}[/tex]  

           sin θ₂ = [tex]\frac{B_y}{B}[/tex]  

           Bₓ = B cos θ₂

           [tex]B_y[/tex] = B sin θ₂

           Bₓ = 75 cos 50 = 48.21 m

           [tex]B_y[/tex] = 75 sin 50 = 57.45 m

we add the components.

           Cₓ = Aₓ + Bₓ

           [tex]C_y = A_y + B_y[/tex]  

           Cₓ = 37.59 + 48.20 = 85.8 m

           Cy = -13.68 + 57.45 = 43.8 m

b) We use the Pythagorean theorem to find the modulus of the resultant vector.

          C² = Cₓ² + [tex]C_y^2[/tex]  

          C = [tex]\sqrt{85.8^2 + 43.8^2 }[/tex]  

         C = 96.3 m.

c) We use trigonometry to find the angle of the resultant vector.

         tan θ =[tex]\frac{C_y}{C_x}[/tex]  

         θ = tan⁻¹ [tex]\frac{C_y}{C_x}[/tex]  

         θ = tan⁻¹ [tex]\frac{43.8}{85.8}[/tex]  

         θ = 27.0º

In conclusion, using the analytical method of the components we can find the results for the questions about the sum vector are:

     a) In the attached we have a scheme of the vectors

     b) the modulus is C = 96.3 m

     c) The angle is θ = 27.0º

Learn more about vector addition here:  brainly.com/question/25681603

Answer:

265.9Hz

Explanation:

In an open pipe, both ends of the pipes are opened. The fundamental frequency in an open pipe is expressed as fo = V/2L where;

f is the frequency of the wave

V is the velocity of the wave = 343m/s

L is the length of the pipe = 1.29m

Substituting the value to get the fundamental frequency in the open pipe we have;

Fo = 343/2(1.29)

Fo = 343/2.58

Fo = 132.95Hz

Harmonics are integral multiples of the fundamental frequency e.g 2fo, 3fo, 4fo, 5fo...

The first harmonic in the open pipe will be f1 = 2fo

Since f1 =2(132.95)

f1 = 265.9Hz

The frequency of the first harmonic if the pipe is open at each end is 265.9Hz

Answer:

132.95 Hz.

Explanation:

Given:

v = 343 m/s

L = 1.29 m.

Since the pipe is open at both ends,

L = λ/2

λ = v/f = 2L

= 2 × 1.29

= 2.58 m

f = 343/2.58

= 132.95 Hz.

Answer:

The orientation b has the largest magnetic torque.

Explanation:

As the complete question is not given, the complete question is attached herewith

From the diagram

[tex]|\mu_a|=|\mu_b|=|\mu_c|=|\mu|[/tex]

Also the angles for the 3 orientations are given as

[tex]\theta_a>90\\\theta_b=90\\\theta_c<90\\[/tex]

Now as the torque τ is given as

[tex]\tau=|\mu||B|sin\theta[/tex]

As the value of μ and B is same so value of τ is maximum for sin θ is maximum so

[tex]sin \theta_{max}=1\\\theta_{max}=90[/tex]

So the orientation b has the largest magnetic torque.

The orientation with the largest magnetic torque on the dipole is Orientation B (See attached image).

The torque on the dipole is defined as:

τ = µ×B,

where B is the external magnetic field.

The magnitude of this torque is µB sinθ, where θ is the angle between B and µ

Magnetic Torque is highest when;

→     →

µ ⊥ β

That is when θ  = 90°. Hence B is the correct answer. Please see attached image.

Learn more about Magnetic Torque at:
https://brainly.com/question/6585693

Answer: c. they will hit the ground at the same time

Explanation:

The volume of both objects is almost the same, so the force of friction will be the same in each one, so we can discard it.

Now, when yo drop an object, the acceleration of the object is always g = 9.8m/s^2 downwards, independent of the mass of the object.

So if you drop two objects with the same volume but different mass, because the acceleration is the same for both of them, they will hit the ground at the same time, this means that the density of the object has no impact in how much time the object needs to reach the floor.

So the correct option is c

Answer:

The answer for this is that they each have the same number of protons.

Explanation:

Plutonium is an element that is also used in nuclear power plants, because of the same amount of protons in it, plutonium is used in nuclear power plants.

The fact about all uranium atoms is that they have the same number of protons that make them very large and can be harmful if not carefully treated.

Answer:

same number of protons

Explanation: