A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up from chest height to full arm extension, a distance of 0.53 m .
Part A: How much work does the weightlifter do to lift the barbell one time?
Part B: If the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body?
Part C: How many 500 Calorie donuts can she eat a day to supply that energy?

Answers

Answer 1

Answer:

A) Workdone = 223.57 N-m

B) 22357 J of energy

C) Number of donuts = 10.7 donuts

Explanation:

A) The work done is calculated from the formula;. Work done = Force × Distance

We are given;

Mass; m = 43 kg

Distance = 0.53 m

Force(weight) = mg = 43 × 9.81

Thus;

Work done = 43 × 9.81 × 0.53

Workdone = 223.57 N-m

B) We are told she does 23 repetitions a day.

Thus, we assume 23% efficiency.

So, Work = Energy

Thus;

At 100% efficiency;

Energy = (223.57/100%) × 23 repetitions = 5142.11 J

Now, since she is only 23% efficient, she will expend; 5142.11/0.23 J = 22357 J of energy to do 5390 J of work.

C) from conversions; 4.18 J = 1 calorie

Thus;

22357 J ÷ 4.18 J/cal = 5348.565 calories

We how many 500 calorie donuts she can eat in a day to supply that energy.

Thus;

Number of donuts = 5348.565 cal ÷ 500 cal /donut

Number of donuts = 10.7 donuts


Related Questions

A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the coefficient of contraction is 0.68 and the coefficient of velocity is 0.92, what is the discharge?

Answers

Answer:

The discharge rate is [tex]Q = 0.0192 \ m^3 /s[/tex]

Explanation:

From the question we are told that

   The  diameter is  [tex]d = 60 \ mm = 0.06 \ m[/tex]

    The  head is  [tex]h = 6 \ m[/tex]

     The  coefficient of contraction is  [tex]Cc = 0.68[/tex]

     The  coefficient of  velocity is  [tex]Cv = 0.92[/tex]

The radius is mathematically evaluated as

         [tex]r = \frac{d}{2}[/tex]

substituting values

        [tex]r = \frac{ 0.06 }{2}[/tex]

        [tex]r = 0.03 \ m[/tex]

The  area is mathematically represented as

      [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.03)^2[/tex]

      [tex]A = 0.00283 \ m^2[/tex]

 The  discharge rate is mathematically represented as

        [tex]Q = Cv *Cc * A * \sqrt{ 2 * g * h}[/tex]

substituting values

       [tex]Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}[/tex]

       [tex]Q = 0.0192 \ m^3 /s[/tex]

Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.

Answers

Answer:

they must be affordable because they have to pay for it or they wont get the stuff they are bying.

Explanation:

need a brainliest please.

Answer: B, they must be affordable.

Explanation:

The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge?

Answers

Answer:

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r²...................... Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k.................. Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = 180000(9×10⁹)/0.028²

q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

What is the density of the unknown fluid in Figure below? ρwater = 1000 kgm−3

Answers

Answer:

2500 kg/m³

Explanation:

P = P

ρgh = ρgh

ρh = ρh

(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)

ρ ≈ 2500 kg/m³

Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.75
Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.23×1030kg
. Find the radius of the exoplanet's orbit.

Answers

Answer:

[tex]r=4.24\times 10^{11}\ m[/tex]

Explanation:

Given that,

Orbital time period, T = 3.75 earth years

Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]

We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.

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

r is the radius of the exoplanet's orbit.

Solving for r we get :

[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]

We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]

So,

[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]

So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].

A resistance heater having 20.7 kW power is used to heat a room having 16 m X 16.5 m X 12.3 m size from 13.5 to 21 oC at sea level. The room is sealed once the heater is turned on. Calculate the amount of time needed for this heating to occur in min. (Write your answer in 3 significant digits. Assume constant specific heats at room temperature.)

Answers

Answer:

t = 23.6 min

Explanation:

First we need to find the mass of air in the room:

m = ρV

where,

m = mass of air in the room = ?

