¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?

Answer:

Manometer is a measures to the atmospheric measure.

Barometer is a measure to the atmospheric pressure.

Explanation:

Answer:

Explanation:

momentum =mass*speed

p=1500 kg*6 m/s

p=9000 kgm/s

Answer:

False

Explanation:

This is because according to newtons second law which says the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. So take for example a net a net force in opposite direction will cause an object to slow down.

velocity vector here is not the same as acceleration vector

Answer:

916.9C

Explanation:

Using Q= It

Q= 0.54A x 1698s

= 916.9C

The charge delivered by a current with an average of 0.54 A over 28.3 minutes is 916.9C.

Electric charge is defined as the physical property of matter that causes that matter to experience a force when placed in an electromagnetic field that may be positive or negative such that charges repel each other and unlike charges attract each other. An object with the absence of net charge is said to be neutral.

The SI unit of electric charge is the coulomb which is a derived SI unit and is represented by the symbol C.

Electric Charge is represented as

Q= It

where, Q= Electric charge

I= Electric current

t= time

Q= 0.54A x 1698s

= 916.9C

Thus, the charge delivered by a current with an average of 0.54 A over 28.3 minutes is 916.9C.

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Answer:

4.8E-19C

Explanation:

Charge is calculated by getting the net sum of protons and electrons

So

87protons - 84electrons= +3

So the charge of the ion is +3

But 1proton= 1.6*10^-19C

So charge will be 3 x 1.6E-19C

4.8x10^-19C

The question is missing. Here is the complete question.

A point charge Q moves on the x-axis in the positive direction with a speed of 160 m/s. A point P is on the y-axis at y = + 20mm. The magnetic field produced at point P, as the charge moves through the origin is equal to  -0.6μT k. What is the charge Q? [tex]\mu_{0}=4.\pi.10^{-7}[/tex]T.m/A)

Answer: Q = [tex]-0.015.10^{-3}[/tex]C

Explanation: Magnetic Field (B) is a vector field, i.e., has magnitude and direction, and describes the distribution of magnetic force in the space around. It happens when an electrical charge is in movement.

Its magnitude is determined by the formula:

[tex]B = \frac{\mu_{0}}{4.\pi}\frac{Qvsin(\theta)}{r^{2}}[/tex]

where

[tex]\mu_{0}[/tex] is the vacuum permeability constant;

r is the distance between charge and a point you want to know the magnetic field;

θ is the angle between velocity and distance r;

For this question, magnetic field has that magnitude when charge is passing through the origin. So, angle between velocity and distance is 90°.

Calculating to determine charge:

[tex]-0.6.10^{-6} = \frac{4.\pi.10^{-7}}{4.\pi}\frac{Q.160.sin(90)}{(2.10^{-2})^{2}}[/tex]

[tex]-0.6.10^{-6} = 10^{-7}\frac{Q.160.1}{(2.10^{-2})^{2}}[/tex]

[tex]-2.4.10^{-10} = Q.160.10^{-7}[/tex]

[tex]Q = \frac{-2.4.10^{-10}}{160.10^{-7}}[/tex]

[tex]Q = -0.015.10^{-3}[/tex]

Charge Q is [tex]-0.015.10^{-3}[/tex]C or [tex]-0.015[/tex]mC.

Answer:

A) f = 1.28x10¹⁵ Hz

B) y = 0.20 m

Explanation:

A) The frequency of light can be found as follows:

[tex] f = \frac{c}{\lambda} [/tex]

Where:

c: is the speed of light = 3.0x10⁸ m/s

λ: is the wavelength

The wavelength can be calculated using the following equation:

[tex] sin(\theta) = \frac{\lambda(m - 1/2)}{d} [/tex]

Where:

m = 3, d = 6 μm, θ = 5.6°        

[tex] \lambda = \frac{d*sin(\theta)}{m - 1/2} = \frac{6 \cdot 10^{-6} m*sin(5.6)}{3 - 1/2} = 2.34 \cdot 10^{-7} m [/tex]      

Now, the frequency is:

[tex] f = \frac{c}{\lambda} = \frac{3.0 \cdot 10^{8} m/s}{2.34 \cdot 10^{-7} m} = 1.28 \cdot 10^{15} Hz [/tex]    

Hence, the frequency of light used for this experiment is 1.28x10¹⁵ Hz.

B) The distance between the second bright fringe and central fringe (y) is:

[tex] tan(\theta) = \frac{y}{D} [/tex]        

[tex] y = D*tan(\theta) = 2 m*tan(5.6) = 0.20 m [/tex]

Therefore, the distance between the second bright fringe and central fringe is 0.20 m.

I hope it helps you!

Answer:

Gravitational forces and drag forces

Explanation:

All objects that are above a certain distance above the ground has gravitational force acting on them. A falling body like the iclicker has gravitational force pulling it down with a magnitude proportional to its mass and the acceleration due to gravity. As the iclicker fall, it experiences air resistance due to the drag forces on it from the air molecules around. This drag force acts upwards so as to oppose the motion of the iclicker downwards, and the magnitude is minimal when compared to the gravitational force. The resultant force is therefore downwards

Answer:The answer is option A.

Momentum=mass×velocity.

Orange train=mass is 200 and velocity is 1.

which is 200×1=200kgm/s.

The initial momentum of the orange train is 200 kgm/s.

The orange train is having mass of 200 kg and it is travelling with speed of 1 m/s.

                     So, the orange trains momentum will be ,

                               P = mv

                            ∴ P = 200 x 1

                            ∴ P = 200 kgm/s

Therefore, the initial momentum of the orange train is 200kgm/s.

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Answer:

[tex]1.12056\times 10^{-8}\ N[/tex].

Kramer could try blowing air on the bowling pin.

Explanation:

[tex]m_1[/tex] = Mass of bowling ball = 7 kg

[tex]m_2[/tex] = Mass of bowling pin = 1.5 kg

[tex]r[/tex] = Distance between them = 0.25 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of gravity is given by

[tex]F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 7\times 1.5}{0.25^2}\\\Rightarrow F=1.12056\times 10^{-8}\ N[/tex]

The force of gravity between the ball and the pin is [tex]1.12056\times 10^{-8}\ N[/tex].

Kramer could try blowing air on the bowling pin.

Answer:

18.24 seconds

Explanation:

First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.

You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

Answer:

(a) The magnetic energy density in the field is 6.366 J/m³

(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

magnitude of magnetic field inside solenoid, B = 4 T

inner diameter of solenoid, d = 6.2 cm

inner radius of the solenoid, r = 3.1 cm = 0.031 m

length of solenoid, L = 26 cm = 0.26 m

(a) The magnetic energy density in the field is given by;

[tex]u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3[/tex]

(b) The energy stored in the magnetic field within the solenoid

[tex]U_B = u_B V\\\\U_B = u_B AL[/tex]

[tex]U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\[/tex]

0.37 dg

1 dg ....... 100 mg

x dg .........37 mg

x = 1×37/100 = 37/100 = 0.37 dg

Answer:

m₂ = 15660 kg

Explanation:

Given that,

Before collision,

Mass of box car 1, m₁ = 8700 kg

Speed of box car 1, u₁ = 14 m/s

Speed of box car 2, u₂ = 0 (at rest)

After collision,

The two stick together and move off with a speed of 5 m/s

Let m₂ is the mass of the second car. As cars stick together, it is a case of inelastic collision. Using the conservation of momentum as follows :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\8700\times 14+0=(8700+m_2)5\\\\121800=43500+5m_2\\\\121800-43500=5m_2\\\\m_2=\dfrac{78300}{5}\\\\m_2=15660\ kg[/tex]

So, the mass of the second car is 15660 kg.

Answer:

15000 kg

Explanation:

m1v1+m2v2=(m1m2)Vf

8700(15)=(8700+m)5.5

130500=8700+m(5.5)

130500/5.5=8700+m

130500-8700=m

m=15027.3 kg

Answer:

Pressure, [tex]P=1.84\times 10^7\ Pa[/tex]

Explanation:

Given that,

Force with which pickaxe strikes is 3600 N

The end of the pickaxe measure 15mm by 13mm.

We need to find the pressure exerted on the pipe by the pickaxe. Pressure exerted is equal to the force acting per unit area. So,

[tex]P=\dfrac{F}{A}[/tex]

A = 15 mm × 13 mm = 195 mm² = 0.000195 m²

So,

[tex]P=\dfrac{3600\ N}{0.000195\ m^2}\\\\P=18461538.46\ Pa\\\\\text{or}\\\\P=1.84\times 10^7\ Pa[/tex]

Answer:

299.4 MeV

Explanation:

For 40 Ar

Number of neutrons = 22

Number of protons= 18

Mass of each proton = 1.007277 amu

Mass of each neutron= 1.008665 amu

Total mass of protons= 18 × 1.007277 amu = 18.130986 amu

Total mass of neutrons = 22 × 1.008665 amu = 22.19063 amu

Total mass of protons and neutrons= 18.130986 + 22.19063 = 40.321616 amu

Mass defect = 40.321616 amu - 40 amu

Mass defect = 0.321616 amu

Since 931 is the conversion factor from amu to MeV

Binding energy = 0.321616 amu × 931 = 299.4 MeV

Binding energy = 299.4 MeV

Answer:

