A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.44 radians with the horizontal. The frictional coefficients between the crate and plane are (\muμs = 0.61, \muμk = 0.23 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?

Answers

Answer 1

Answer:

The  acceleration is  [tex]a = 6.2 m/s^2[/tex]

Explanation:

From the question we are told that

    The  angle which the inclined plane make with horizontal is  [tex]\theta = 0.44 \ rad[/tex]

     The frictional coefficients are  [tex]\mu_{\mu s} = 0.61[/tex] and  [tex]\mu_{\mu k} = 0.23[/tex]

     

The force acting on the  crate  is mathematically represented as  

        [tex]f = F_w + F_N[/tex]

Here f is the net force at which the crate is sliding down the plane which is mathematically represented as

      [tex]f = ma[/tex]

        [tex]F_w[/tex] is the force due to weight which is mathematically represented as

        [tex]F_w = mg sin (\theta)[/tex]

       and  [tex]F_N[/tex] the force due to friction which is mathematically represented as

       [tex]F_N = \mu_{\mu k } * mg cos(\theta )[/tex]

So  

     [tex]ma = mgsin(\theta ) + \mu_{\mu k} mg cos(\theta )[/tex]

      [tex]a = gsin(\theta ) + \mu_{\mu k } * g cos(\theta)[/tex]

substituting values

      [tex]a = 9.8 sin(0.44 ) + 0.23 * 9.8* cos(0.44)[/tex]

      [tex]a = 6.2 m/s^2[/tex]


Related Questions

Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?
Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=

Answers

Answer:

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V

A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is

Answers

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

If d is 10% smaller than D, how do the pressure drops experienced by the fluid across region I and region II compare?

Answers

Answer:

   ΔP = - 0.262 ρ v₁²

   

Explanation:

This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

suppose the pipe is horizontal, therefore

           y₁ = y₂

         P₁ -P₂ = ½ ρ (v₁² -v₂²)             1

 

Now let's use the continuity equation

         v₁ A₁ = v₂ A₂                              2

           

The area of ​​a circle is

         A = π r²

also the radius is half the diameter r = d /2

         A₁ = π d₁² / 4

let's substitute in equation 2

         v1 π d₁² / 4 = v2 π d₂² / 4

          v₁ d₁² = v₂ d₂²

          v₂ = v₁ d₁² / d₂²

let's substitute in equation 1

          P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )

Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1

           d₂ = D -0.1D = 0.9 D

we substitute in the previous equation

          P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)

          P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )

          P1 - P2 = ½ ρ v12 (- 0.524)

          ΔP = - 0.262 ρ v₁²

   

in this solution we assume that the data in zone I is known

a bear has a mass of 500kg and 100,000 J of mechanical kinetic energy. What is the speed of the bear? (velocity) Can someone please answer with the formula included? Step by step pls.

Answers

Answer:

Velocity of bear (v) = 20 m/s

Explanation:

Given:

Mass of bear (m) = 500 kg

Mechanical kinetic energy (K.E) = 100,000 J

Find:

Velocity of bear (v) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2(m)(v)²

100,000 = 1/2(500)(v)²

200,000 = 500(v)²

400 = (v)²

Velocity of bear (v) = 20 m/s

Which of the following is true of children with chronic illness? a.) They are all eligible to recievie special education services. b.) Very few such children are enrolled in public schools c.) Their eligibility for soeical education services depends on whether their conditions adversely affect their educational functioning. d.) They represent a large proportion of the children eligible for speical education services.

Answers

Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.

The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum

Answers

Answer:

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

Explanation:

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

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

Where:

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

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

The impact experimented by the ball due to collision is:

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

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

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

Where:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

In a nutshell, the right choice is option D.

A system consists of two particles. The first particle has mass m1 = 6.60 kg and a velocity of (4.20i + 2.00j) m/s, and the second particle has mass m2 = 2.00 kg and a velocity of (2.00i − 3.60j) m/s. (Express your answers in vector form.)

Required:
a. What is the velocity of the center of mass of this system?
b. What is the total momentum of this system?

Answers

Answer:

a. 8.465 m/s

b.22.3659

Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed.
In this final state
a) spheres #1 and #2 are still uncharged.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 both carry positive charge.
d) spheres #1 and #2 both carry negative charge.
e) sphere #1 carries positive charge and #2 carries negative charge

Answers

Answer:

sphere #1 carries positive charge and #2 carries negative charge

This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged

At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit is:

Answers

Answer:

Explanation:

The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point

Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima  is π radian .

At the scene of an accident, there are skid marks which are 200 feet long showing where a car put on the brakes. The driver swears that he was obeying the speed limit. If we assume that the car braked with a constant deceleration of 16ft/sec^2, we can determine whether or not the driver is telling the truth. how fast was he going when he applied the brakes?

Answers

Answer:

He was going at 54.5mph

Explanation:

Measuring distances in feet and time (starting with the brakes being applied) in seconds,

we have a(t) = −16, s(0) = 0, and v(T) = 0, s(T) = 200,

where T is the time of braking.

So Finding antiderivatives, v(t) = −16t + A and so s(t) = −8t²+ At +B. Since s(0) = 0,

we have B = 0. Since v(T) = 0,

we have T = A /16.

Putting this into s() gives

200 = −8(A/ 16)² +A· (A /16) = A²/32.

This gives A2 = 6,400, so A = 80 = v(0). That is, the initial speed was 80 ft/s ≈ 54.5 mph.

The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. If a cell of e.m.f. 1.2 volts is balanced by it, the balancing length will be:
 
A)  90 cm
B)  60 cm
C)  50 cm
D)  75 cm​

Answers

Answer:

Potential gradient = 1.6 v/m  

E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l

l=1.2\1.6

=3\4

3\4=0.75m

convert m into cm 0.75x10

0.75m=75 cm

ans=75 cm

Explanation:

Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, and thus the magnetic field it produces. The second coil is connected only to an ammeter. The ammeter will indicate that a current is flowing in the second coil:______.
a. only if the second coil is connected to the power supply by rewiring it to be in series with the first coil.
b. only when the current in the first coil changes.
c. whenever a current flows in the first coil.
d. only when a steady current flows in the first coil.

Answers

Answer:

B. only when the current in the first coil changes

Explanation:

This is because for current to be induced in the second coil they must be an change in current inyhe first coil in line with Faraday's first law of electromagnetic induction. Which state that whenever their is a change in magnetic lines of flux an emf is induced

Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 232 and 194 ​newtons, forming an angle of 67

Answers

Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°

A drum rotates around its central axis at an angular velocity of 19.4 rad/s. If the drum then slows at a constant rate of 8.57 rad/s2, (a) how much time does it take and (b) through what angle does it rotate in coming to rest

Answers

Answer:

Explanation:

Using equations of motion:

(a)

v=u+at

∴0=19.4−8.57t

∴t=19.4/8.57

=2.3s

B. Using s= ut + 1/2 at²

19.4(2.3)-1/28.57(2.3)²

= 21.92rad

A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.

Required:
What is the maximum emf induced in the loop?

Answers

Answer:

The maximum emf induced in the loop is 0.132 Volts

Explanation:

Given;

radius of the circular loop, r = 9.5 cm

intensity of the wave, I = 0.0295 W/m²

wavelength, λ = 6.40 m

The intensity of the wave is given as;

[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]

where;

B₀ is the amplitude of the field

c is the speed of light = 3 x 10⁸ m/s

μ₀ is permeability of free space = 4π x 10⁻⁵ m/A

[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]

Area of the circular loop;

A = πr²

A = π(0.095)²

A = 0.0284 m²

Frequency of the wave;

f = c / λ

f = (3 x 10⁸) / (6.4)

f = 46875000 Hz

Angular velocity of the wave;

ω = 2πf

ω = 2π(46875000)

ω = 294562500 rad/s

The maximum induced emf is calculated as;

emf = B₀Aω

       = (1.572 x 10⁻⁸)(0.0284)(294562500)

       = 0.132 Volts

Therefore, the maximum emf induced in the loop is 0.132 Volts

a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all directions. What is the waves intensity 25 km from the antenna

Answers

Answer:

[tex]I=2.67\times 10^{-6}\ W/m[/tex]

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m[/tex]

So, the wave intensity 25 km from the antenna is [tex]2.67\times 10^{-6}\ W/m[/tex].

As more energy from fossil fuels and other fuels is released on Earth, the overall temperature of Earth tends to rise. Discuss how temperature equilibrium explains why Earth’s temperature cannot rise indefinitely.

Answers

Answer:

processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Explanation:

The temperature of planet Earth is due to two main types of process, internal and external.

Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect

External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.

These are the two terms that heat the Earth

In addition there are several processes so the planet loses energy,

* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission

* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system

* liquid (water) system that absorbs large amounts of heat to change state and temperature.

These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value

A stone is thrown vertically upward with a speed of 21.0 m/s . How fast is it moving when it is at a height of 11.0 m ?

Answers

Answer:

15.01 m/s

Explanation:

Applying newtons equation of motion,

v² = u²+2gs.................. Equation 1

Where v = final velocity, u = initial velocity, s = distance, g = acceleration due to gravity.

Given: u = 21 m/s, s = 11 m

Constant: g = -9.8 m/s² ( g is negative because the stone against the force of gravity).

Substitute these values into equation 1

v² = 21²-2(11)(9.8)

v² = 441-215.6

v² = 225.4

v = √225.4

v = 15.01 m/s

A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward

Answers

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s

speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.

A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.

Answers

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Explanation:

From the question we are told that

   The  width of the slit is  [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]

     The angle is  [tex]\theta = 11^o[/tex]

The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as

        [tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]

Where [tex]\beta[/tex] is mathematically represented as

        [tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]

substituting values

      [tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]

     [tex]\beta = 708.1 \ rad[/tex]

So

  [tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low-frequency sound is originating from.
(a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 20 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear?
(b) Assume the same person was scuba diving and a low-frequency sound source was to the right of the scuba diver. How long (in ) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s? S
(c) What is significant about the time interval of the two situations?

Answers

Answer:

(a) 0.59 ms

(b) 0.15 ms

(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.

Explanation:

(a) distance between ears = 20 cm = 0.2 m

speed of sound generated = 340 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                        = [tex]\frac{0.2}{340}[/tex]

                        = 5.8824 × [tex]10^{-4}[/tex]

                        = 0.59 ms

The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.

(b) distance between ears = 20 cm = 0.2 m

speed of sound in water = 1530 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                         = [tex]\frac{0.2}{1530}[/tex]

                         = 1.4815 × [tex]10^{-4}[/tex]

                         = 0.15 ms

The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.

(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.

A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds

B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds

C) The significance about the time interval of the two situations is that;

Transmission of sound varies with different mediums.

A) We are given;

Distance between the two ears; d = 20 cm = 0.2 m

Speed of sound; v = 340 m/s

Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;

t = d/v

t = 0.2/340

t = 0.588 × 10⁻³ seconds

B) We are now told that the speed of sound in water is 1530 m/s. Thus;

t = 0.2/1530

t = 0.131 × 10⁻⁵ seconds

C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.

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Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The moment about the origin O of the force F is Mo

Answers

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.

1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?

Express your answer numerically in kilowatts to two significant figures.

2) What is the total force F on the panels exerted by radiation pressure from the sunlight?

Express the total force numerically, to two significant figures, in units of newtons.

Answers

Answer:

1) 14 kW

2) 4.67 x 10^-5 N

Explanation:

Area of solar panel = 10 m^2

Intensity of sun's radiation incident on earth = 1.4 kW/m^2

Solar power absorbed = ?

We know that the intensity of radiation on a given area is

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where I is the intensity of the radiation

P is the power absorbed due to this intensity on a given area

A is the area on which this radiation is incident

From the equation, we have

P = IA

P = 1.4 kW/m^2  x  10 m^2 = 14 kW

b) For a perfect absorbing surface, the radiation pressure is given as

p = I/c

where p is the radiation pressure

I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2

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

substituting values, we have

p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa

we know that Force = pressure x area

therefore force on the solar panels is

F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N

Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. The amount of charge that passes per unit time is called

Answers

The amount of charge that passes per unit time is called electric current .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids

Answers

Explanation:

It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.

As they are in equilibrium, the force of gravity due to each other is same. So,

[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]

So, the ratio of masses X/Y is 0.0847

A 4.7 kg solid sphere made of metal, whose density is 4000 kg /m3, is suspended by a cord. The density of water is 1000 kg/m3.When the sphere is immersed in water, the tension in the cord is closest to:

Answers

Answer:

34.58 N

Explanation:

From the question,

Tension in the cord =  Weight of the solid metal- Upthrust

T = W-U................... Equation 1

But,

W = mg

Where m = mass of solid, g = acceleration due to gravity.

Given: m = 4.7 kg, g = 9.8 m/s²

W = 4.7(9.8) = 46.1 N.

U = (Density of water×Volume of water displaced.)gravity

U = D×V×g.

But,

V = Mass of solid/density of solid

V = 4.7/4000

V = 1.175×10⁻³ m³.

Therefore,

U = 1.175×10⁻³(1000)(9.8)

U = 11.52 N

Therefore,

T = 46.1-11.52

T = 34.58 N

An inductor is connected to a 18 kHz oscillator. The peak current is 70 mA when the rms voltage is 5.4 V What is the value of the inductance L

Answers

Answer:

The value of the inductance is 0.955 mH

Explanation:

Given;

frequency of the oscillator, f = 18 kHz = 18,000 Hz

the peak current, I₀ = 70 mA = 0.07 A

the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V

The root mean square current is given as;

[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]

[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]

Inductive reactance is given by;

[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]

Inductance is given by;

[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]

L = 0.955 mH

Therefore, the value of the inductance is 0.955 mH

The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.

Given the following data:

Oscillator frequency = 18 kHzPeak current = 70 mARms Voltage = 5.4 V

To determine the value of the inductance (L):

First of all, we would find the root mean square (rms) current by using the formula:

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]

Next, we would calculate the inductive reactance of the oscillator by using the formula:

[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]

Now, we can solve for the value of the inductance (L):

[tex]L = \frac{X_L}{2\pi f}[/tex]

Where:

L is the inductance.f is the frequency.[tex]X_L[/tex] is the inductive reactance.

Substituting the parameters into the formula, we have;

[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]

L = [tex]9.55 \times 10^{-4}[/tex] Henry.

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have an electrical charge of +1, while
have an electrical charge of -1.
A. Neutrons, electrons
B. Protons, electrons
C. Electrons, neutrons
D. Electrons, protons

Answers

Answer:

B

Explanation:

Protons have a positive electrical charge of +1,

Electrons have a negative charge of -1,

Neutrons have a neutral charge of about 0.

If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is

Answers

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.

Answers

Answer:

3.26m

Explanation:

See attached file

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