When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time

Answer:

the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J

Explanation:

mass of block A = 2 kg

mass of block B = 4 kg

distance the blocks were pushed = 100 m

NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.

Work done in moving the 2 kg mass along the 100 m distance is,

work = force x distance moved

force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)

force = 19.62 N

therefore,

work done = 19.62 x 100 = 1962 Joules of work.

According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.

Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J

Answer:

130.528779365 degrees

Explanation:

The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.

n1/n2 = sin(theta2)/sin(theta1)

let theta1 be 30 degrees, and n1 be the refractive index of air = 1

1/1.5 = sin(theta2)/sin(30deg)

solve:

sin(theta2) = 2/3 sin(30deg) = 1/3

theta2 = arcsin (1/3) = 19.4712206345 degrees

The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.

Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.

Angle between = 180-30-19.4712206345 = 130.528779365 degrees

The angle between the reflected and transmitted rays 130.5287 degrees

The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.

[tex]\dfrac{n_1}{n_2} = \dfrac{sin(\theta_2)}{sin(\theta_1)}[/tex]

let [tex]\theta_1[/tex] be 30 degrees, and n1 be the refractive index of air = 1

[tex]\dfrac{1}{1.5} = \dfrac{sin(\theta_2)}{sin(30)}[/tex]

solve:

[tex]sin(\theta_2) = \dfrac{2}{3} sin(30) = \dfrac{1}{3}[/tex]

[tex]\theta_2 = sin ^{-1}\dfrac{1}{3} = 19.4712 \ degrees[/tex]

The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.

Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.

Angle between = 180-30-19.4712206345 = 130.528779365 degrees

Hence the angle between the reflected and transmitted rays 130.5287 degrees

To know more about the Refraction of light follow

https://brainly.com/question/10729741

Explanation:

The relationship between electric force and distance between charged objects is given by the formula as follows :

[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]

k is electrostatic constant and d is distance between charges

The electric force between charges is inversely proportional to the square of distance between them.

Answer:

A) Fb = 671.3 N

B) The diver will sink.

Explanation:

A)

The buoyant force applied on an object by a fluid is given by the following formula:

Fb = Vρg

where,

Fb = Buoyant Force = ?

V = Volume of the water displaced by the object = 68.5 L = 0.0685 m³

ρ = Density of Water = 1000 kg/m³

g = 9.8 m/s²

Therefore,

Fb = (0.0685 m³)(1000 kg/m³)(9.8 m/s²)

Fb = 671.3 N

B)

Now, in order to find out whether the diver sinks or float, we need to find weight of the diver with gear.

W = mg = (71.8 kg)(9.8 m/s²)

W = 703.64 N

Since, W > Fb. Therefore, the downward force of weight will make the diver sink.

The diver will sink.

Answer:

The angular momentum is  [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Explanation:

From the question we are told that

    The mass of the woman is  [tex]m = 50 \ kg[/tex]

     The angular  speed of the rim is  [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]

      The mass of the disk is  [tex]m_d = 110 \ kg[/tex]

       The radius of the disk is [tex]r_d = 4.0 \ m[/tex]

The moment of inertia of the disk is mathematically represented as

        [tex]I_D = \frac{1}{2} m_d r^2_d[/tex]

substituting values

          [tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]

          [tex]I_D = 880 \ kg \cdot m^2[/tex]

The moment of inertia of the woman is  

          [tex]I_w = m * r_d^2[/tex]

substituting values

        [tex]I_w = 50 * 4^2[/tex]

       [tex]I_w =800\ kg[/tex]

The moment of inertia of the system (the woman + the large disk ) is  

        [tex]I_t = I_w + I_D[/tex]

substituting values  

      [tex]I_t = 880 +800[/tex]

     [tex]I_t =1680 \ kg \cdot m^2[/tex]

The angular momentum of the system is

      [tex]L = I_t w[/tex]

substituting values  

      [tex]L = 1680 * 5.024[/tex]

      [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .

Answer:

10 metres

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this particular problem or Question efficiently.

=> The speed of the space ship can be found from this relationship: ✓(1 - [v^2/c^2] ) = 1/2.

=> The length of the space ship = 20 m.

=> Assumption = '' If the length of the ship is measured by the inertial earth-bound observer".

Thus, from the speed of the space ship can be found from this relationship we can determine the value;

✓(1 - [v^2/c^2] ) = 1/2.

V = 20 × 1/2 = 10 metres.

Note that we use the contraction formula to solve for V.

Answer:

it represents a fundamental difference. (more info below)

Explanation:

Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.

hope this helped!

Complete question is;

A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

Answer:

The second dart leaves the gun two times faster than the first one.

Explanation:

If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

Potential energy = kinetic energy

Thus;

½kx² = ½mv²

Making velocity "v" the subject, we have;

v = √(kx²/m)

Since the initial distance is "x", thus initial launching velocity is;

v1 = √(kx²/m)

Since next distance is 2x, thus, second launch velocity is;

v2 = √(k(2x)²/m)

Expanding, we have;

v2 = √(4kx²/m)

v2 = 2√(kx²/m)

Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

So, v2 = 2v1

Hence, The second dart leaves the gun two times faster than the first one.

Answer:

a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]

Where:

[tex]m_{w}[/tex] - Mass of water, measured in kilograms.

[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].

[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].

Then,

[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]

[tex]Q_{needed} = 180748800\,J[/tex]

The energy needed in kilowatt-hours is:

[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]

[tex]Q_{needed} = 50.208\,kWh[/tex]

The electric energy required to heat up the water is:

[tex]E = \frac{50.208\,kWh}{0.75}[/tex]

[tex]E = 66.944\,kWh[/tex]

Lastly, the cost of heating a hot tub is: (USD - US dollars)

[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]

[tex]E = 6.024\,USD[/tex]

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]

[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]

[tex]i = 88.200\,A[/tex]

Answer:

A) 4D

Explanation:

The distance traveled by the cars before coming to rest can be determined by 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,

s = distance traveled

Vf = Final Speed = 0 m/s

Vi = Initial Speed

a = deceleration rate

First, we consider Car B and we assign a subscript 2 for it:

Vf₂ = 0 m/s  (As, car finally stops)

s₂ = D

a₂ = - a  (due to deceleration)

D = (0² - Vi₂²) /(-2a)

D = Vi₂²/2a    -------- equation (1)

Now, we consider Car A and we assign a subscript 1 for it:

Vf₁ = 0 m/s  (As, car finally stops)

s₁ = ?

a₁ = - a  (due to deceleration)

Vi₁ = 2 Vi₂  (Since, car A was initially traveling at twice speed of car B)

s₁ = (0² - Vi₁²) /(-2a)

s₁ = (2Vi₂)²/2a

s₁ = 4 (Vi₂²/2a)

using equation (1), we get:

s₁ = 4D

Therefore, the correct option is:

A) 4D

Answer:

11.8 Joules

Explanation:

Given:-

- The height of the target ball, si = 0.860 m

- The mass of target and steel ball, m = 0.012 kg

- The target ball travels a distance ( x ) after being struck = 1.40 m

Find:-

What is the kinetic energy (in joules) of the target ball just after it is struck?

Solution:-

- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.

- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:

                        [tex]vf^2 = vi^2 - 2*g*x[/tex]

Where,

                  vf: The final velocity of target ball at maximum height = 0

                  g: The gravitational acceleration constant = 9.8 m/s^2

- Plug in the required parameters and evaluate the ( vi ) as follows:

                      [tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]

- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:

                       [tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]

Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.

Answer:

C

Explanation:

The intermolecular forces between the water molecule is less binding than that of the copper molecule. Hence the water would take a shorter time to be converted to vapour where the temperature of boiling is constant however the temperature of that of the copper molecule keeps increasing.

Answer:

[tex]T_{1}[/tex] = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about [tex]T_{1}[/tex] we have:

(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0

⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)

[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

[tex]T_{2}[/tex]= 59.535 ÷ 2.4

[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23

[tex]T_{1}[/tex] = 14.88 N

Answer:

Increased blood circulation to the body.

Explanation:

plato/edmentum

Answer

The rate at which the magnetic field is changing is  [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]

Explanation

From the question we are told that

   The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]

    The radius is  [tex]r = 1.5 \ m[/tex]

The rate of change of the  magnetic  field  is mathematically represented as

        [tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]

Where [tex]dl[/tex] is change of a unit length

     [tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]

Where A is the area which is mathematically represented as

     [tex]A = \pi r^2[/tex]

    So

    [tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

  [tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

where L is the circumference of the circle which is mathematically represented as

     [tex]L = 2 \pi r[/tex]

So

     [tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]

      [tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]

       [tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]

substituting values

      [tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]

      [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]    

Answer:

Explanation:

The kinetic energy possessed by the block when sliding will be equal to the energy needed to compress the string.

Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²

m = mass of the block (in kg) = 2.8 kg

v = speed of the block (in m/s) = 2.4 m/s

k = force constant of the spring

e = extension (in metres) = 0.056m

Since KE = energy stored in the spring

1/2 mv² = 1/2 ke²

mv² = ke²

2.8(2.4)²  = k(0.056)²

16.128 = 0.003136k

k = 16.128/0.003136

k =  5,142.86

The force constant of the spring is 5,236.36

The force that constant of the spring is 5,142.86.

The kinetic energy that should be possessed by the block at the time when sliding will be equivalent to the energy required to compress the string.

Here

Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²

m = mass of the block (in kg) = 2.8 kg

v = speed of the block (in m/s) = 2.4 m/s

k = force constant of the spring

e = extension (in metres) = 0.056m

Since KE = energy stored in the spring

So,

1/2 mv² = 1/2 ke²

mv² = ke²

Now

2.8(2.4)²  = k(0.056)²

16.128 = 0.003136k

k = 16.128/0.003136

k =  5,142.86

Learn more about force here: https://brainly.com/question/3398162

Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.

While outside the shell, the electric field is given by: k(q + Q)/r²

Where;

K= is a constant which is given as, 8.99 x 10^9 N m² / C².

Q= source charge which creates the electric field

q= is the test charge which is used to measure the strength of the electric field at a given location.

r= is the radius

Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.

Answer:

Explanation:

12.)

A. Opposite poles attract

B. Same poles repel

13.)

IDK

Answer:

(a) 0.0885 Hz

(b) 0.21 m/s

Explanation:

(a) Frequency: This can  be defined as the number of cycle completed in one seconds.

From the question,

Note: 2 crest = one cycle,

If four crest = 22.6 s,

Then two crest = (22.6/2) s

= 11.3 s.

T = 11.3 s

But,

F = 1/T

F = 1/11.3

F = 0.0885 Hz.

(b)

Using,

V = λF...................... Equation 1

Where V = speed of wave, F = Frequency of wave, λ = wave length.

Given: F = 0.0885 Hz, λ = 2.37 m.

Substitute these values into equation 1

V = 2.37(0.0885)

V = 0.21 m/s.

Answer:

Explanation:

Let the point source have power P .

At distance r , the intensity I

I = P / 4πr² . If intensity at 6 m and 2 m be I₁ and I₂

I₁ = P / 4π x 6²

I₂ =  P / 4π x 2²

I₁ / I₂ = 2² / 6²

= 1 / 9

I₂ = 9 I₁

Intensity will be 9 times that at 6 m .

Complete Question

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .

Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm

Part B

If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?

Answer:

Part A  

    [tex]R = 1081 \ J[/tex]

Part B  

     [tex]h = 169.7 \ m[/tex]

Explanation:

From the question we are told that

  The mass of the blade is  [tex]m_b = 0.65 \ kg[/tex]

   The length is  [tex]l = 0.55 \ m[/tex]

   The angular speed is  [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]

Generally the moment of inertia of the of this mower is mathematically evaluated as

         [tex]I = \frac{m_b * l^2 }{12}[/tex]

substituting values

         [tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]

         [tex]I = 0.016 \ kg m^2[/tex]

Generally the rotational kinetic energy of the bland is  

        [tex]R = \frac{1}{2} * I * w^2[/tex]

substituting values

       [tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]

     [tex]R = 1081 \ J[/tex]

At point where the gravitational potential energy is equal to the rotational kinetic energy  we have that

       [tex]P = R = m_b * h * g[/tex]

Where P is the  gravitational potential energy

substituting values

          [tex]1081 = 0.65 * 9.8 * h[/tex]

=>       [tex]h = 169.7 \ m[/tex]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex]     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]

B. The energy of the emitted photon is given by the following formula:

[tex]E=h\frac{c}{\lambda}[/tex]   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

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

λ: wavelength of the photon

You first convert the energy from eV to J:

[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]

Next, you use the equation (2) and solve for λ:

[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]

C. The radius of the orbit is given by:

[tex]r_n=n^2a_o[/tex]   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

[tex]r_5=5^2(2.380)=59.5[/tex]

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

A) The energy E of the emitted photon in electron volts is; E = 2.856 eV

B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm

C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm

A) Formula for the energy E of the emitted photons is;

E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])

We are given;

n₂ = 5

n₁ = 2

Thus;

E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])

E = 2.856 eV

B) The formula for the wavelength is;

λ = hc/E

where;

h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s

c is speed of light = 3 × 10⁸ m/s

E is energy of photon

λ is wavelength of the photon

Earlier we saw that E = 2.856 eV. Converting to Joules gives;

E = 4.5758 × 10⁻¹⁹ J

Thus;

λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)

λ = 4.344 × 10⁻⁷ m

Converting to nm gives;

λ = 434.4nm

C) Formula for the radius of the hydrogen atom is;

rₙ = n²a₀

where;

a₀ is bohr's radius = 5.292 × 10⁻¹¹ m

n = 5

Thus;

rₙ = 5² × 5.292 × 10⁻¹¹

rₙ = 1.323 × 10⁻⁹

rₙ = 1.323 nm

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Answer:

Loudness: decreases

Amplitude: decreases

Pitch: stays the same

Frequency: stays the same

Explanation:

1.

An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.

2.

Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.

3 and 4.

The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.

Hope this helps!

Before the bus starts moving, the bus and the skater are both standing still.

When the bus starts moving and pulls away from the bus-stop, the skater stays right where she is.  

The people outside on the sidewalk see her standing still, and they see the bus moving out from under her.  

The other passengers on the bus see her rolling backwards down the aisle, toward the back of the bus.

Answer:

a.  4 V

b. 0.697 A

Explanation:

Magnetic field strength B =  0.732 T

length of rod l = 0.362 m

velocity of rod v = 15.1 m/s

a.  EMF can be calculated as

E = Blv = 0.732 x 0.362 x 15.1 = 4 V

b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω

current I = V/R = 4/5.74 = 0.697 A

the current flows in a clockwise direction

Answer:

Explanation:

Sensorimotor Infants "think" by acting on the world with their eyes, ears, hands, and mouth.

Preoperational. Development of language and make-believe play takes place.

Concrete Operational children think in a logical, organized fashion only when dealing with concrete information they can perceive directly.

Formal Operational.  Adolescences  can also evaluate the logic of verbal statements without referring to real-world circumstances.

Sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.

The way youngsters think, investigate, and figure things out is referred to as cognitive development.

Piaget defined four stages of cognitive development:

1. Sensorimotor. From birth through the age of 18-24 months.

2. Preoperational.From infancy (18-24 months) until toddlerhood (age 7)

3. Operational concrete. 7 to 11 years old

4. Formal operational. From adolescence to adulthood

Hence, sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.

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Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

[tex]U=(M+m)gh[/tex]       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]

The potential energy is:

[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

[tex]Mv_1+mv_2=(M+m)v[/tex]    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

[tex]M(0)+mv_2=(M+m)v[/tex]    

[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]    

The force on the cord after the impact is 2.59N

Answer:

Explanation:

Apply the law of conversion of energy

[tex]V_f = \sqrt{2gh}[/tex]

where,

h = height of which the bullet and block rise after impact

[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]

Therefore,

[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]

From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]

[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]

C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system

[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]

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Answer:

B. At the instant when the block is at the highest point, directed at the block.

Explanation:

Motion of an object is the change in the position of the object with respect to time. On the earth, gravity has a great influence on the motion of an object (especially in a vertical direction).

When the block is projected up in the air, it moves with a varying velocity until the velocity becomes zero due to gravity. Which make the object to rest a little in the air (when velocity = gravity) and starts to fall freely.

To ensure hitting the block by the ball, it is thrown at the block when the block is at its highest point in the air. Since the block would be at rest at this instant before it start to fall at a constant acceleration under gravity.