ρ = density of air at room temperature = 1.2041 kg/m³

V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³

Therefore,

m = (1.2041 kg/m³)(3247.2 m³)

m = 3909.95 kg

Now, we find the amount of energy consumed to heat the room:

E = m C ΔT

where,

E = Energy consumed = ?

C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C

ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C

Therefore,

E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)

E = 29324.62 KJ

Now, the time period can be calculated as:

P = E/t

t = E/P

where,

t = Time needed = ?

P = Power of heater = 20.7 KW

Therefore,

t = 29324.62 KJ/20.7 KW

t = (1416.65 s)(1 min/60 s)

t = 23.6 min

Monochromatic light falls on two very narrow slits 0.047 mm apart. Successive fringes on a screen 6.60 m away are 8.9 cm apart near the center of the pattern.
Determine the wavelength and frequency of the light.

Answers

Answer::

   [tex]\lambda = 634 nm[/tex]

  [tex]f = 4.73 *10^{14} \ Hz[/tex]

Explanation:

From the question we are told that

      The  distance of separation is  [tex]d = 0.047 \ mm = 0.047 *10^{-3} \ m[/tex]

       The  distance of the screen is  [tex]D = 6.60 \ m[/tex]

      The  width of the fringe is [tex]y = 8.9 \ cm = 0.089 \ m[/tex]

     

Generally the width of the width of the fringes is mathematically represented as

          [tex]y = \frac{\lambda * D }{d }[/tex]

=>       [tex]\lambda = \frac{y * d }{D }[/tex]

=>      [tex]\lambda = \frac{ 0.089 * (0.047 *10^{-3}) }{6.60 }[/tex]

=>    [tex]\lambda = 634 *10^{-9}[/tex]

=>  [tex]\lambda = 634 nm[/tex]

Generally the speed of light is mathematically represented as

         [tex]c = f * \lambda[/tex]

=>       [tex]f= \frac{ c}{\lambda }[/tex]

=>         [tex]f= \frac{ 3.0 *10^{8}}{634 *10^{-9}}[/tex]

=>     [tex]f = 4.73 *10^{14} \ Hz[/tex]

It is just as difficult to accelerate a car on a level horizontal surface on the Moon as it is here on Earth because

Answers

Answer:

Mass of the car is independent of gravity

Explanation:

Here, we want to state the reason why even though we have the acceleration due to gravity absent on the moon, it is still difficult to accelerate a car on a level horizontal level on the moon.

The answer to this is that the mass of the car that we want to accelerate is independent of gravity.

Had it been that gravity has an effect on the mass of the said car, then we might conclude that it will not be difficult to accelerate the car on a horizontal surface on the moon.

But due to the fact that gravity has no effect on the mass of the car to be accelerated, then the problem we have on earth with accelerating the car is the same problem we will have on the moon if we try to accelerate the car on a horizontal level surface.

The hydrogen spectrum has a red line at 656 nm, and a blue line at 434 nm. What is the first order angular separation between the two spectral lines obtained with a diffraction grating with 5000 rulings/cm?

Answers

Answer:

Explanation:

grating element or slit width a = 1 x 10⁻² / 5000

= 2 x 10⁻⁶ m

angular width of first order spectral line of wavelength λ

= λ / a

for blue line angular width

= 434 x 10⁻⁹ / 2 x 10⁻⁶ radian

= 217 x 10⁻³ radian

for red line angular width

= 656 x 10⁻⁹ / 2 x 10⁻⁶ radian

= 328 x 10⁻³ radian

difference of their angular width

= 328 x 10⁻³  - 217 x 10⁻³

= 111 x 10⁻³ radian

Ans .

3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from the central maxima (m=3) is at an angle of 45°?

Answers

Answer:

Wavelength is 471 nm

Explanation:

Given that,

Lines per unit length of diffraction grating is 500 lines/mm.

