1/​A current of 0.5A flows through a resistor of resistance of 20 Ω.
​Calculate the potential difference across the resistor.


Please help:)

Answer:

It is always less than 1 because the load arm is always longer than effort arm.

Explanation:

In the formula, MA= effort divided by load which makes it less than 1.

This helps by reducing the applied force(effort). It is a speed multiplier.

 Hope it helps.

Answer:

a)    d = 60m (distance)

b)    D = 60m (displacement)

c)    v = 20 m/min

d)   |v| = 20 m/min

Explanation:

a) The distance traveled by Jorge is 60m

d = 60m

b) The module of the displacement D, is equal to the values of the distance d, because Jorge walked in a straight line.

D = d = 60m

c) The speed of Jorge is given by the following formula:

[tex]v=\frac{d}{t}[/tex]

d: distance = 60m

t: time of the walk = 3min

[tex]v=\frac{60m}{3min}=20\frac{m}{min}[/tex]

The speed is 20 m/min

The module of the Jorge's velocity is:

[tex]|v|=\frac{D}{t}[/tex]

D: displacement = d = 60m

t: time = 3 min

[tex]|v|=\frac{60m}{3min}=20\frac{m}{min}[/tex]

The module of Jorge's velocity is 20 m/min

Jorge recorre una distancia de 60 m en 3 min. Su desplazamiento de 60 m, lo hace con una rapidez, y módulo de su velocidad, de 20 m/min.

Jorge camina en linea recta por una plaza recorriendo 60 metros en 3 minutos (t).

La distancia recorrida (d) es 60 m.

Como se mueve en linea recta, el desplazamiento (D) coincide con la distancia recorrida, es decir, es de 60 m.

Podemos calcular la rapidez (s) de Jorge usando la siguiente fórmula.

[tex]s = \frac{d}{t} = \frac{60m}{3min} = 20m/min[/tex]

La rapidez es una magnitud escalar y coincide con el modulo de la velocidad, dado que la velocidad es vectorial.

Jorge recorre una distancia de 60 m en 3 min. Su desplazamiento de 60 m, lo hace con una rapidez, y módulo de su velocidad, de 20 m/min.

Aprende mas: https://brainly.com/question/24773606

Answer:

circular motion

Explanation:

As a motorcycle takes a sharp turn, the type of motion that occurs is called circular motion.

Circular motion is a movement of an object along a circular path. As this motorcycle makes the sharp turn, it is acted upon by a centripetal force which directs the motorcycle towards the center.

Therefore, circular motion is the correct answer to the question.

Answer:

Circular Motion

Answer:

(i) Si device are coated with B or Li for neutron detection

(ii) Fast neutrons are normally first slowed down  to thermal energies before measurement

(iii) The strip detectors consists of p⁺ and n implants in the region of a very thin (300 to 400 μm) depletion zone  though which the neutrons traverse producing a readout based on the generated charges being directed to the cathode, p⁺, material  which then transmits the charge to the device electronics

Explanation:

(i) Detection of neurons with only an Si device in not possible due to the large neutrons path in Si such that the silicon needs to be coated with B or Li which readily interact with neutrons. The neutron interaction with the reactive coating produces an alpha particle which can be detected by the semi conductor and a nucleus

(ii) Neutrons having a kinetic energy that is more than 1 MeV which as such has a velocity of more than 15,000 km/s is known as fast neutrons or fission neutrons. The fast neutrons are slowed in a nuclear reactor by neutron moderation to thermal energies

Due their high speed, fast neutrons are normally slowed down which however results in the loss of some vector properties of the neutron

Techniques for fast neutron detection includes,

1) Recoil detectors which are capable of fast neutron detection without moderation

2) Bonner spheres detector first converts the fast moving electron to slow down before detection

3) Scintillation counter are widely used but require the conversion of the neutron to a charged particles before detection

(iii) Strip detectors provide high precision measurement of a particle's crossing point which can be further improved by use of low noise electronics

Applications of silicon strip detectors include

1) Particle tracking in research in particle physics

2) Particle tracking in researches in x-spectroscopy nuclear research

3) Imaging in x-talography

4) Medical research imaging

5) Particle tracking and imaging in astrophysics.

Answer:

(i)    t = 5s

(ii)    x = 3.75*10^5 m

Explanation:

(i) To calculate the time that the electron takes to reach twice the value of its initial velocity, you use the following formula:

[tex]v=v_o+at[/tex]         (1)

vo: initial velocity of the electron = 5*10^4 m/s

v: final velocity of the electron = 2vo = 1*10^5 m/s

a: acceleration of the electron = 1*10^4 m/s^2

You solve the equation (1) for t, and replace the values of the parameters:

[tex]t=\frac{v-v_o}{a}=\frac{1*10^5m/s-5*10^4m/s}{1*10^4m/s^2}=5s[/tex]

The electron takes 5s to reach twice its initial velocity.

(ii) The distance traveled by the electron in such a time is:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]          (2)

you replace the values of the parameters in the equation (2):

[tex]x=(5*10^4m/s)(5s)+\frac{1}{2}(1*10^4m/s^2)(5s)^2\\\\x=3.75*10^5m[/tex]

The distance traveled by the electron is 3.75*10^3m/s

The catch in this one is:  We don't know how much force the student used to push the bike.  

It wasn't necessarily the 100N.  That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium.  The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push.  That's what's so great about wheels !  That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel".  I looked at it, I turned it over and I looked on all sides. I thought it was too simple.  I didn't know then it was elegant. I threw him out.  I was so dumb.  I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ?  Stay tuned:

-- The bike covered 15 meters in 5 seconds.  Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.  

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

Power = 36.73 watts

Answer:

white

Explanation:

white is less absorbent then black

Answer:

The third charge must be placed 0.548 m from q₁.

Explanation:

Let r = 3m  be the distance between charge q₁ and q₂.

Let x be the distance between charge q₁ and charge q₃ (the third positive charge)

Then r - x is the distance between charge q₂ and q₃

Let the electrostatic force between q₁ and q₃ be F = kq₁q₃/x²

Let the electrostatic force between q₂ and q₃ be F' = kq₂q₃/(r - x)²

Since F + (-F') = 0 (the signs on the forces are different since the forces are in opposite directions)which is required when the net force on q₃ is zero, then

F - F' = 0

F = F'

So, kq₁q₃/x² = kq₂q₃/(r - x)²

q₁/x² = q₂/(r - x)²

[(r - x)/x]² = q₂/q₁

taking square-root of both sides,

(r - x)/x = ±√q₂/q₁

r/x - 1 = ±√q₂/q₁

r/x = 1 ±√q₂/q₁

x = r/(1 ±√q₂/q₁)

substituting the values of the variables r = 3 m, q₁ = 0.50 nC and q₂ = 10 nC

x = 3 m/(1 ±√10 nC/0.5 nC)

x = 3 m/(1 ±√20)

x = 3 m/(1 ± 4.472)

x = 3 m/5.472 or 3 m/-3.472

x = 0.548 m or -0.864 m

So the third charge must be placed 0.548 m to the right of q₁ or 0.864 m to the left of q₁.

Since we are concerned about the line of charge that connects q₁ and q₂, the third charge must be placed 0.548 m from q₁.

Answer:

(a)     T = 4.09*10^-8 s

(b)     r = 0.19m

Explanation:

(a) In order to determine the rotational period of the protons, you first calculate the angular speed of the proton. You use the following formula for the radius of the trajectory described by the electron in a region with a constant magnetic field:

[tex]r=\frac{mv}{qB}[/tex]               (1)

m: mass of the proton = 1.67*10^-27 kg

q: charge of the proton = 1.6*10⁻19C

B: magnitude of the magnetic field = 1.6T

v: speed of the proton = 0.1c = 0.1(3*10^8m/s) = 3*10^7m/s

The angular speed of the proton is v/r, then, you solve for w in the equation (1) and replace the values of the parameters:

[tex]\omega=\frac{v}{r}=\frac{qB}{m}\\\\\omega=\frac{(1.6*10^{-19}C)(1.6T)}{1.67*10^{-27}kg}=1.53*10^8\frac{rad}{s}[/tex]

The rotational period T is:

[tex]T=\frac{2\pi}{1.53*10^8rad/s}=4.09*10^{-8}s[/tex]

(b) The radius of the protons' orbit is calculated by using the equation (1):

[tex]r=\frac{(1.67*10^{-27}kg)(3*10^7m/s)}{(1.6*10^{-19}C)(1.6T)}\\\\r=0.19m[/tex]

The radius of the protons' orbit is 0.19m

Answer:

The resultant of two vectors:

P

  and    

Q

​  

 is given by  

R=  

P  

2

+Q  

2

+2PQcosθ

​  

where, P and Q are magnitudes of two vectors and θ is the angle between two vectors.

Here θ=?  P=8N  Q=15N  R=17N

Therefore,

17=  

8  

2

+15  

2

+2×8×15cosθ

​  

289=64+225+240cosθ

289=289+240cosθ

289−289=240cosθ

cosθ=0

cosθ=cos90

θ=90  

0

Explanation:

Answer:

i. rubbing the material against a magnet

ii. placing the material in a magnetic field of opposite polarity

iii. placing the material near a strong magnet

Explanation:

Ferromagnetic materials are majorly metals which can be easily attracted by a magnet. ferromagnetic materials are made up of domains, behaves as minute pieces of magnet. They can be rearranged to align when under the influence of an external magnetic field.

The alignment of the domains in a ferromagnetic material can be caused by either of the following: rubbing the material against a magnet, placing the material in a magnetic field of opposite polarity, placing the material near a strong magnet.

Answer:

B, D, E

Explanation:

edg2020

Answer:

Explanation:

Answer:

The person should exercise 150 minutes

Explanation:

Recall that the Lorentz factor must be applied to the relationship between the elapsed time (T) of the clock at rest (time measured on Earth), and the elapsed time (T') measured by the clock in the frame moving at 0.8 c. The equation becomes:

[tex]T'=\frac{T}{\sqrt{1-(\frac{0.8\,c}{c})^2 } } \\T'=\frac{90\,min}{\sqrt{1-0.8^2} } \\T'=150\,\,min[/tex]

Answer:William Herschel found the seventh planet now known as uranus

<!> Brainliest is appreciated! <!>

Explanation:

Answer:

William Herschel

Explanation:

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

[tex]\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\[/tex]

Δtm is  time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the  sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

[tex]\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr[/tex]

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of [tex]kgm^{2 }\ / s[/tex].

Answer:

70p per week

Explanation:

Power * time = energy

1 kw * 1 hour = 1 kwh per day

For 7 days,

Energy used = 1 kwH/day * 7 days = 7kwh

Total cost per week = 7 kwh * 10p

= 70p

Answer:

A opção A está correta.

O sistema formado pela garrafa térmica e a água perde 400 cal de calor para o meio ambiente.

Option A is correct.

The system formed by the thermos and the water loses 400 cal of heat to the environment.

Explanation:

Quando a temperatura de um sistema reduz, fica claro que o sistema perdeu calor ou energia térmica. Como a temperatura é um dos indicadores mais claros disso, esta conclusão é hermética e correta.

Mas, para saber a quantidade de calor perdida para o meio ambiente, agora fazemos alguns cálculos de energia térmica.

Transferência de calor de ou para o sistema de água e garrafa térmica = c × ΔT

c = capacidade térmica do sistema de água e garrafa térmica = 80 cal /°C

ΔT = Alteração da temperatura do sistema de água e garrafa térmica = (temperatura final) - (temperatura inicial) = 55 - 60 = -5°C

Calor transferido = 80 × -5 = -400 cal.

