An action that has the ability to change an object’s state of motion is

Answer:

N/C = V/m.

Explanation:

The SI unit of electric field is N/C. Sometimes the units are written as V/m.

We know that,

1 V = 1 J/C

Using dimensional analysis,

The dimensional formula for Joules is [M¹L² T⁻²].

The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].

So,

J/C = [M¹L² T⁻³I¹] ...(1)

The dimensional formula of Newton is [M¹ L¹ T⁻²]

The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].

N/C= [M¹L² T⁻³I¹] ....(2)

From (1) and (2) it is clear that the N/C is equal to V/m.

Answer:

[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

Explanation:

The pressure at a depth of a fluid is

       P = ρ g y

where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.

In this case the pressure for fluid A is

      Pa = ρₐ g y

the pressure for fluid B is

      P_b = ρ_b g y

depth y not changes as the gauge is stationary

if we look for the relationship between these pressures

       [tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]

        [tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

therefore we see that the pressure measured for fluid B is different from the pressure of fluid A

if  ρₐ < ρ_b B the pressure P_b is greater than the initial reading

   ρₐ>  ρ_b the pressure in B decreases with respect to the reading in liquid A

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

Answer:

Can't understand the language

Answer:

A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.

Explanation:

This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.

Answer:

Increasing until terminal velocity is reached

Explanation:

Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.

Answer:

Magnification = 1

Explanation:

given data

radius of curvature r = - 0.983 m

image distance u = - 0.155

solution

we get here first focal length that is

Focal length, f = R/2     ...................1

f = -0.4915 m

we use here formula that is

[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex]      .................2

put here value and we get

[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]  

so

Magnification will be here as

m = [tex]- \frac{v}{u}[/tex]

m =  [tex]\frac{0.155}{0.155}[/tex]

Answer:

The magnification is 1.5.

Explanation:

radius of curvature, R = - 0.983 m

distance of object, u = - 0.155 m

Let the distance of image is v.

focal length, f = R/2 = - 0.492 m

Use the mirror equation

[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]

The magnification is given by

m = - v/u

m = 0.226/0.155

m = 1.5

Answer:

d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

Explanation:

That is a problem of electric forces, given by Coulomb's law

          F = [tex]k \frac{ q1q2}{r^2}[/tex]

We use that charges of the same sign repel and charges of different signs do not attract, so the net force is

           ∑ = F₁₃ + F₂₃

          F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]

a) the charge is placed at the midpoint between the other two

         F_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]

calculate us

          F_ {net} = 9 10⁹ / 0.5²   2 10⁻⁶ (50 -25) 10⁻⁶

          F_ {net} = 1,800 N

b) where must be placed q3 so that the force is zero

for this case the charge q3 is outside the spheres

          ∑ F = 0

          F₁₁₃ = F₂₃

          k q_1 / r_{13}² = k q₂ q₃ / r₂₃²

          q₁/ r₁₂²   = q₂ / r₂₃²

suppose the distance

          r₁₂ = d

the he other sphere is

          r₂₃ = d + 1

we substitute

          q₃ / d² = q₂ / (d + 1) ²

          (d + 1) ² = q₂ / q₃ d²

           d² (1 - q₂/ q₃) + 2d + 1 = 0

we solve the equation of a second

            d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2

             d = -2 /2

             d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

Answer:

to adjust from distant to the near objects

Explanation:

Answer:

The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.

Explanation:

The human eye is the optical instrument which works on the refraction of light.

The ability of eye lens to change its focal length is called accommodation power of eye.

The focal length of eye lens is changed by the action of ciliary muscles.

When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.

Answer:

433.15K

Explanation:

(320°F − 32) × 5/9 + 273.15 = 433.15K

i. [tex]T = 36.8\:\text{N}[/tex]

ii. [tex]a = 2.45\:\text{m/s}^2[/tex]

iii. [tex]x = 1.23\:\text{m}[/tex]

Explanation:

Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]

Forces acting on m1:

[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]

Forces acting on m2:

[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]

Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us

[tex]m_2g - m_1g = m_2a + m_1a[/tex]

or

[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]

Solving for a,

[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]

[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]

We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).