The third maxima from the central maxima (m=3) is at an angle of 45°

We need to find the color of laser light shines through a diffraction grating.

The condition for maxima is :

[tex]d\sin\theta=m\lambda[/tex]

d = 1/N, N = number of lines per mm

[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]

Convert 76.2 kilometers to meters?

Answers

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

A student wants to create a 6.0V DC battery from a 1.5V DC battery. Can this be done using a transformer alone

Answers

Answer:

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

Explanation:

A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.

Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.

So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

1. Rank the transformers on the basis of their rms secondary voltage. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
2. 100 A of rms current is incident on the primary side of each transformer. Rank the transformers on the basis of their rms secondary current. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns

Answers

Answer:

1. Transformer 3> Transformer 1 and 2 > Transformer 4

2. Transformer 2,3,5 > Transformer 1 > Transformer 4

Explanation:

From;

Vs/Vp = Ns/Np

Where;

Vp= voltage in primary coil

Vs= voltage in secondary coil

Ns= number of turns in secondary coil

Np= number of turns in primary coil.

Vs= Ns×Vp/Np

Vs= 480 ×2000/4000

Vs= 240 V

Vs= 480 ×1000/2000

Vs=240 V

Vs= 120 × 2000/500

Vs= 480 V

Vs= 240 × 500/1000

Vs= 120 V

2. Ns/Np= Ip/Is

Is= Np×Ip/Ns

Is= 1000 × 100/2000

Is= 50 A

Is= 2000 × 100/1000

Is= 200 A

Is= 1000 × 100/500

Is= 200 A

Is= 500 × 100/2000

Is= 25 A

Is= 4000 × 100/2000

Is= 200 A

An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency of this engine?

Answers

Answer:

10%

Explanation:

Efficiency = work done / energy used

e = (10 m × 100 N) / (10,000 J)

e = 0.1

The efficiency is 0.1, or 10%.

1. A 0.430kg baseball comes off a bar and goes straight up in the air. At a height of 10.0m, the baseball has a speed of 25.3m/s. Determine the mechanical energy at the height. Show all your work. 2. What is the baseball's mechanical energy when it is at a height of 8.0m? Explain?

Answers

Answer:

180 J

Explanation:

Mechanical energy = kinetic energy + potential energy

ME = KE + PE

ME = ½ mv² + mgh

ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)

ME = 180 J

Mechanical energy is conserved, so it is 180 J at all points of the trajectory.

The baseball's mechanical energy when it is at a height of 8.0m is 180 J.

What is mechanical energy?

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time. Mechanical energy is always conserved.

Mechanical energy = kinetic energy + potential energy

Given is the mass of baseball m= 0.430 kg, height h =10m, speed v= 25.3m/s.

ME = KE + PE

ME = ½ mv² + mgh

Substitute the values, we get

ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)

ME = 180 J

Thus, the baseball's mechanical energy when it is at a height of 8.0m is 180 J.

Learn more about mechanical energy.

https://brainly.com/question/13552918

#SPJ2

Which scientist proposed a mathematical solution for the wave nature of light?

Answers

Answer:

Explanation:

Christian Huygens

Light Is a Wave!

Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.

y=k/x, x is halved.
what happens to the value of y

Answers

Answer:

y is doubled

Explanation:

If x is halved, that means the value is doubled. Here is an exmaple:

y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?

Answers

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      [tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

       

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.00 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?

Answers

Answer:

B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction

Explanation:

In an electromagnetic wave the electric and magnetic fields are in phase

         c = E / B

         B = E / c

let's calculate

          B = 4.00 / 3 10⁸

          B = 1.33 10⁻⁸ T

To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.

 If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz

Answers

Answer:

380 kHz

Explanation:

The speed of sound is taken as 1500 m/s

The length of the fetus is 1.6 cm long

The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.

