A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.

Answer:

0.46N

Explanation:

See attached file

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

Answer:

Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.

Electric field strength is also known as Electric Field Intensity .

Explanation:

Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.

Please mark me as the brainliest!!!

Thanks!!!

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

Answer:

Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)

Where the charge of a single proton is:

C = 1.6x10^-19 C

Now, you need to remember that when we are working with charges, we are working with discrete math:

What means that?

If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)

So the consecutive charge will be:

C = 2*1.6x10^-19 C = 3.2x10^-19 C.

So, because we are working in discrete math, we can not have any object that has charge between  1.6x10^-19 C and 3.2x10^-19 C.

Particularly, 2.0x10^-19 C is in that range, so we can conclude that:

No, an object can not carry a charge of 2.0x10^-19 C.

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

Answer:

The  total energy is  [tex]T = 169.02 \ J[/tex]

Explanation:

From the question we are told that

    The  Poynting vector (energy flux ) is  [tex]k = 0.939 \ W/m^2[/tex]

    The length of the rectangle is  [tex]l = 1.5 \ m[/tex]

    The  width of the rectangle is  [tex]w = 2.0 \ m[/tex]

    The time taken is [tex]t = 1 \ minute = 60 \ s[/tex]

The total electromagnetic energy falls on the area is mathematically represented as

      [tex]T = k * A * t[/tex]

Where A  is the area of the rectangle which is mathematically represented as

           [tex]A= l * w[/tex]

 substituting values

         [tex]A= 2 * 1.5[/tex]

        [tex]A= 3 \ m^2[/tex]

substituting values

        [tex]T = 0.939 * 3 * 60[/tex]

        [tex]T = 169.02 \ J[/tex]

Answer:

The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

Answer:

The approximate displacement of the object is  23  m.

Explanation:

Given that:

v = 4t + 5 (m/s)  for 3< t< 7; n= 4

The approximate displacement of the object can be calculated as follows:

The velocities at the intervals of t are :

3

4

5

6

the velocity at the intervals of t =  7 will be left out due the fact that we are calculating the left endpoint Reimann sum

n = 4 since there are 4 values for t, Then there is no need to divide the velocity values

v(3) = 4(3)+5

v(3) = 12+5

v(3) = 17

v(4)= 4(4)+5

v(4) = 16 + 5

v(4) = 21

v(5)= 4(5)+5

v(5) = 20 + 5

v(5) = 25

v(6) = 4(6)+5

v(6) = 24 + 5

v(6) = 29

Using Left end point;

[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]

= 23 m

Answer:

Zack should direct his throw outward and toward the back of the car.

Explanation:

As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.

The solution is throw 3.

I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.

The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.

When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.

The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.

Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.

Learn more about skydiver on:

https://brainly.com/question/29253407

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Answer:

Pencil is to pen

Step by step explanation:

They are similar items, as they are both tools, but are different as to how they function.

Answer:

(a) f

(b) r

(c) s

Explanation:

There are two forces on the sphere: weight and buoyancy.

Sum of forces in the y direction:

∑F = ma

B − mg = 0

B = mg

Buoyancy is equal to the weight of the displaced fluid, or ρVg, where ρ is the density of the fluid and V is the displaced volume.

ρVg = mg

ρV = m

V = m/ρ

(a) The mass decreases, so the displaced volume decreases.

(b) The sphere's density is constant and its radius increases, which means its mass increases, so the displaced volume increases.

(c) The mass stays the same, so the displaced volume is the same.

Answer:

D &B

Explanation:

Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force

Answer:

Explanation:

Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction ,  so their relative velocity

= u + v = 2.3 m /s ( given )

We shall apply conservation of momentum law for the movement of astronaut and metal plate

mu  = M v where m is mass of astronaut , M is mass of metal plate

71 u = 230 x v

71 ( 2.3 - v ) = 230 v

163.3 = 301 v

v = .54 m / s

u = 1.76 m / s

honeycomb will be at rest  because honeycomb surface  is frictionless . Plate will slip over it . Over plate astronaut is walking .

a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s

b ) velocity of astronaut relative to honeycomb will be + .54 m /s

Here + ve direction is assumed to be the direction of astronaut .  

Answer:

The longer the cord, the lower the illumination

Explanation:

The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.

Long wires have more electrical resistance than shorter ones.

Let us consider this formula:

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

From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

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

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

Answer:

The magnitude of the magnetic field is 8.333 x 10⁻⁷ T

Explanation:

Given;

charge on the lightening bolt, C = 15.0 C

time the charge passes by, t = 1.5 x 10⁻³ s

Current, I is calculated as;

I = q / t

I = 15 / 1.5 x 10⁻³

I = 10,000 A

Magnetic field at a distance from the bolt is calculated as;

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

where;

μ₀ is permeability of free space = 4π x 10⁻⁷

I is the current in the bolt

r is the distance of the magnetic field from the bolt

[tex]B = \frac{\mu_o I}{2\pi r} \\\\B = \frac{4\pi *10^{-7} 10000}{2\pi *24} \\\\B = 8.333 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field is 8.333 x 10⁻⁷ T

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

Answer:

Option (A)

Explanation:

A 20 kg boy chases the butterfly with a speed of 2 meter per second.

Angle at which he runs is 70° North of West.

Therefore, Horizontal component (Vx) directing towards West will be,

Vx = v(Cos70°)

Vy = v(Sin70°)

Since momentum of a body is defined by,

Momentum = Mass × Velocity

Therefore, Westerly component of the momentum will be,

Momentum = 20 × (v)(Cos70°)

                   = 20 × 2Cos70°

                   = 13.68

                   ≈ 13.7 kg-meter per second

Therefore, Option (A) will be the answer.

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium