A 1500 kg car travelling at 25 m/s collides with a 2500 kg van which had
stopped at a traffic light. As a result of the collision the two vehicles become
entangled. Which of the following pairs given in the table below shows the
initial speed the entangled mass will move and the type of collision​

t Answer:

1) the time of the pigeon 1 is less, so it comes first

2) v = - 6,997 m / s ,  3)     v = 10.15 m / s ,

4) the displacement of the second point in greater

5)     x = 0.883 m

Explanation:

For this exercise we will use the kinematics equations

1) ask which chick reaches the ground first

we calculate for the first chick that has zero initial velocity

          y = y₀ + v₀ t - ½ g t²

          0 = yo - ½ g t²

          t = √ 2 y₀ / g

let's calculate

          t = √ (2 2.50 / 9.8)

          t = 0.714 s

We calculate the time it takes for the second chick that has velocity v = (1 i ^ - 1.5 j⁾ m / s

           y = y₀ + v₀t - ½ g t²

           0 = 2.5 - 1.5 t - ½ 9.8 t²

           4.9 t² + 1.5 t - 2.5 = 0

            t² + 0.306 t - 0.510 = 0

we solve the quadratic equation

            t = [0.306 ± √ (0.306² - 4 (-0.510))] / 2

            t = [0.306 ± 1.46] / 2

The results are

            t₁ = -0.577 s

            t₂ = 0.883 m / s

we take positive time as correct

the time of the pigeon 1 is less, so it comes first

2) the speed of the first chick is

              v = v₀ - g t

we can see that

              v = -gt

              v = - 9.8 0.714

              v = - 6,997 m / s

the negative sign indicates that the speed is down

3) the speed of the other bird is

              v = -1.5 - 9.8 0.883

               v = 10.15 m / s

4) which chick has the greatest displacement. The first point falls vertically and its displacement is y₀

The second point describes a parabola and its displacement is

           d = √ (x² + y₀²)

therefore we see that the displacement of the second point in greater

5) calculate the horizontal displacement of the second point

           x = vx t

           x = 1 0.883

           x = 0.883 m

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

[tex]f'=f(\frac{v-v_o}{v+v_s})[/tex]    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

[tex]v=\lambda f\\\\[/tex]

v: speed of sound

λ: wavelength = 0.700 m

[tex]f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz[/tex]

Next, you replace the values of all parameters in the equation (1):

[tex]f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz[/tex]

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

[tex]I=\frac{P}{4\pi r^2}[/tex]

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

[tex]I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}[/tex]

hence, the intensity is 8.12*10^⁻4 W/m^2

Answer:

protons

Explanation:

Answer:

D. all of the above

Explanation:

Answer:

Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.

Answer:

look at the data for the control group to see how much the fish grew without the new food.

Explanation:

i got it right

Complete Question

The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate of electric energy consumption in one house is 1.0 kW .

If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.

Answer:

The area is  [tex]A = 1.26 *10^{7} m^2[/tex]

Explanation:

From the question we are told that

      The efficiency is  [tex]\eta =[/tex]12%

      The number of houses is  [tex]N = 350000[/tex]

       The light energy per day is [tex]E = 20 \ MJ[/tex]

       The average rating of electric energy for a house is  [tex]E_h = 1.0 \ k W = 1000W[/tex]

Generally the electric energy which the solar cells covering [tex]1 \ m^2[/tex] produces in a day is

       [tex]E_s = \eta * E[/tex]

           [tex]E_s = 0.12 * 20*10^{6}[/tex]

          [tex]E_s = 2.4 MJ m^{-2}[/tex]

Energy for required by one house for one day is  

        [tex]E_H = E_h * 1 \ day[/tex]  

       [tex]E_H = 1000 * 24 * 3600[/tex]  

        [tex]E_H = 86.4 MJ[/tex]

Energy needed for 350000 house is

      [tex]E_z = 86.4 *10^{6} * 350000[/tex]  

     [tex]E_z = 3.02 *10^{7} MJ[/tex]

The area covered is mathematically represented as

            [tex]A = \frac{3.02*10^{7} \ MJ}{2.4 \ MJ m^{-2}}[/tex]

           [tex]A = 1.26 *10^{7} m^2[/tex]

Answer:

608.4m/s

Explanation:

We are given that

Mass of Sleigh,M=1200 kg

Speed of Sleigh,u=322 m/s

Speed of jet,u'=680 m/s

Mass of jet,m=4800 kg

Total mass=M+m=1200+4800=6000 kg

We have to find the final velocity of the two objects after the collision.

The collision is inelastic .

By using law of conservation of momentum

[tex]Mu+mu'=(m+M)v[/tex]

Using the formula

[tex]1200\times 322+4800\times 680=6000v[/tex]

[tex]6000v=3650400[/tex]

[tex]v=\frac{3650400}{6000}[/tex]

[tex]v=608.4m/s[/tex]

Hence, the final  velocity of two objects after the collision=608.4m/s

Answer:

Mobility of the minority carriers, [tex]\mu_{n} =1184.21 cm^{2} /V-sec[/tex]

Diffusion coefficient for minority carriers,[tex]D_{n} = 29.20 cm^2 /s[/tex]

Verified from Einstein relation as  [tex]\frac{D_{n} }{\mu_{n} } = 25 mV[/tex]

Explanation:

Length of sample, [tex]l_{s} = 2 cm[/tex]

Separation between the two probes, L = 1.8 cm

Drift time, [tex]t_{d} = 0.608 ms[/tex]

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), [tex]\mu_{n} = \frac{V_{d} }{E}[/tex]

Where the drift velocity, [tex]V_{d} = \frac{L}{t_{d} }[/tex]

[tex]V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s[/tex]

and the Electric field strength, [tex]E = \frac{V}{l_{s} }[/tex]

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

[tex]\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec[/tex]

The electron diffusion coefficient, [tex]D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }[/tex]

[tex]\triangle x = (\triangle t )V_{d}[/tex], where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

[tex]\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm[/tex]

[tex]D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s[/tex]

For the Einstein equation to be satisfied, [tex]\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V[/tex]

[tex]\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV[/tex]

Verified.

Answer:

The ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is 0.93

Explanation:

The formula for the refractive index of a medium is given as:

n = c/v

where,

n = refractive index of the medium

c = speed of light in vacuum

v = speed of light in that medium

FOR CARBON TETRACHLORIDE:

n = n₁ = 1.461

v = v₁

Therefore,

1.461 = c/v₁  

v₁ = c/1.461     ----- equation (1)

FOR ETHYL ALCOHOL:

n = n₂ = 1.361

v = v₂

Therefore,

1.361 = c/v₂  

v₂ = c/1.361     ----- equation (2)

Now, dividing equation (1) by equation (2), we get:

v₁/v₂ = (c/1.461) / (c/1.361)

v₁/v₂ = 1.361/1.461

v₁/v₂ = 0.93

Hence, the ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is 0.93

Answer:

(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ  (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.

Note: find an attached copy of a part of the solution to the given question below.

Explanation:

Solution

Recall that:

d = 6.6 um

λ₀ =d/10

λ₀ = 6.6 um

Now,

(a) We find the wavelength λ of the light in water.

Thus,

λ water = (λ₀ )/n

= 0.66/1.33

So,

λ water = λ = 0.496 um

(b) We find the phase difference between the waves from slit 1 and 2

Now,

if a <<d  and a<<λ

Then the path difference between the rays will be

Δ S₂N = Δ d sinθ

Thus, the phase difference becomes,

S = 2π Δ/λ is S= 2π Δ d sinθ/ λ

(c) The next step is to derive an expression for the intensity  I as function of O and other relevant parameters.

Now,

Let p be the point where these two rays interfere with each other.

Thus,

The electric field vector coming out from slot and and slot 2 is

E₁= E₀₁ cos (ks₁ p - wt) i

E₂ = E₀₂ cos (ks₂ p - wt) i

Note: Kindly find an attached copy of a part of the solution to the given question below.

Answer:

The angle of deflection will be "1.07 × 10⁻⁷°".

Explanation:

The given values are:

Mass of a cow,

m = 1100 kg

Mass of bob,

mb = 1 kg

The total distance between a cow and bob will be,

d = 2 m

Let,

The tension be "t".

The angle with the verticles be "[tex]\theta[/tex]".

Now,

Vertically equating forces

⇒  [tex]T\times Cos \theta =mb\times g[/tex] ...(equation 1)

Horizontally equating forces

⇒  [tex]T\times Sin \theta = G\times M\times \frac{mb}{d^2}[/tex] ...(equation 2)

From equation 1 and equation 2, we get

⇒  [tex]tan \thata=\frac{G\times M}{g\times d^2}[/tex]

O putting the estimated values, will be

⇒  [tex]\theta = tan(\frac{6.674\times 10^{-11}\times 1100}{9.8\times 2^2} )[/tex]

⇒  [tex]\theta = 1.07\times 10^{-7}^{\circ}[/tex]

Answer:

A.  Crest

Explanation:

Longitudinal wave is a type of wave that is characterized by the particles of the medium's movement in a parallel direction in comparison to the direction in which wave travels, such that, in compression of longitudinal wave, the density of the wave medium is at its highest due to its closeness together than natural state, while in rarefaction, the density is at its lowest due to wave medium spread apart than normal.

Similarly, in Transverse wave, the crest of a wave implies the medium has reached the highest point while the trough of the wave depicts the lowest point the wave medium has reached.

Therefore, longitudinal wave's compression and rarefaction equates accordingly to the crest and trough of a transverse wave.

Hence, a compression of a longitudinal wave is like a CREST of a transverse wave.

Answer:

I = ΔVA[1 - α (T₀ - T)]/Lρ₀

Explanation:

We have the following data:

ΔV = Battery Terminal Voltage

I = Current through wire

L = Length of wire

A = Cross-sectional area of wire

T = Temperature of wire, when connected across battery

T₀ = Reference temperature

ρ = Resistivity of wire at temperature T

ρ₀ = Resistivity of wire at reference temperature

α = Temperature Coefficient of Resistance

From OHM'S LAW we know that;

ΔV = IR

I = ΔV/R

but,  R = ρL/A   (For Wire)

Therefore,

I = ΔV/(ρL/A)

I = ΔVA/ρL

but,   ρ = ρ₀[1 + α (T₀ - T)]

Therefore,

I = ΔVA/Lρ₀[1 + α (T₀ - T)]

I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹

using Binomial Theorem:

(1 +x)⁻¹ = 1 - x + x² - x³ + ...

In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).

Since, α generally has very low value. Thus, its higher powers can easily be neglected.

Therefore, using this Binomial Approximation, we can write:

[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]

Thus, the equation becomes:

I = ΔVA[1 - α (T₀ - T)]/Lρ₀

Answer:

Explanation:

The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression

ω²R where ω is angular velocity and R is radius of the circular path .

angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.

ω = 2π / T

= 2π / 60

= .1047

angular acceleration =  .1047² x .22

= 2.41 x 10⁻³ rad / s² .

So, the required radial acceleration is [tex]a=0.21 cm/s[/tex]

The rate of change of velocity with respect to time.

Acceleration is a vector quantity.

The formula for the acceleration is,

[tex]a=\frac{V^2}{R}[/tex]

It is given that the radius is 22 cm

Now, substituting the given values into the above formula we get,

[tex]a=\frac{V^2}{22}[/tex]

Here, 1 second is equal to [tex]\frac{2\pi}{60}[/tex] then,

[tex]\\w=\frac{2\pi}{60}\\v=wR\\v=\frac{2\pi}{60}\times 19\\a=(\frac{2\pi}{60}\times 19)^2\times 19\\a=0.21 cm/s[/tex]

Learn more about the topic acceleration:

https://brainly.com/question/11021097

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

Answer:

T =  4200N

Explanation:

When the submersible craft is at rest, the tension in the cable is 6000N.

With this information you can calculate the weight of the craft by summing the forces (the summation of the force is zero because the craft is at rest):

[tex]T-W=0\\\\W=T=6000N[/tex]

When the craft is going down with a constant speed, there is a drag force of 1800N. Then, by using the second Newton law you have:

[tex]T-W+F_d=0[/tex]   (1)

Fd: drag force

The summation of the forces is zero because the craft moves with constant velocity, that is, there is no acceleration.

You calculate the new tension on the cable by solving the equation (1) for T:

[tex]T=W-Fd=6000N-1800N=4200N[/tex]

hence, the tension is 4200N

Answer:

It represents battery

Answer: I think #1 represent lights and #2 represent the battery

Answer:

Explanation:

The speed in meters per second is ...

  (90 km/h)(1000 m/km)(1 h/(3600 s)) = 25 m/s

The braking acceleration can be found from ...

  a = v²/(2d) = (25 m/s)²/(2×50 m) = 6.25 m/s²

Then the braking force is ...

  F = Ma = (1800 kg)(6.25 m/s²) = 11,250 N

The torque on the center of gravity is ...

  T = (11,250 N)(0.90 m) = 10,125 N·m

__

If we let x and y represent the normal forces on the front and rear wheels, respectively, then we have ...

  x + y = (10 m/s²)(1800 kg) = 18000 . . . . . newtons

  1.7x -1.3y = 10,125 . . . . . . . . . . . . . . . . . . . newton-meters

The latter equation balances the torque due to the wheel normal forces with the torque due to braking forces.

Multiplying the first equation by 1.3 and adding that to the second, we have ...

  3.0x = (1.3)(18,000) + 10,125

  x = 33,525/3 = 11,175 . . . . . . . . . . . newtons normal force on front tyres

  y = 18000 -11175 = 6,825 . . . . . . . .newtons normal force on back tyres

The least coefficient of friction is the ratio of horizontal to vertical acceleration, 6.25/10 = 0.625.

Answer:

Protons contribute the most to the size of the atom

Answer:

Protons

Explanation:

Protons contribute more to both the mass and size of an atom. Protons contribute more to an atom's mass while electrons contribute more to its size.

Answer:

The charge in each ball will be 3 * 10^-12 C

Explanation:

(Assuming the correct charge of the second ball is 8 * 10^-12)

When the balls are brought in contact, all the charges are split evenly among then.

So first we need to find the total charge combined:

(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C

Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:

(9 * 10^-12) / 3 = 3 * 10^-12 C

So the charge in each ball will be 3 * 10^-12 C

Answer:

uniformly accelerated motion

Explanation:

The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

[tex]I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)[/tex]   (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

[tex]I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}[/tex]

Hence, the fraction of the maximum intensity is I = 0.636*Imax

Answer:a

Explanation: they are all positive

The equivalent vectors of both the vectors lie in the first quadrant. So, the right choice is a. first quadrant.

The resultant of the vectors can be calculated using the Parallelogram theorem of vector addition. In this theorem, a parallelogram is formed using the vectors, having the vectors as the adjacent side and the diagonal of this parallelogram is the resultant of both the vectors.

The vectors A and B are drawn at the given angle, angles of 20° and 80°  respectively. Using the vectors, A and B, the dotted parallelogram is drawn, and the vector C is the resultant.

Learn more about vectors here:

https://brainly.com/question/25705666

Answer:

a) [tex]a_{T}=8.50m/s^{2}[/tex], [tex]v_{T}=2.78 m/s[/tex] and [tex]N=279.83N[/tex]

b) [tex]a_{T}=0m/s^{2}[/tex], [tex]v_{T}=5.68 m/s[/tex] and [tex]N=795.2N[/tex]

Explanation:

a)

In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:

[tex]h_{0}=2.5m-h_{1}[/tex]

so we get:

[tex]h_{1}=2.5m(sin (20^{o}))[/tex]

[tex]h_{1}=0.855m[/tex]

so

[tex]h_{0}=2.5m-h_{1}[/tex]

[tex]h_{0}=2.5m-0.855m[/tex]

[tex]h_{0}=1.645m[/tex]

once we got this value, we can find the final height the same way. This time the angle is 30° so we get:

[tex]h_{f}=2.5m-h_{2}[/tex]

[tex]h_{f}=2.5m-2.5m(sin 30^{o}))[/tex]

[tex]h_{f}=1.25m[/tex]

Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:

[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]

the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:

[tex]U_{0}=U_{f}+K_{f}[/tex]

so now we substitute with the corresponding formulas:

[tex]mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}[/tex]

if we divided both sides of the equation by the mass, then the equation simplifies to:

[tex]gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}[/tex]

and now we can solve for the final velocity so we get:

[tex]v_{f}=\sqrt{2g(h_{0}-h_{f}}[/tex]

and we can now substitute values:

[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}[/tex]

which solves to:

[tex]v_{f}=2.78m/s[/tex]

which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:

We can build a free body diagram (see attached  picture) and do a balance of forces so we get:

[tex]\sum{F_{x}}=ma_{T}[/tex]

in this case we only have one x-force which is the x-component of the weight, so we get that:

[tex]w_{x}=ma_{T}[/tex]

so we get:

[tex]mgcos\theta=ma_{T}[/tex]

and solve for the tangential acceleration so we get:

[tex]a_{T}=gcos\theta[/tex]

[tex]a_{T}=9.81m/s^{2}*cos(30^{o})[/tex]

[tex]a_{t}=8.50m/s^{2}[/tex]

Now we can find the centripetal acceleration by using the formula:

[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]

[tex]a_{c}=\frac{(2.78m/s)^{2}}{2.5m}[/tex]

so we get:

[tex]a_{c}=3.09m/s^{2}[/tex]

Next, we can find the normal force which is found by doing a sum of forces on y, so we get:

[tex]\sum{F_{y}}=ma_{c}[/tex]

so we get:

[tex]N-w_{y}=ma_{c}[/tex]

and we solve for the normal force so we get:

[tex]N=ma_{c}+w_{y}[/tex]

and substitute:

[tex]N=ma_{c}+mg sin\theta[/tex]

when factoring we get:

[tex]N=m(a_{c}+g sin\theta)[/tex]

and we substitute:

[tex]N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]

which yields:

N=279.83N

b)

The procedure for part b is mostly the same with some differences due to the angle. First:

[tex]h_{0}=1.645m[/tex]

[tex]h_{f}=0m[/tex]

so

[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]

in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:

[tex]U_{0}=K_{f}[/tex]

so we get:

[tex]mgh_{0}=\frac{1}{2}mv_{f}^{2}[/tex]

so we solve for the final velocity so we get:

[tex]v_{f}=\sqrt{2gh_{0}}[/tex]

and we substitute:

[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}[/tex]

[tex]v_{f}=5.68m/s[/tex]

according to the free body diagram we get that:

[tex]a_{T}=gcos\theta[/tex]

[tex]a_{T}=9.81m/s^{2}(cos 90^{o})[/tex]

which yields:

[tex]a_{T}=0[/tex]

we can also find the centripetal acceleration, so we get:

[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]

[tex]a_{c}=\frac{(5.68m/s)^{2}}{2.5m}[/tex]

so we get:

[tex]a_{c}=12.91m/s^{2}[/tex]

and we can do a sum of forces on y to find the normal force:

[tex]\sum{F_{y}}=ma_{c}[/tex]

so we get:

[tex]N-w_{y}=ma_{c}[/tex]

and we solve for the normal force so we get:

[tex]N=ma_{c}+w_{y}[/tex]

and substitute:

[tex]N=ma_{c}+mg [/tex]

when factoring we get:

[tex]N=m(a_{c}+g)[/tex]

and we substitute:

[tex]N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]

which yields:

N=795.2N

Following are the calculation to the angles:

For angle 30°:

consider the forces in tangential in direction:  

[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 30^{\circ} \\\\a_t = 8.496 \ \frac{m}{s^2}\\\\[/tex]

calculate the speed of the child:

[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]  

Integrating the equation:

[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{30} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 30-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 30-\sin 20) = \frac{V^2}{2}\\\\v=2.79 \ \frac{m}{s}\\\\[/tex]

consider the forces in the normal declaration:  

[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 30) = \frac{35 \times 2.78^2}{2.5}\\\\N= 279.87\ N[/tex]

For angle 90°:

consider the forces in tangential in direction:  

[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 90^{\circ} \\\\a_t = 0\ \frac{m}{s^2}\\\\[/tex]

calculate the speed of the child:  

[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]  

Integrating the equation:

[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{90} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 90-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 90-\sin 20) = \frac{V^2}{2}\\\\V=\sqrt{\frac{9.81 \times 2.5 (\sin 90-\sin 20) }{2}}[/tex]

   [tex]=\sqrt{\frac{9.81 \times 2.5 (1 -0.342)}{2}}\\\\=\sqrt{8.06}\\\\=2.83[/tex]

consider the forces in the normal declaration:  

[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 90) = \frac{35 \times 2.78^2}{2.5}\\\\N= \frac{35 \times 2.78^2}{2.5 \times 35\times 9.81\times 1}\\[/tex]

    [tex]= \frac{270.494}{858.375}\\\\=0.315[/tex]

Learn more:

brainly.com/question/13077544

Answer:

(a)   ΔФ = -0.109W

(b)  emf = 28.43V

(c)   Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

[tex]\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)[/tex]   (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

[tex]\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W[/tex]

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

[tex]emf=-N\frac{\Delta \Phi}{\Delta t}[/tex]         (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

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

[tex]emf=-(60)(\frac{-0.109W}{0.230s})=28.43V[/tex]

The induced emf is 28.43V

(c) The induced current in the coil is given by:

[tex]I_{in}=\frac{emf}{R}[/tex]      (3)

R: resistance of the coil     (it is necessary to have this value)

emf :induced emf  = 28.43V

Answer:

A. mass

Explanation:

Mass determines the quantity of inertia for an object. Mass is the quantity that depends upon the inertia of an object. The inertia that an object has is directly proportional to the mass of the object.

An object that has more mass has a greater tendency as compared to the object that has less mass to resist changes in its state of motion.

Answer:

The current is  [tex]I = 0.4595 \ A[/tex]

Explanation:

From the question we are told that

    The length of the wire is  [tex]L = 64.7 \ cm = 0.647 \ m[/tex]

    The magnetic field is  [tex]B = 0.370 \ T[/tex]

      The magnetic force is  [tex]F = 0.110 \ N[/tex]

Generally the magnetic force exerted by the field is mathematically represented as

      [tex]F = IL B sin \theta[/tex]

Making I the subject we have

      [tex]I = \frac{F}{LB \ sin\theta}[/tex]

The angle here is  90° since the wire is perpendicular  to the direction of the magnetic field

      Substituting values

     [tex]I = \frac{0.110}{ (0.370) * 0.647 \ sin(90)}[/tex]

   [tex]I = 0.4595 \ A[/tex]

Explanation:

45 m/s × (3.28 ft/m) × (1 mi / 5280 ft) × (3600 s/h) = 101 mph

This is actually 3 significant figures.  To write in 2 significant figures, you must use scientific notation:

1.0×10² mph