O sinal de menos mostra que o calor é transferido para fora do sistema, ou seja, o calor é perdido no sistema.

Espero que isto ajude!!!

English Translation

The thermos (also known as "Dewar vase") is an extremely useful device to conserve bodies (essentially liquid) at high temperatures, minimizing energy exchanges with the environment, which is generally colder. A thermos contains water at 60 o C. The thermos + water set has a thermal capacity of C = 80 cal / o C. The system is placed on a table and, after a considerable period of time, its temperature decreases to 55 o C. In this case, it is concluded that the system formed by the thermos and the water inside:

a) lost 400 cal. B) gained 404cal. C) lost 4 850 cal. D) gained 4 850 cal. E) did not exchange heat with the external environment.

Solution

When a system's temperature reduces, it is clear to conclude that the system has lost heat or thermal energy. Since temperature is one of clearest indicators of this, this conclusion is airtight and correct.

But, to know the amount of heat lost to the environment, we now do some thermal energy calculations.

Heat transferrred from or to the water and thermos system = c × ΔT

c = heat capacity of the water and thermos system = 80 cal/°C

ΔT = Change in temperature of the water and thermos system = (final temperature) - (initial temperature)

= 55 - 60 = -5°C

Heat transferred = 80 × -5 = -400 cal.

The minus sign shows that the heat is transferred out of the system, that is, the heat is lost from the system.

Hope this Helps!!!

Answer:

(i have corrected the answers to 3 significant figures)

When dealing with projectile motion, we should consider its vertical and horizontal components.

Vertical velocity = 50sin60° m/s

Horizontal velocity = 50cos60° m/s

a) when it reaches the peak, meaning it can no longer travel further upwards, indicating the final vertical velocity is 0.

Take g=9.80665

using formula a = (v-u) /t

9.80665= (0-50sin60°) / t

t= 4.42 s

b) Consider the vertical component.

using formula t = 2u / g

t = 2(50sin60°) / 9.80665

t = 8.83s

c) Consider the vertical component again,

using formula H = u² / 2g

H = (50sin60°)² / 2(9.80665)

H = 95.6m

d) This is also the range of the projectile motion.

Using formula R = u²sin2θ / g

This time u should just be the initial velocity (neither horizontal/vertical)

R = 50²sin2(60) / 9.80665

= 221 m

- you can also do this by using s = ut formula, using the time calculated from b), but consider the u as the horizontal component.

The correct answer is D. Amount of time and area of physical contact between the substances.

Explanation:

Heat transfer refers to the flow of thermal energy or heat between two or more objects. This process involves multiple factors and implies heat from the hottest object goes to the coldest one until there is an equilibrium. To begin, heat transfer depends on the amount of thermal energy in the objects because objects must have a different amount of thermal energy for heat to flow.

Besides this, the amount of energy that flows depends on the time and the contact between the substances of objects. Indeed, objects need to be in contact or close to each other for heat to transfer, and the time needs to be enough for the process to occur. For example, if you place a pot over the fire just for a few seconds it is likely the heat transferred is minimal, which does not occur if you leave the pot more time. At the same time if the pot is in close contact with fire more heat will be transferred.-

Answer:

the answer is D on edginuity.

Explanation:

Answer:

Explanation:

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Answer:

The thrust is [tex]6\times 10^6\ N[/tex]

Explanation:

Given that,

Mass of gas, [tex]m=1.5\times 10^3\ kg[/tex]

The rate at which the gas is expelling, [tex]\dfrac{dv}{dt}=4\times 10^{3}\ m/s[/tex]

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

[tex]F=\dfrac{p}{t}[/tex]

Also, p = mv

[tex]F=\dfrac{mv}{t}[/tex]

So,

[tex]F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N[/tex]

So, the thrust is [tex]6\times 10^6\ N[/tex]

Answer:

4500 joules

Explanation:

since work(joules) = force(newtons) x distance(meters),

150N x 30M = 4500 Joules

Answer:

