A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

A) The speed of the scooter at which the driver will feel weightlessness is;

v = 68.586 m/s

B) The apparent weight of both the driver and the scooter at the top of the hill is;

F_net = 779.1 N

We are given;

Mass; m = 159 kg

Radius; r = 480 m

A) Since it's motion about a circular hill, it means we are dealing with centripetal force.

Formula for centripetal force is given as;

F = mv²/r

Now, we want to find the speed of the scooter if the driver feels weightlessness.

This means that the centripetal force would be equal to the gravitational force.

Thus;

mg = mv²/r

m will cancel out to give;

v²/r = g

v² = gr

v = √(gr)

v = √(9.8 × 480)

v = √4704

v = 68.586 m/s

B) Now, he is travelling with speed of;

v = 68.586 m/s

And the radius is 2r

Let's first find the centripetal acceleration from the formula; α = v²/r

Thus; α = 4704/(2 × 480)

α = 4.9 m/s²

Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;

F_app = m(g - α)

F_net = 159(9.8 - 4.9)

F_net = 779.1 N

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Answer:

F = F₀ 0.2

Explanation:

For this exercise we apply Coulomb's law with the initial data

     F₀ = k q_A q_B / d²

indicate several changes

q_A ’= ½ q_A

q_B ’= 1/10 q_B

d ’= ½ d

let's substitute these new values ​​in the Coulomb equation

          F = k q_A ’q_B’ / d’²

          F = k ½ q_A 1/10 q_B / (1/2 d)²

          F = (k q_A q_B / d2) ½ 1/10 2²

          F = F₀ 0.2

Answer:

Torque = 8.38Nm

Explanation:

Time= 8.00s

angular speed (w) =400 rpm

Moment of inertia (I)= 1.60kg.m2 about its rotation axis

We need to convert the angular speed from rpm to rad/ sec for consistency

2PI/60*n = 0.1047*409 = 41.8876 rad/sec

What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?

Then we need to use the formula below for our torque calculation

from basic equation T = J*dω/dt ...we get

Where : t= time in seconds

W= angular velocity

T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm

Therefore, constant torque that is required is 8.38 Nm

Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.

Given-

Inertia of the flywheel is 1.60 kg m squared.

Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,

[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]

[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]

[tex]\omega = 41.89[/tex]

Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.

When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,

[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]

[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]

[tex]\tau=8.38[/tex]

Hence, the required constant torque is 8.38 N-m.

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Answer:

a) h = 13,205.4 m

b)  r_f = 2.12 106 m

c)        e% = 0.68%

Explanation:

a) This is an exercise we are asked to use energy conservation,

Starting point. On the surface of Mimas

        Em₀ = K = ½ m v²

Final point. Where the ball stops

       [tex]Em_{f}[/tex] = U = m g h

        Em₀ = Em_{f}

        ½ m v² = m g h

         h = ½ v² / g

let's calculate

         h = ½ 41² / 0.0636

         h = 13,205.4 m

b) For this part we are asked to use the law of universal gravitation, write the energy

starting point. Satellite surface

           Em₀ = K + U = ½ m v² - GmM / r_o

final point. Where the ball stops

            [tex]Em_{f}[/tex]= U = - G mM / r_f

          Em₀ = Em_{f}

          ½ m v² - G m M / r_o = - G mM / r_f

In this case all distances are measured from the center of the satellite

         1 / rf = 1 / GM (-½ v² + G M / r_o)

let's calculate

         1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)

         1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)

          1 / r_f = 4,714 10⁻⁷

          r_f = 1 / 4,715 10⁻⁷

          r_f = 2.12 106 m

to measure this distance from the satellite surface

          r_f ’= r_f - r_o

          r_f ’= 2.12 106 - 1.98 105

         r_f ’= 1,922 106 m

c) the percentage difference is

          e% = 13 205.4 / 1,922 106 100

          e% = 0.68%

The estimate of part a is a little low

Answer:

(a) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a          --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r          [v = linear velocity of particle, r = radius of circular path]

Therefore, equation (i) becomes;

F = m v²/ r             --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ          -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

Combine equations (ii) and (iii) as follows;

m (v² / r) = qvBsinθ         [divide both side by v]

m v / r = qBsinθ              [make r subject of the formula]

r = (m v) / (qBsinθ)              ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v          --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

Answer:

W = 1413.75 J

Explanation:

It is given that,

Mass of car, m = 650 kg

Initial speed of the car, u = 0.7 m/s

Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s

Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.

[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]

So, the work done by the car is 1413.75 J.

Answer:

The intensity is [tex]I = 500 mW/m^2[/tex]

Explanation:

From the question we are told that

    The  intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]

Generally the intensity of the light emerging from the polarizer is  mathematically represented as

          [tex]I = \frac{I_o}{2}[/tex]

substituting values

         [tex]I = \frac{1000 *10^{-3}}{2}[/tex]

         [tex]I = 500 *10^{-3} W/m^2[/tex]

         [tex]I = 500 mW/m^2[/tex]

Answer:

32

Explanation:

Answer:

therefore horizontal displacement changes increasing with linear velocity

Explanation:

Since the plane flies horizontally, the only speed that exists is

              v₀ₓ = 55.0 m / s

the time is the time it takes to reach the floor, which we can find because the speed on the vertical axis is zero

               y =y₀ + v₀ t - ½ g t2

               0 = I₀ + 0 - ½ g t2

               t = √ 2y₀o / g

time is that we use to calculate the x-axis displacement

 The distance it travels to reach the floor is

              x = v t

              x = 55 12

              x = 660 m

When the speed horizontally the time remains the same and 120

             x ’= v’ 12

therefore horizontal displacement changes increasing with linear velocity

Answer:

Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.

In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

Answer:

(b) Increases

Explanation:

The potential energy between two point charges is given as;

[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]

Where;

k is the coulomb's constant

q₁  ans q₂ are the two point charges

r is the distance between the two point charges

Since the two charges have opposite sign;

let q₁ be negative and q₂ be positive

Substitute in these charges we will have

[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]

The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.

Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.

Answer:

4°C

Explanation:

Water is densest at 4°C.  Since dense water sinks, the bottom of the lake will be 4°C.

Answer:

The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.

Explanation:

Answer:

The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.

Explanation:

Edmentum sample answer

Answer:

The frequency is    [tex]f = 0.221 \ Hz[/tex]

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]

Let the voltage of the capacitor when it is fully charged be  [tex]V_o[/tex]

Then the voltage of the capacitor at time t is  said to be  [tex]V = \frac{V_o}{2}[/tex]

   Now  this voltage can be  mathematical represented as

      [tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]

Where  RC  is the time constant

   substituting values  

    [tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]- \frac{0.5}{RC} = ln (0.5)[/tex]

     [tex]-\frac{0.5}{RC} = -0.6931[/tex]

     [tex]RC = 0.721[/tex]

Generally the cross-over frequency for a low pass filter is mathematically represented as

          [tex]f = \frac{1}{2 \pi * RC }[/tex]

substituting values  

           [tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]

           [tex]f = 0.221 \ Hz[/tex]

Answer:

Explanation:

We know that the equivalent resistance of resistors connected in parallel is expressed as

[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]

the L.C.M is 6R

[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]

Answer:

The new acceleration would be 9 m/s².

Explanation:

Acceleration of an object is 6 m/s²

Net force is equal to the product of mass and acceleration i.e.

F = ma

[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]

If the net force was tripled and the mass were doubled, it means,

F' = 3F

m' = 2m

Let a' is new acceleration. So,

[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]

So, the new acceleration would be 9 m/s².

✴ Case - I

⟶ Force = F

⟶ Mass = m

⟶ Acceleration = 12m/s²

✴ Case - II

⟶ Force = 2F

⟶ Mass = 3m

➳ Acceleration in second case.

⇒ This question is completely based on the concept of newton's second law of motion.

⇒ As per this law, Force is defined as the product of mass and acceleration.

Mathematically, F = ma

[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]

New acceleration would be 12 m/s²

Given that;

Acceleration of object = 12 m/s²

New net force = 2f

New mass = 3m

Find:

New acceleration

Computation:

[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]

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Answer:

Explanation:

a )

current in the wire = potential diff / resistance

= 23 / (15 x 10⁻³ )

= 1.533 x 10³ A .

b )

For resistance of a wire , the formula is

R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area of wire

putting the given values

15 x 10⁻³ = 4ρ / π x .003²

ρ = 106  x 10⁻⁹ ohm. m

= 10.6 x 10⁻⁸ ohm m

The metal wire appears to be platinum .

(a) The current in the wire is 1.533 x 10³ A

(b) The resistivity of the wire material is 10.6 x 10⁻⁸ Ωm

(c) The material of the wire is Platinum

(a) According to the Ohm's Law:

V = IR

where V is the potential difference

I is the current

and R is the resistance

So,

I = V/R

I = 23 / (15 x 10⁻³ )

I = 1.533 x 10³ A

(b) The resistance of a wire is expressed as:

R = ρ L / A

where ρ is the resistivity,

L is length

and A is the cross-sectional area

15 x 10⁻³ = 4ρ / π x .003²

ρ = 106  x 10⁻⁹ Ωm

ρ = 10.6 x 10⁻⁸ Ωm

The metal from which the wire is made is platinum.

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Answer:

a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

b) Your velocity after collision is 2.64 m/s

c) The force you felt is 7392 N

d) you and your brother undergo an equal amount of acceleration

Explanation:

Your mass [tex]m_{y}[/tex] = 60 kg

your brother's mass [tex]m_{b}[/tex] = 30 kg

mass of the car [tex]m_{c}[/tex] = 80 kg

your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)

your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s

your final speed [tex]v_{y}[/tex] after collision = ?

your brother's final speed [tex]v_{b}[/tex] after collision = ?

a) equations you need to use to figure out how fast you and your brother are moving after the collision is

[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

but [tex]u_{y}[/tex] = 0 m/s

the equation reduces to

[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

b) if your little brother reverses with velocity of 0.36 m/s it means

[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)

then, imputing values into the equation, we'll have

[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)

330 = 140[tex]v_{y}[/tex] - 39.6

369.6 = 140[tex]v_{y}[/tex]

[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s

This means you will also reverse with a velocity of 2.64 m/s

c) your initial momentum = 0  since you started from rest

your final momentum = (total mass) x (final velocity)

==>  (60 + 80) x 2.64 = 369.6 kg-m/s

If the collision lasted for 0.05 s,

then force exerted on you = (change in momentum) ÷ (time collision lasted)

force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N

d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s

your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2

your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s

his deceleration is (3 - 0.36)/0.05 = 52.8 m/s

you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost

Answer:

T = 2689.6N

Explanation:

Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree

while both weight are acting perpendicular to the length. Hence we have :

Torque ( clockwise) = Torque ( anticlockwise)

m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1

Where m1 = 36kg

m2 = 20kg

g = 9.81m/s^2

Theta = 12

Substituting into equation 1

36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)

353.16L/2+196.2L = T ×0.2079(2L/3)

176.58L+196.2L = T × 0.1386L

372.78L = 0.1386LT

T = 372.78L/0.1386L

T = 2689.6N

Answer:

Amplitude, A = 5 m

Explanation:

Let a progressive wave is given by equation :

[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)

The general equation of a progressive wave is given by :

[tex]y=A\sin (\omega t-kx)[/tex] ....(2)

Here,

A is the amplitude of the wave

[tex]\omega[/tex] is the angular frequency

k is propagation constant

We need to find the amplitude of the wave.

If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.

Answer:

The  heat loss is  [tex]H = 8400\ W[/tex]

Explanation:

From the question we are told that

   The thickness is  [tex]t = 10 \ mm = 0.01 \ m[/tex]

    The inner temperature is  [tex]T_i = 25 ^oC[/tex]

    The outer temperature is [tex]T_o = 5 ^oC[/tex]

    The length of the window is  L  = 1 m  

    The  width of the window is  w  =  3 m  

Generally the heat loss is mathematically represented as

      [tex]H = \frac{k * A * \Delta T}{t}[/tex]

Where  k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]

   and A  is the area of the window with value

           [tex]A = 1 * 3[/tex]

            [tex]A = 3 \ m^2[/tex]

substituting values

       [tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]

       [tex]H = 8400\ W[/tex]

Answer:

Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.

If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.

Answer:

450°C

Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .

η = 1 - T2 / T1

η = 1/6 initially

when T2 is reduced by 65°C then η becomes 1/3

Solution

η = 1/6

1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)

When η = 1/3 :

η = 1 - ( T2 - 75 ) / T1

1/3 = 1 - (T2 - 75)/T1.........................(2)

T2 - T1 = -75 [ because T2 is reduced by 75°C ]

T2 = T1 - 75...........................(3)

Put this in (2) :

> 1/3 = 1 - ( T1 - 75 - 75 ) / T1

> 1/3 = 1 - (T1 - 150 ) /T1

> (T1 - 150) / T1 = 1 - 1/3

> ( T1 -150 ) / T1 = 2/3

> 3 ( T1 - 150 ) = 2 T1

> 3 T1 - 450 = 2 T1

Collecting the like terms

3 T1- 2 T1 = 450

T1 = 450

The temperature initially was 450°C

Explanation:

1.  Impulse = change in momentum

J = Δp

J = mΔv

In the x direction:

Jₓ = mΔvₓ

Jₓ = (0.40 kg) (30 m/s cos 45° − (-20 m/s))

Jₓ = 16.5 kg m/s

In the y direction:

Jᵧ = mΔvᵧ

Jᵧ = (0.40 kg) (30 m/s sin 45° − 0 m/s)

Jᵧ = 8.49 kg m/s

The magnitude of the impulse is:

J = √(Jₓ² + Jᵧ²)

J = 18.5 kg m/s

The average force is:

FΔt = J

F = J/Δt

F = 1850 N

2. Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

In the x direction:

(1000 kg) (0 m/s) + (1500 kg) (-12 m/s) = (1000 kg + 1500 kg) vₓ

vₓ = -7.2 m/s

In the y direction:

(1000 kg) (20 m/s) + (1500 kg) (0 m/s) = (1000 kg + 1500 kg) vᵧ

vᵧ = 8 m/s

The magnitude of the final speed is:

v = √(vₓ² + vᵧ²)

v = 10.8 m/s

3. Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

In the x direction:

(0.8 kg) (18 m/s cos 45°) + (0.36 kg) (9.0 m/s) = (0.8 kg + 0.36 kg) vₓ

vₓ = 11.6 m/s

In the y direction:

(0.8 kg) (-18 m/s sin 45°) + (0.36 kg) (0 m/s) = (0.8 kg + 0.36 kg) vᵧ

vᵧ = -8.78 m/s

The magnitude of the final speed is:

v = √(vₓ² + vᵧ²)

v = 14.5 m/s

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

[tex] W = F*d [/tex]

Where:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

[tex]F = W_{x} = mgsin(\theta)[/tex]

Where:

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]    

Now, we can find the work:

[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

Answer:

The voltage induced in the coil  is 1.25 V.

Explanation:

Given;

number of turns, N = 100 turns

cross sectional area of the copper coil, A = 0.1 m²

initial magnetic field, B₁ = 0.5 T

final magnetic field, B₂ = 1.00 T

duration of change in magnetic field, dt = 4 s

The induced emf in the coil is calculated as;

[tex]emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts[/tex]

Therefore, the voltage induced in the coil  is 1.25 V.