For this 1.6 cm baby, the wavelength must not exceed

λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =

0.4 cm = 0.004 m   this is the wavelength of the required ultrasonic sound.

we know that

v = λf

where v is the speed of a wave

λ is the wavelength of the wave

f is the frequency of the wave

f = v/λ

substituting values, we have

f = 1500/0.004 = 375000 Hz

==> 375000/1000 = 375 kHz ≅ 380 kHz

Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved

Answers

Answer:

1. Elastic collision

2. Inelastic collision    

Explanation:

Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision

the collision is described by the equation bellow

[tex]m1U1+ m2U2= m1V1+m2V2[/tex]

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity

the collision is described by the equation bellow

[tex]m1U1+ m2U2= V(m1+m2)[/tex]

What is the maximum wavelength of incident light for which photoelectrons will be released from gallium

Answers

Answer:

292 nm

Explanation:

The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J

at maximum wavelength, the energy of the photons is equal to its work function

Energy of the electron = hf

but hf = hc/λ

where h is the planck's constant = 6.63 × 10-34 m^2 kg/s

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength that this occurs, which is the maximum wavelength

Equating, we have

hc/λ =  ∅

substituting, we have

(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19

(1.989 x 10^-25)/(6.81 x 10^-19) = λ

λ = 292.07 x 10^-9 = 292 nm

"In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to"

Answers

Answer:

To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D

A mass M slides downward along a rough plane surface inclined at angle \Theta\:Θ= 32.51 in degrees relative to the horizontal. Initially the mass has a speed V_0\:V 0 = 6.03 m/s, before it slides a distance L = 1.0 m down the incline. During this sliding, the magnitude of the power associated with the work done by friction is equal to the magnitude of the power associated with the work done by the gravitational force. What is the coefficient of kinetic friction between the mass and the incline?

Answers

Answer: μ = 0.8885

Explanation: Force due to friction is calculated as: [tex]F_{f} = \mu.N[/tex]

At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.

Power associated with work done by friction is [tex]P=F_{f}.x[/tex]. The variable x is displacement the object "spent its energy".

Power associated with work done by gravitational force is P = mghcosθ, where h is height.

The decline forms with horizontal plane a triangle as draw in the picture.

To determine force due to friction:

[tex]F_{f}.x=mghcos(\theta)[/tex]

[tex]F_{f}=\frac{mghcos(\theta)}{x}[/tex]

Replacing force:

[tex]\frac{m.g.h.cos(\theta)}{x} = \mu.m.g.cos(\theta)[/tex]

[tex]\mu=\frac{h}{x}[/tex]

Calculating h using trigonometric relations:

[tex]sin(32.51) = \frac{h}{1}[/tex]

h = sin(32.51)

Coefficient of Kinetic friction is

[tex]\mu=\frac{sin(32.51)}{1}[/tex]

μ = 0.8885

For these conditions, coefficient of kinetic friction is μ = 0.8885.

A small glass bead charged to 5.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 910 N. What is the total charge on the rod?

Answers

Answer:

Explanation:

Let B= bead

Q = rod

the electric field at the glass bead pocation is

(Gauss theorem)

E = Q / (2 π d L εo)

the force is

F = q E = q Q / (2 π d L εo)

then

Q = 2 π d L εo F / q

Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC

Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact

Answers

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height

Answers

Answer:

The hoop

Explanation:

Because it has a smaller calculated inertia of 2/3mr² compares to the disc

P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.

Answers

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower

Answers

Answer:

The projectile strikes the tower at a height of 354.824 meters.

Explanation:

The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:

Horizontal motion

[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]

Vertical motion

[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.

[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

The time spent for the projectile to strike the tower is obtained from first equation:

[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]

If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:

[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]

[tex]t \approx 7.071\,s[/tex]

Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])

[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]

[tex]y \approx 354.824\,m[/tex]

The projectile strikes the tower at a height of 354.824 meters.

A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?

Answers

Answer:

There is no induced current on the coil.

Explanation:

Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.

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