Explanation:

a )

one rotation per 24 hours

Time period of rotation of the earth T =  time / no of rotation

T = 24 x 60 x 60 s / 1

= 86400 s .

b )

angular velocity = angle of rotation in radian / time

in one rotation , angle made at the centre = 2π radian

= 2 x 3.14 radian

angular velocity ω = 2π / T

= 2 x 3.14 / 86400  radian / s

= 72.68 x 10⁻⁶ radian / s

c )

Relation between linear and angular velocity is as follows

v = ω  x R where R is radius of the earth and v is linear velocity .

linear velocity = 72.68 x 10⁻⁶ x 6.4 x 10⁶ m /s

= 465.152 m /s

=

Answer:

The correct answer is d

Explanation:

In this exercise they ask us which statement is correct, for this we plan the solution of the problem, this is a Doppler effect problem, it is the frequency change due to the relative speed between the emitter and the receiver of sound.

The expression for the Doppler effect of a moving source is  

f ’= (v / (v- + v_s) f  

From this expression we see that if the speed the sound source is different from zero feels a change in the  frequency.

The correct answer is d

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

Answer:

a) 500 N

b) 250 J

c) 0.87 m

d) 1.58 m/s, at 0.6 m from the wall

Explanation:

The mass of the object m = 100 kg

the spring constant k = 1000 N/m

length of the the spring = 1 m

extension of the string = 50 cm = 0.5 m

a) Force used to pull the mass is gotten from Hooke's law equation

F = -kx

where F is the force used to pull = ?

k is the spring constant = 1000 N/m

x is the extension = 0.5 m

substituting, we have

F = 1000 x 0.5 = 500 N  this force is used to pull the mass

b) The work done in moving the mass = Fx

==> 500 x 0.5 = 250 J

c) The energy stored up in the spring U = [tex]\frac{1}{2}kx^{2}[/tex]

U = [tex]\frac{1}{2}*1000*0.5^{2}[/tex] = 125 J

energy available for the mass from its equilibrium position = 250 - 125 = 125 J

this energy is equivalent to the work done by the spring on the mass by moving it closer to the wall

Work W = (weigh of the mass) x distance moved

weight = mg

where m is the mass = 100 kg

g is acceleration due to gravity = 9.81 m/s^2

substituting, we have

W = mgd

where d is the distance the mass moves closer to the wall

W = 100 x 9.81 x d

but W = 125 J

125 = 981d

d = 125/981 = 0.13 m

closeness to the wall = L - d

where L is the natural length of the spring = 1 m

closeness to the wall = 1 - 0.13 = 0.87 m

d) The maximum kinetic energy of the object  will be halfway between the extended length and the final resting place.

extended length = 1 + 0.5 m = 1.5 m

distance from resting place = 1.5 - 0.87 = 0.63 m from the wall

At this point, all the mechanical energy on the mass and spring system is converted to kinetic energy of motion.

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

substituting,

125 = [tex]\frac{1}{2}*100*v^{2}[/tex] = [tex]50v^{2}[/tex]

[tex]v^{2}[/tex] = 125/50 = 2.5

v = [tex]\sqrt{2.5}[/tex] = 1.58 m/s

Answer:

The detector temperature doesn't affect retention time

Explanation:

Retention time is one of the chromatographic parameters. Is defined as the time of a compound spends from injection to detection.

A solute in GC is added to the injector where is volatilized. When volatilized, it pass through a column until the detector.

The detector temperature doesn't affect retention time. To change retention time you must change injector temperature or column temperature. An increase in column or injector temperature results in a decrease in retention time.

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

a

 [tex]\epsilon = 7.57 \ V[/tex]

b

 [tex]\epsilon =0.0757 \ V[/tex]

c

  [tex]\epsilon =0.00699 \ V[/tex]

Explanation:

From the question we are told that  

  The number of turns is  N  =  40  turns  

   The  diameter is  [tex]d = 19 \ cm = 0.19 \ m[/tex]

    The initial  magnetic field is [tex]B_i = 0.40 \ T[/tex]

    The final magnetic field is [tex]B_f = 0 \ T[/tex]  

Generally the cross-sectional area of the coil is mathematically represented as

       [tex]A = \pi \frac{d^2}{4}[/tex]

=>    [tex]A = 3.142 * \frac{0.19^2}{4}[/tex]

=>    [tex]A = 0.0284 \ m^2[/tex]

At [tex]\delta t = 0.60 \ s[/tex]

The  induced emf is  

        [tex]\epsilon = - N * \frac{[B_f - B_i ] * A }{\delta t }[/tex]

=>    [tex]\epsilon = - 40 * \frac{[- 0.40 ] * 0.0284 }{0.06}[/tex]

=>   [tex]\epsilon = 7.57 \ V[/tex]

At [tex]\delta t = 6 \ s[/tex]

The  induced emf is  

        [tex]\epsilon = - N * \frac{[B_f - B_i ] * A }{\delta t }[/tex]

=>    [tex]\epsilon = - 40 * \frac{[- 0.40 ] * 0.0284 }{6}[/tex]

=>   [tex]\epsilon =0.0757 \ V[/tex]

At [tex]\delta t = 65 \ s[/tex]

The  induced emf is  

        [tex]\epsilon = - N * \frac{[B_f - B_i ] * A }{\delta t }[/tex]

=>    [tex]\epsilon = - 40 * \frac{[- 0.40 ] * 0.0284 }{65}[/tex]

=>   [tex]\epsilon =0.00699 \ V[/tex]

Answer:

C) 0.5 m/s^2

Explanation:

[tex]Mass = 20kg\\Force = 10 N\\Acceleration = ?\\\\Force = Mass \times Acceleration \\\\Acceleration = \frac{Force}{Mass}\\\\ Acceleration = \frac{10}{20}\\ \\Acceleration = 1/2 = 0.5m/s^2[/tex]

Answer:

that there would be more clouds.

Explanation:

one of the cycles is precipitation. without water there would be no rain and clouds are what causes the rain to come down

Answer:

it's D or 4 that there would be more clouds

Explanation:

hoped it helped!!

Answer:

Nitrogen 13

Explanation:

We need fill in the blank the nuclide symbol for the missing particle in the following nuclear equation.

[tex]X\rightarrow ^0e_1+^{13}C_6[/tex]

In this reaction, the emission of positron takes place. The atomic number in the LHS and the RHS should be same. So,

[tex]^{13}_7 N \rightarrow ^0e_1+^{13}C_6[/tex]

So, the LHS have nitrogen 13, a radioactive isotope.

Answer:

and the forces acting on the particle as a function of time.

Explanation:

The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. hope this helps you :)

The linear impulse and momentum equation is obtained by integrating the force acting on the object with respect to time.

The Linear Impulse is defined as the momentum exerted by a force during a time interval.

The linear impulse and momentum equation is obtained by integrating the force with respect to time.

From the equation of the impulse-momentum theorem:

∆p = J

where,

∆p is the change in momentum of the object,

J is the impulse applied to it.

The impulse is defined as the product of force and the time over which the force is applied.

Thus, we can write the impulse as:

J = ∫ F dt

where,

F is the force acting on the object,

dt is the time interval during which the force acts.

By substituting this expression for J in the momentum equation,

we get:

∆p = ∫ F dt

which is the linear impulse-momentum equation.

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Answer:

A.46.7mph

B.50mph

Explanation:

We know that from the data given

Julie drove 50 miles at a speed of 38 mph, and another 50 miles for 62mph. Thus , for the first 50 miles, she drove for the following time

T1= distance/speed

====> 50/38= 1.3hrs

the next 50 miles, she drove for

T2= 50/62

= 0.8hrs

So average speed is

Totaldistance/ total time

100/2.1

= 46.7mph

B. Because the time she used driving at 35 mph is the same amount of time she used driving at 65 mph, the average speed is just the average of the two speeds given which is

(50+50)/2 = 50mph

Explanation:

Answer:

46.95 mph

50 mph

Explanation:

first, we start by finding the average speed. The average speed if given by the relation

Average Speed = Total distance / Total Time

from the question, we know that

total distance is 100 mi

total time is time of the first distance + time of the second distance

t = (50mi / 38) + (50mi / 62)

t = 1.32 + 0.81

t = 2.13 hrs

Then, the average speed going to grandmothers house will be

Speed = 100mi / 2.13 hrs

Speed = 46.95 mph

on her return trip

speed on the first distance * half the time + speed on the second distance * half the time

x = V1 * t/2 + V2 * t/2

100 = 38 * t/2 + 62 * t/2

100 = 100 * t/2

t/2 = 100/100

t/2 = 1

t = 2 hrs

The average speed then will be, = 100 mi / 2 hrs = 50mph

Answer:

The  depth of stream to the student is  [tex]d_1 = 0.2932 \ m[/tex]

Explanation:

From the question we are told that

   The actual  depth of the stream is [tex]d = 39 \ cm = 0.39 \ m[/tex]  

    The  refractive index of the water is  [tex]n = 1.33[/tex]

Generally the apparent depth of the stream is mathematically represented as

         [tex]d_1 = \frac{d}{1.33}[/tex]

substituting values

        [tex]d_1 = \frac{ 0.39}{1.33}[/tex]

        [tex]d_1 = 0.2932 \ m[/tex]