Magnetic orientation in rocks that is opposite to the current orientation of Earth's magnetic field is called ____ magnetic polarity.

Explanation:

If an object is thrown downward with an initial velocity of v₀​, then the distance it travels is given by s = 4.9 t²+v₀t

Now an object is thrown downward from a cliff 400 m high and it travels 138.3 m in 3 sec. We need to find initial velocity of the​ object.

           s = 4.9 t²+v₀t

           138.3 = 4.9  x 3²+ v₀ x 3

             3v₀ = 94.2

               v₀ = 31.4 m/s

Initial velocity of the​ object = 31.4 m/s

Operation of a heat engine is best described as using input heat to perform work and rejects excess thermal energy to a lower temperature reservoir.

This is defined as the form of energy that is transferred between two substances at different temperatures.

Heat engine uses heat to perform work which results in the rejection of excess thermal energy to a lower temperature reservoir which was why option C was chosen as the most appropriate choice.

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The correct description of a heat engine's operation is that it uses heat to perform work, while also rejecting a portion of the heat to a cooler reservoir; it is a cyclical process where energy is partially converted from thermal to mechanical form according to the Second Law of Thermodynamics.

The statement that best describes the operation of a heat engine is (c) A heat engine uses input heat to perform work and rejects excess thermal energy to a lower temperature reservoir. In thermodynamics, a heat engine is a system that converts heat or thermal energy to mechanical work. This process involves transferring energy from a high-temperature source, the hot reservoir (Qh), and partly converting this energy into work (W), while the rest is rejected as waste heat into a lower temperature sink, the cold reservoir (Qc).

Based on the Second Law of Thermodynamics, no heat engine can convert 100% of the input thermal energy into work because there is always some waste heat that must be expelled to a colder environment. The efficiency of a heat engine is measured as the ratio of work done to the heat input from the hot reservoir. The theoretical upper limit of this efficiency depends on the temperatures of the hot and cold reservoirs and is given by the Carnot efficiency, which is always less than 100%.

Answer:

a) Q = [tex]9*10^{-10}[/tex] C; E = [tex]2.03*10^5[/tex] V/m; V = 406.8 V

b) dE/dt = [tex]4.07*10^{11}[/tex] V/(m.s); No

c) [tex]J_d[/tex] = 3.6 [tex]A/m^2[/tex]; Equal

Explanation:

Given parameters are:

Area, A = 5 cm^2

Separation, d = 2 mm

Changing current, [tex]i_c[/tex] = 1.8 mA

At time t = 0 the charge [tex]Q_0[/tex] = 0

a) Here, we are asked to find charge, Q, electric field, E, and potential difference, V at time t = 0.5 [tex]\mu s[/tex]

[tex]Q = i_ct = 1.8*10^{-3}*5*10^{-7} = 9*10^{-10}[/tex] C

[tex]E = \sigma/\epsilon_0 = (Q/A)/\epsilon_0 = (9*10^{-10}/5*10^{-4})/(8.85*10^{-12}) = 2.03*10^5[/tex] [tex]\frac{V}{m}[/tex]

[tex]V = Ed = 2.03*10^{-5}*2*10^{-3} = 406.8[/tex] V

b) [tex]E = (Q/A)/\epsilon_0[/tex]

⇒ [tex]\frac{dE}{dt} = \frac{dQ}{dt} \frac{1}{\epsilon_0 A} = \frac{i_c}{\epsilon_0 A} = \frac{1.8*10^{-3}}{5*10^{-4}*8.85*10^{-12}} = 4.07*10^{11}[/tex] V/(m.s)

No, it is constant that does not vary in time because [tex]i_c[/tex] is constant.

c) the displacement current density, [tex]J_d = \epsilon_0\frac{dE}{dt} = \epsilon_0\frac{i_c}{\epsilon_0 A} = i_c/A[/tex]

⇒ [tex]J_d = 1.8*10^{-3}/(5*10^{-4}) = 3.6[/tex] [tex]A/m^2[/tex]

[tex]i_d =J_dA = 3.6*5*10^{-4} = 1.8*10^{-3}[/tex] A

So, [tex]i_c[/tex] and [tex]i_d[/tex] are equal.

Answer:

[tex]a)[/tex]The charge on the plates[tex]$Q=9 * 10^{-10} C ; E=2.03 * 10^{5} \mathrm{~V} / \mathrm{m} ; \mathrm{V}=406.8 \mathrm{~V}$[/tex]

[tex]b)[/tex]The time rate of change of the electric field between the plates[tex]$\mathrm{dE} / \mathrm{dt}=4.07 * 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{s}) ; \mathrm{No}$[/tex]

[tex]c)[/tex]The displacement current density jD between the plates[tex]$J_{d}=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]Equals

Explanation:

Given parameters are:

Area, [tex]$A=5cm^{2}[/tex]

Separation, [tex]$\mathrm{d}=2 \mathrm{~mm}$[/tex]

Changing current, [tex]$i_{c}=1.8 \mathrm{~mA}$[/tex]

At time [tex]$\mathrm{t}=0$[/tex] the charge [tex]$Q_{0}=0$[/tex]

a) Here, we are asked to find charge, [tex]$Q$[/tex], electric field, [tex]$E$[/tex] and potential difference, [tex]$\mathrm{V}$[/tex] at time [tex]$\mathrm{t}=0.5 \mu \mathrm{s}$[/tex]

[tex]$Q=i_{c} t=1.8 \times10^{-3} \times 5 \times10^{-7}=9 \times10^{-10} \mathrm{C}$[/tex]

[tex]$E=\sigma / \epsilon_{0}=(Q / A) / \epsilon_{0}=\left(9 \times10^{-10} / 5 \times10^{-4}\right) /\left(8.85\times 10^{-12}\right)=2.03 \times10^{5} \frac{\mathrm{V}}{m}$[/tex]

[tex]$V=E d=2.03 \times10^{-5} \times2\times 10^{-3}=406.8 \mathrm{~V}$[/tex]

b) [tex]$E=(Q / A) / \epsilon_{0}$[/tex]

[tex]$\Rightarrow \frac{d E}{d t}=\frac{d Q}{d t} \frac{1}{\epsilon_{0} A}=\frac{i_{c}}{\epsilon_{0} A}=\frac{1.8 \times10^{-3}}{5\times 10^{-4} \times 8.85 \times 10^{-12}}=4.07 \times 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{S})$[/tex]

No, it is constant that does not vary in time because is constant.

c) the displacement current density, [tex]$J_{d}=\epsilon_{0} \frac{d E}{d t}=\epsilon_{0} \frac{i_{c}}{\epsilon_{0} A}=i_{c} / A$[/tex]

[tex]$\Rightarrow J_{d}=1.8 \times 10^{-3} /\left(5 \times10^{-4}\right)=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]

[tex]$i_{d}=J_{d} A=3.6 \times5 \times10^{-4}=1.8\times 10^{-3} \mathrm{~A}$[/tex]

So, [tex]$i_{c}$[/tex] and [tex]$i_{d}$[/tex] are equal.

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Answer:

14days, after registration

Explanation:

When you move into new jeey from another state you must have your vehicle inspected within ___.

answer: 14days after registration. To register a vehicle in New Jersey, one must have the following:

A valid driver license

A valid probationary license or a validated New Jersey Permit

Valid Insurance

Vehicle registration cards

Vehicles are inspected every two years,  and five years for a new vehicle. There are checklist used by the inspector to ascertain the conditions of the vehicle and ensure if it is road worthy. Inspection are carried out to reduce road accidents which can endanger the life of the driver and other road users.

Answer:

[tex]n=101.7\times 10^6[/tex]

Explanation:

It is given that,

Frequency of the radio wave, [tex]f=101.7\ MHz=101.7\times 10^6\ Hz[/tex]

We know that the number of vibrations per second is called frequency of an object. We need to find the number of vibrations per second. Clearly, the number of vibrations per second represented in a radio wave is [tex]101.7\times 10^6[/tex]. Hence, this is the required solution.

Answer:

1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )

Explanation:

This is used to describe Discrimination Training  in learning and behavior. The 1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )

The 1100 Hz tone acts as a conditioned stimulus (CS), predicting food, while the 1300 Hz tone serves as an unconditioned stimulus (UCS), with no associated response. Understanding pitch and loudness is key in differentiating stimuli in conditioning.

The question relates to classical conditioning concepts in behavioral psychology, where the 1100 Hz tone is conditionally associated with the provision of food. In this context, the 1100 Hz tone functions as a conditioned stimulus (CS), since it predicts the arrival of food, prompting the rat to press the lever. On the other hand, the 1300 Hz tone does not forecast the coming of food and, thus, operates as an unconditioned stimulus (UCS), having no effect on the rat's lever-pressing behavior. It's also important to note the perception of frequency, which we refer to as pitch, and the perception of intensity, which is understood as loudness. The ability to discern differences in pitch allows for different responses to the 1100 Hz and 1300 Hz tones.

Explanation:

Minimum of aerobic activity 150 minutes, or a mix of moderates and intensive exercise 75 minutes of vigorous aerobic activity a week. We can  extend this practice throughout the week with the instructions. Mild aerobic workouts might include practices like quick strolling or swimming, while activity like running can include strong aerobic activity.

Answer:

2.71207 Hz

Explanation:

v = Speed of sound in air = 343 m/s

[tex]v_r[/tex] = Relative speed between the speakers and the student = 1.02 m/s

f' = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

[tex]f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.02}{343}\\\Rightarrow f_1=454.64396\ Hz[/tex]

Frequency of sound heard as the student moves closer to the other speaker

[tex]f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.02}{343}\\\Rightarrow f_2=457.35603\ Hz[/tex]

The difference in the frequencies is

[tex]f_2-f_1=457.35603-454.64396=2.71207\ Hz[/tex]

The student hears 2.71207 Hz

Answer:100 K

Explanation:

We know that Freezing Point of water is [tex]0^{\circ}C[/tex] at 1 atm

Converting it to Kelvin we get 273.15 K

Boiling Point of water is [tex]100^{\circ}C[/tex]

converting it to Kelvin we get 373.15 K

Difference in the number we get =373.15-273.15=100 K

Temperature difference is independent of degree and will remain same for Celsius and Kelvin

Final answer:

The normal force on a 25.2 kg block is 246.96 N on a level surface, 212.66 N on a 30.8° incline, and 315.36 N in an elevator accelerating upward at 2.78 m/s².

Explanation:

To calculate the magnitude of the normal force on a 25.2 kg block under various circumstances, we use different physics principles for each scenario:

For the block on a horizontal surface, N = (25.2 kg)(9.8 m/s2) = 246.96 N.

For the block on a 30.8° incline, N = (25.2 kg)(9.8 m/s2)(cos(30.8°)) = 212.66 N.

For the block in an accelerating elevator, N = (25.2 kg)(9.8 m/s2 + 2.78 m/s2) = 315.36 N.

Answer:

0 N

Explanation:

According to Newton's first law of motion, an object in motion stays in motion until acted upon by an unbalanced force.  With no friction in space to unbalance the cannonball, it will continue to keep going.

The  amount of force needed to keep it going equals​ to 0 N

The following information should be considered:

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Answer:

a. before

Explanation:

Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?

from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .

displacement is distance in a specific direction.

before the displacement was maximum at 2mm was instant at time=0.04s.

But later was constant at 0.06s at a displacement point of 1mm

Answer:

Option B

0.3 m/s2 South

Explanation:

Acceleration, [tex]a=\frac {v-u}{t}[/tex] where v and u are final and initial velocities respectively, t is the time taken

Substituting 14.1 m/s for v, 17.7 m/s for u and 12 s for t then

[tex]a=\frac {14.1 m/s- 17.7 m/s}{12}=-0.3 m/s^{2}[/tex]

Since this is negative acceleration, it's direction is opposite hence 0.3 m/s2 South

[tex]1 \times 10^{-10.1} \mathrm{Wm}^{-2}[/tex] is the intensity of the sound.

Answer: Option B

Explanation:

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about [tex]1 \times 10^{-12} \mathrm{Wm}^{-2}[/tex]). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     [tex]\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)[/tex]

Where,

I = Intensity of the sound produced

[tex]I_{0}[/tex] = Standard Intensity of sound of 60 decibels = [tex]1 \times 10^{-12} \mathrm{Wm}^{-2}[/tex]

So for 19 decibels, determine I as follows,

                   [tex]19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)[/tex]

                  [tex]\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}[/tex]

                  [tex]\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9[/tex]

When log goes to other side, express in 10 to the power of that side value,

                  [tex]\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}[/tex]

                  [tex]I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}[/tex]

Answer:

y=39.057 m

Explanation:

Using Kinematic relation

[tex]s=ut+ \frac{1}{2}at^2[/tex]

given u= 5m/s

a=g= -9.81 [tex]m/s^2[/tex]( directed downward)

[tex]s=5t- \frac{1}{2}(9.80)t^2[/tex]

Also, we know that

v=u+at

v=5-9.80t

at time t= 0.250 sec

[tex]s=5\times0.25- \frac{1}{2}(9.80)0.25^2[/tex]

s=0.94375 m

now position of sandbag

y= 40-0.94375

y=39.057 m

Answer:

[tex]E = 86.4 \times 10^6 J[/tex]

Explanation:

Given data:

light energy = 20 MJ

Electric consumption is 1.0 kW

Duration of energy consumption is 24 hr

Energy consumption is given as

[tex]E = Power \times time[/tex]

[tex]E = 1 \times 10^3 \times 24 \times 3600[/tex]

[tex]E = 8.64 \times 10^6 J = 86.4 MJ [/tex]

[tex]E = 86.4 \times 10^6 J[/tex]

Answer:

The kinetic energy is the same as in space, which in general is very large

Explanation:

If the Earth had no atmosphere we can use the conservation of kinetic energy in two points

Initial. In space far from the planet

      Em₀ = k = ½ m v₀²

Final. Just before touching the surface of the Earth

      [tex]Em_{f}[/tex] = K = ½ m v²

As there is no rubbing

      Em₀ = [tex]Em_{f}[/tex]

      [tex]Em_{f}[/tex]= ½ m v₀²

The kinetic energy is the same as in space, which in general is very large

When the Earth has an atmosphere we must use the energy work theorem

      W = ΔK

The work is done by the friction forces when the meteor enters the atmosphere, increases in density as it approaches the surface, so the work also increases.

       W =[tex]K_{f}[/tex] - K₀

       [tex]K_{f}[/tex] = K₀ - W

       [tex]K_{f}[/tex] = ½ m v₀² - W

We see that the kinetic energy decreases as the work increases, this makes the impact is higher and part of the meteor also evaporates by friction at the entrance

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

[tex]v^{2}=u^{2}+2as[/tex] and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

[tex]u=\sqrt {v^{2}-2gs}[/tex]

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

[tex]u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s[/tex]

Final answer:

To find the ball's initial speed, use the kinematic equation [tex]v^2[/tex] = [tex]u^2[/tex] + 2*g*d, where u is the initial velocity and v is the final velocity. The initial speed calculated is approximately 3.07 m/s.

Explanation:

To find the ball's initial speed when thrown straight up into the air, we can use the principles of kinematics under the influence of gravity. The known variables are the final velocity (v = 6.79 m/s), the distance fallen (d = 1.87 m), and the acceleration due to gravity (g = 9.81 [tex]m/s^2[/tex]).

We need to remember that the ball is moving upward against gravity, so we will consider g to be negative in our calculations. We will use the following kinematic equation which relates velocity, acceleration, and displacement:

[tex]v^2[/tex] = [tex]u^2[/tex] + 2*g*d

Where u is the initial velocity, v is the final velocity, g is the acceleration due to gravity, and d is the displacement.

Let's solve for u:

[tex]v^2 = u^2[/tex] + 2 * (-g) * d

[tex]u^2 = v^2[/tex] - 2 * g * d

[tex]u^2[/tex] = [tex](6.79 m/s)^2[/tex] - 2 * (9.81 [tex]m/s^2[/tex]) * (1.87 m)

[tex]u^2[/tex] = 46.1041 [tex]m^2/s^2[/tex] - 36.6894 [tex]m^2/s^2[/tex]

[tex]u^2[/tex] = 9.4147 [tex]m^2/s^2[/tex]

u = sqrt(9.4147 [tex]m^2/s^2[/tex])

u ≈ 3.07 m/s

Therefore, the initial speed of the ball was approximately 3.07 m/s.

Explanation:

It is given that,

Mass of the car 1, [tex]m_1=900\ kg[/tex]

Initial speed of car 1, [tex]u_1=15i\ m/s[/tex] (east)

Mass of the car 2, [tex]m_2=750\ kg[/tex]

Initial speed of car 2, [tex]u_1=20j\ m/s[/tex] (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]

[tex]900\times 15i +750\times 20j=(900+750)V[/tex]

[tex]13500i+15000j=1650V[/tex]

[tex]V=(8.18i+9.09j)\ m/s[/tex]

The magnitude of speed,

[tex]|V|=\sqrt{8.18^2+9.09^2}[/tex]

V = 12.22 m/s

(b) Let [tex]\theta[/tex] is the direction the wreckage move just after the collision. It is given by :

[tex]tan\theta=\dfrac{v_y}{v_x}[/tex]

[tex]tan\theta=\dfrac{9.09}{8.18}[/tex]

[tex]\theta=48.01^{\circ}[/tex]

Hence, this is the required solution.

To solve this problem, we can use the principle of conservation of momentum. We calculate the momentum of each car before the collision, then find the total momentum before the collision. Since the cars stick together after the collision, we can calculate the velocity of the wreckage and determine its direction using trigonometry.

To solve this problem, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.  

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Answer:

28.6 m/s

Explanation:

the verified expert clearly isnt an expert no shade tho

Answer:

α = 2  rad/s²

Explanation:

Newton's second law for rotation:

τ = I * α   Formula  (1)

where:

τ : It is the torque applied to the body.  (N*m)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

I =  8.0 kg * m²   :moment of inertia of the disk

R =  1.6 m : radius of the disk

F = 10.0 N : tangential force applied to the disk

Torque applied to the disk

The torque is defined as follows:

τ = F*R

τ = 10.0 N* 1.6 m

τ = 16 N*m

Angular acceleration of the disk ( α  )

We replace data in the formula (1):

τ = I * α

16 = 8 *α

α = 16 / 8

α = 2  rad/s²

Answer:

D

Explanation:

Gene flow is the transfer of genetic variation from one population to another

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

Answer:

D. 43° 52`  

Explanation:

A bearing is an angle, measured clockwise from the north direction. When solving a bearing problem, it is good to represent the bearings in the given question with diagram.

The diagrammatically representation of the bearing of lines A and B, 16° 10` and 332° 18` respectively given in the question is shown in the figure attached.

At Point A, we will calculate angle ∠BAO.

Calculating the angle ∠BAO

∠BAO = 90° - 16° 10`

          = 73° 50`

At Point B, we will calculate angle ∠ABO.

Calculating the angle ∠ABO

∠ABO = 332° 18` - 270° 0`

          = 62° 18`

At Point O, we will calculate the include angle ∠BOA.

Calculating the angle ∠BOA

∠BAO + ∠ABO + ∠BOA = 180°  (sum of angles in a triangle)

73° 50` + 62° 18` + ∠BOA = 180°

136° 8` + ∠BOA = 180°

∠BOA = 180° - 136° 8`

∠BOA = 43° 52`

The value of the included angle BOA is 43° 52

Answer:

The compression in the spring is 0.012 meters.

Explanation:

It is given that,

Mass of the child, m = 25 kg

Spring constant of the spring, [tex]k=2\times 10^4\ N/m[/tex]

When the child makes a nice big bounce, she finds that at the bottom of the bounce she is accelerating upward at [tex]9.8\ m/s^2[/tex]. Let x is the compression in the spring. The force of gravity is balanced by the force of the spring as :

[tex]mg=kx[/tex]

[tex]x=\dfrac{mg}{k}[/tex]

[tex]x=\dfrac{25\ kg\times 9.8\ m/s^2}{2\times 10^4\ N/m}[/tex]

x = 0.012 meters

So, the compression in the spring is 0.012 meters. Hence, this is the required solution.

Final answer:

The spring is compressed by 2.45 cm when the child is accelerating upward at 9.8 m/s² at the bottom of the bounce, as calculated using the forces acting on the child and Hooke's Law.

Explanation:

To determine how much the spring is compressed, we must consider the forces acting on the child at the bottom of the bounce. The upward force the spring exerts must be equal to the sum of the gravitational force on the child and the force required to accelerate the child upward at 9.8 m/s².

The gravitational force acting on the child is Fg = m × g, where m is the mass of the child (25 kg) and g is the acceleration due to gravity (9.8 m/s2). Thus, Fg = 25 kg × 9.8 m/s² = 245 N.

The additional force needed to accelerate the child upward at 9.8 m/s² is also Fa = m × a, yielding Fa = 245 N. The total force exerted by the spring then is Fs = Fg + Fa = 490 N.

To find the compression of the spring, we use Hooke's Law, Fs = k × x, where k is the spring constant (2.0 × 104 N/m) and x is the compression.

Solving for x, we get x = Fs / k = 490 N / (2.0 × 104 N/m) = 0.0245 m or 2.45 cm.

Final answer:

The crate has an acceleration of 0.43 m/s^2. It travels 21.5 m in 10.0 s and has a speed of 4.3 m/s at the end of 10.0 s.

Explanation:

To solve this problem, we can use Newton's second law, which states that force is equal to mass times acceleration (F = ma).

(a) We are given that the mass of the crate is 32.5 kg and the net horizontal force acting on it is 14.0 N. Plugging these values into the equation, we get:

F = ma

14.0 N = 32.5 kg * a

a = 14.0 N / 32.5 kg

a = 0.43 m/s^2

So, the acceleration produced is 0.43 m/s^2.

(b) To find the distance traveled by the crate in 10.0 s, we can use the equation of motion: distance = initial velocity * time + (1/2) * acceleration * time^2.

Since the crate starts at rest, the initial velocity is 0 m/s:

distance = 0 * 10.0 s + (1/2) * 0.43 m/s^2 * (10.0 s)^2

distance = 0 + (1/2) * 0.43 m/s^2 * 100.0 s^2

distance = 21.5 m

So, the crate travels 21.5 m in 10.0 s.

(c) To find the speed of the crate at the end of 10.0 s, we can use the equation of motion: final velocity = initial velocity + acceleration * time.

Since the crate starts at rest, the initial velocity is 0 m/s:

final velocity = 0 + 0.43 m/s^2 * 10.0 s

final velocity = 4.3 m/s

So, the speed of the crate at the end of 10.0 s is 4.3 m/s.

Answer:

False.  the system does not complete the circle movement

Explanation:

For this exercise, we must find the rope tension at the highest point of the path,

          -T - W = m a

The acceleration is centripetal

           a = v² / R

           T = ma - mg

           T = m (v² / R - g)

The minimum tension that the rope can have is zero (T = 0)

          v² / R - g = 0

          v = √ g R

Let's find out what this minimum speed is

          v = √ 9.8 1

          v = 3.13 m / s

We see that the speed of the body is less than this, so the system does not complete the movement.

Answer:

The car is 3.4 s in the air.

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

The vertical position of the car can be obtained by the following equation:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Where:

y = vertical position of car at time t.

y0 = initial vertical position.

v0 = initial velocity.

t = time.

α = launching angle.

g = acceleration of gravity.

The vertical component of the position vector when the car reaches the ground is -28 m (considering the edge of the cliff as the origin of the system of reference) and the initial vertical position is therefore 0 m. The launching angle is 22° below the horizontal (see figure). Then, we only have to find the initial velocity to solve the equation of vertical position for the time of flight.

To find the initial velocity, we have to use two equations: the equation of velocity of the car at the time it reaches the edge of the cliff and the equation of position of the car to find that time:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

If we place the origin of the frame of reference at the point where the car starts rolling, then the initial position is zero. Since the car starts from rest, the initial velocity, v0, is zero. Then, we can find the time it takes the car to travel the 54 m down the incline:

x = x0 + v0 · t + 1/2 · a · t²    (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

54 m = 1/2 · 4.4 m/s² · t²

2 · 54 m / 4.4 m/s² = t²

t = 5.0 s

With this time, we can find the velocity of the car when it reaches the edge of the cliff:

v = v0 + a · t   (v0 = 0)

v = a · t

v = 4.4 m/s² · 5.0 s

v = 22 m/s

Then, the initial velocity of the falling car is 22 m/s. Using the equation of vertical position:

y = y0 + v0 · t · sin α + 1/2 · g · t²   (y0 = 0)

y = v0 · t · sin α + 1/2 · g · t²

-28 m = 22 m/s · t · sin 22° - 1/2 · 9.81 m/s² · t²

0 = 28 m + 22 m/s · t · sin 22° - 4.91 m/s² · t²

Solving the quadratic equation for t using the quadratic formula:

t =3.4 s  (the other values is negative and, thus, discarded).

The car is 3.4 s in the air.

Final answer:

By solving the equation h = 1/2 g t^2 for the time it takes an object to fall 28 m, we find that the car is in the air for approximately 2.39 seconds before it hits the ocean.

Explanation:

To calculate how long the car is in the air, we need to analyze the car's vertical motion separately from its horizontal motion. The car falls 28 m, which we can use with the acceleration of gravity to find the time it takes to hit the water.

We use the equation h = ½ g t^2 where h is the height the car falls (28 m), and g is the acceleration due to gravity (9.81 m/s2). We're looking for t, the time in seconds.

28 m = ½ (9.81 m/s2) t^2

Rearranging for t: t = √(2 * 28 m / 9.81 m/s2)

t = √(5.70 s2)

t = 2.39 s

The car is in the air for approximately 2.39 seconds before hitting the water.

The minimum current rating of the motor disconnecting means for a 40-horsepower, 208-volt, 3-phase squirrel-cage motor is 121 amps.

In order to determine the minimum current rating of the motor disconnecting means, we need to calculate the current drawn by the motor.

The current rating of the motor disconnecting means must be equal to or higher than the calculated current.

First, we need to calculate the current drawn by the motor using the formula:

Current (I) = Power (P) / (Voltage (V) × Power Factor (PF) × √3)

Given that the motor has a power of 40 horsepower and operates at 208 volts with a power factor of 0.85, we can substitute these values into the formula:

I = 40 hp × (746 W/hp) / (208 V × 0.85 × √3)

Solving for I:

I ≈ 121 amps

Therefore, the minimum current rating of the motor disconnecting means for a 40-horsepower, 208-volt, 3-phase squirrel-cage motor is 121 amps.

Final answer:

In this case, the minimum current rating for the motor disconnecting means is approximately 88.5 Amperes, calculated based on the motor's power and voltage specifications.

Explanation:

The minimum current rating of the motor disconnecting means for a 40-horsepower, 208-volt, 3-phase squirrel-cage motor can be calculated using the formula:

Current (A) = Power (W) / (Voltage (V) * sqrt(3))

Plugging in the values:

Current = (40 hp * 746 W/hp) / (208 V * sqrt(3)) = ~88.5 A

n this case, the minimum current rating for the motor disconnecting means is approximately 88.5 Amperes, calculated based on the motor's power and voltage specifications.

Answer:

b. sequencing and allotting time to all project activities

Explanation:

' Project Controls are data collection, data management and predictive methods used to forecast, interpret and proactively control the time and cost results of a project or program; by communicating information in ways that enable effective management and decision-making.

So, a, c and  d are statements are a part of project controlling but b that is

sequencing and allotting time to all project activities  is not a part of project controlling.