4500 joules

Explanation:

150 N x 30 M = 4500 joules

Answer:

[tex]D(t) = 9(0.93)^t cos(30 \pi t)[/tex]

Explanation:

Amplitude begins at 9 cm, [tex]A_0 = 9 cm[/tex]

The amplitude decreases by 7% (0.07) each second

The amplitude function can then be modeled as:

[tex]A(t) = A_0(1 - 0.07)^t\\A(t) = 9(0.93)^t[/tex]

The spring oscillates 15 times each second, the period of oscillation (time to make 1 oscillation) will therefore be calculated as:

T = 1/15

[tex]\frac{2\pi }{B} = \frac{1}{15} \\\\B = 30\pi[/tex]

The graphical equation of the system described is:

[tex]D(t) = A cos ( Bt - C) + D[/tex]

Horizontal shift, C = 0

Vertical shift, D = 0

[tex]D(t) = 9(0.93)^t cos(30 \pi t)[/tex]

Answer:

the labor market.

Explanation:

Dan's plant will require two shifts of skilled workers, seven days a week. As he does his research, he will pay particular attention to the labor market.

Answer:

The wavelength is approximately 611 nm

Explanation:

We can use the formula for the condition of maximum of interference given by:

[tex]d\,sin(\theta)=m\,\lambda\\(0.000250\,\,m)\,\,sin(1.12^o)=8\,\lambda\\\lambda=\frac{1}{8} \,(0.000250\,\,m)\,\,sin(1.12^o)\\\lambda \approx 610.8\,\,nm[/tex]

We can also use the formula for the distance from the central maximum to the 5th minimum by first finding the tangent of the angle to that fifth minimum:

[tex]tan(\theta)=\frac{y}{D}\\ tan(\theta)=\frac{0.0333}{3.02} =0.011026[/tex]

and now using it in the general formula for minimum:

[tex]d\,sin(\theta)\approx d\,tan(\theta)=(m-\frac{1}{2} )\,\lambda\\\lambda\approx 0.00025\,(0.011026)/(4.5)\,\,m\\\lambda\approx 612.55\,\,nm[/tex]

Answer:

The correct answer is  [tex]6.1\times10^{-7}\:m[/tex]

Explanation:

The distance from the central maxima to 5th minimum is:

[tex]x_{5n}-x_{0} =3.33\:cm=0.033\:m[/tex]

The distance between the slits and the screen:

[tex]L = 302\:cm = 3.02\:m[/tex]

Distance between 2 slits: [tex]d = 0.00025\:m[/tex]

[tex](n-\frac{1}{2})\lambda=\frac{d(x_n)}{L}[/tex]

For fifth minima, n = 5... so we have:

[tex]x_{5n}=\frac{9\lambda L}{2d}[/tex]

For central maxima, n = 0... so we have:

[tex]x_{0}=\frac{n\lambda L}{d}=0[/tex]

So the distance from central maxima to 5th minimum is:

[tex]\frac{9\lambda \:L}{2d}-0=0.033[/tex]   (Putting the values, we get):

[tex]\Rightarrow \lambda = 6.1\times 10^{-7}\:m[/tex]

Best Regards!

Answer:

C. 3.2 x 10^8 Ω•m

Explanation:

An insulator is a material that resists the flow of electricity.

In the given data the material with the highest resistivity is the best insulator

3.2 x 10^8 Ω•m

Answer:

I think it is A

Explanation:

Answer:

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

[tex]T=2\pi\sqrt{\frac{L}{g} }[/tex] where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

[tex]T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }[/tex]

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

[tex]T_1=2\pi\sqrt{\frac{2L}{g} }=\sqrt{2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{2}*4=5.7\ sec[/tex]

3) The string length is halved

Given that:

[tex]T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }[/tex]

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

[tex]T_1=2\pi\sqrt{\frac{L}{2g} }=\sqrt{1/2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{1/2}*4=2.8\ sec[/tex]

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.