[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]

Assuming that the two objects start from rest, the distance that they travel after one second is given by

[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]

Being in parallel each device will have an equal voltage drop of 120 V

A. Yes the combination will blow the fuse. See part B for the total current.

B. Toaster = 1800W / 120V = 15A

   Frying Pan = 1400W / 120V = 11.67A

  Lamp = 55W / 120V = 0.458A

Total amps = 15 + 11.67 + 0.458 = 27.128 Amps

27.128A is greater than 15A so the fuse will blow.

285.3 days

Explanation:

The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write

[tex]F_c = F_G[/tex]

or

[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]

where M is the mass of the star and R is the orbital radius around the star. We know that

[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]

where C is the orbital circumference and T is orbital period. We can then write

[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]

Isolating [tex]T^2[/tex], we get

[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]

Taking the square root of the expression above, we get

[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]

which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as

[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]

[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

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

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]

L = 169.5 m

Answer:

elevators

Theatre system

construction pulley

lifts

Answer:

elevator,cargo lift system

We know

[tex]\boxed{\large{\sf Velocity=\dfrac{Distance}{Time}}}[/tex]

[tex]\\ \Large\sf\longmapsto Velocity=\dfrac{6}{\dfrac{1}{2}}[/tex]

[tex]\\ \Large\sf\longmapsto Velocity=6\times 2[/tex]

[tex]\\ \Large\sf\longmapsto Velocity=12km/h[/tex]

Is this the right answer??

We should keep km and min in smallest SI unit

Answer:

The angular acceleration is 26.6 rad/s^2.

Explanation:

Force, F = 314 N

radius, r = 0.281 m

mass, m = 84.2 kg

The grindstone is a disc.

The torque is given by

torque = force x radius

Torque = 314 x 0.281 = 88.234 Nm

The torque is given by

Torque = Moment of inertia x angular acceleration

[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]

Answer:

Sources:

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

Answer:

The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.

Explanation:

Edmentum

The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ[tex]T^{4}[/tex]

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.

So that;

[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]

     = 3.8556 x [tex]10^{-8}[/tex]

     = 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex]

Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

Learn more on energy radiation of objects by visiting: https://brainly.com/question/12550129

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

Answer:

ygguguguhhihhihijijijjojojinjbgy

Answer:

due to production of heat through friction

Explanation:

because of the friction produce between her hands

Answer:

The current in third branch is 11 A.

Explanation:

incoming current in one branch = 18 A

outgoing current in the other branch = 7 A

let the current in the third branch is i.

According to the Kirchoff's fist law in electricity

incoming current = out going current

18 = 7 + i

i = 11 A

The current in third branch is 11 A.

Answer:

[tex]N_f=248N[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=100kg[/tex]

Ladder Length [tex]l=4.0m[/tex]

Mass of Ladder [tex]m_l=25kg[/tex]

Angle [tex]\theta=56 \textdegree[/tex]

Generally the equation for Co planar forces is mathematically given by

[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]

Therefore

[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]

[tex]N_f=248N[/tex]

Answer:

the radius of the bigger loop is 5 cm.

Explanation:

Given;

current in the smaller loop, I₁ = 12 A

current in the larger loop, I₂ = 20 A

radius of the smaller loop, r₁ = 3 cm

let the radius of the larger loop, = r₂

Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.

[tex]B= \frac{\mu_0 I}{2r}[/tex]

The magnetic field at the center of the smaller loop;

[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]

The magnetic field at the center of the bigger loop;

[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]

If the magnetic field at the center is zero, then B₁ = B₂

[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]

Therefore, the radius of the bigger loop is 5 cm.

Answer:

The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.

Explanation: