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The diagram below shows four planets and their distances from the sun.

Light from the sun reaches Earth in 8.3 minutes. In how many minutes does light from the sun reach Mars?

12.62
13.34
15.85
16.60

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:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

 [tex]g = \dfrac{GM}{r^2}[/tex]

 [tex]g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}[/tex]

        g = 8.61 m/s²

Final answer:

The force of gravity on a 1 kg object in low-earth orbit around the Earth depends on the height of the orbit and is smaller than on the Earth's surface.

Explanation:

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. In low-earth orbit around the Earth, the force of gravity on the object is much smaller. This is because the force of gravity decreases as you move further away from the Earth's surface. However, the exact value of the force of gravity in low-earth orbit depends on the height of the orbit. For example, at the height of the International Space Station, the force of gravity on the object would be about 88% of the force on the Earth's surface, which is approximately 8.6 N.

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:

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:

A. DVD-RW

B. BD-R

Explanation:

The RW stands for rewritable.

BD-R uses Blu-ray technology allowing capacities of up to 100GB

The DVD-RW is capable of multiple rewrites, while the Blu-Ray disc holds the largest capacity amongst the options provided, being able to store 25-50 GB of data.

From the options provided (BD-R, Blu-Ray, DVD-RW, and DVD DS [Double-Sided]), the medium capable of multiple rewrites is DVD-RW. 'RW' in DVD-RW stands for 'ReWritable', meaning the media can be written, erased, and rewritten multiple times.

The optical media with the largest capacity is Blu-Ray. A single-layer Blu-Ray disc has a capacity of 25 GB, and a dual-layer Blu-Ray disc can hold 50 GB, more than five times the capacity of a DVD DS (Double Sided) which typically holds about 9 GB.

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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:

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:

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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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

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:

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.

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

[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:

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: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

Answer:

Answer:

a.  1.594 m/s = v

b. 1.274 m/s = v

Explanation:

A) First calculate the potential energy stored in the spring when it is compressed by 0.180 m...

U = 1/2 kx²

Where U is potential energy (in joules), k is the spring constant (in newtons per meter) and x is the compression (in meters)

U = 1/2(13.0 N/m)(0.180 m)² = 0.2106 J

So when the spring passes through the rest position, all of its potential energy will have been converted into kinetic energy.  K = 1/2 mv².

 0.2106 J  = 1/2(0.170 kg kg)v²

0.2106 J  = (0.0850 kg)v²

2.808m²/s² = v²

1.594 m/s = v

(B)  When the spring is 0.250 m from its starting point, it is 0.250 m - 0.180 m = 0.070 m past the equilibrium point.  The spring has begun to remove kinetic energy from the glider and convert it back into potential.  The potential energy stored in the spring is:

U = 1/2 kx² = 1/2(13.0 N/m)(0.070 m)² = 0.031J

Which means the glider now has only 0.2106 J  - 0.031J = 0.1796 J of kinetic energy remaining.

K = 1/2 mv²

0.1796 J = 1/2(0.170 kg)v²

0.138 J = (0.0850 kg)v²

1.623 m²/s² = v²

1.274 m/s = v

To calculate the speed of the glider at different points, we can use the principle of conservation of energy. At 0.180 m from the starting point, the speed is 2.65 m/s. At 0.250 m from the starting point, the speed is 3.89 m/s.

To solve this problem, we can use the principle of conservation of energy. When the glider is released, all of the potential energy stored in the compressed spring is converted into kinetic energy. At the point where the glider has moved 0.180 m from its starting point, the potential energy of the spring is zero, so the total mechanical energy is equal to the kinetic energy. Using the formula for kinetic energy, we can calculate the speed of the glider:

KE = 1/2 mv^2

m = 0.170 kg (mass of the glider)

v = ? (speed of the glider)

At the point where the glider has moved 0.180 m, the potential energy of the spring is zero, so the total mechanical energy is equal to the kinetic energy:

0.5 kx^2 = 0.5 mv^2

k = 13.0 N/m (force constant of the spring)

x = 0.180 m (distance moved by the glider)

By substitute the given values into the equation, we can solve for v:

0.5 * 13.0 N/m * (0.180 m)^2 = 0.5 * 0.170 kg * v^2

Solving for v, we find that the speed of the glider at the point where it has moved 0.180 m from its starting point is 2.65 m/s.

To calculate the speed of the glider at the point where it has moved 0.250 m from its starting point, we can use the same principle of conservation of energy. The initial potential energy of the spring is converted into kinetic energy:

0.5 * 13.0 N/m * (0.250 m)^2 = 0.5 * 0.170 kg * v^2

Solving for v, we find that the speed of the glider at the point where it has moved 0.250 m from its starting point is 3.89 m/s.

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Answer:

28.6 m/s

Explanation:

the verified expert clearly isnt an expert no shade tho

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]

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:

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.

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

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:

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:

When driving in heavy rain, the recommended following distance should be increased to at least 6 seconds due to longer stopping distances on wet pavement. Stopping distances depend on road conditions and driver reaction time, highlighting the need for increased safety margins in poor weather.

Explanation:

The amount of following distance you should allow between you and the vehicle in front of you during heavy rain should be considerably more than what you would maintain in dry conditions. The typical recommendation is to maintain at least a 3-second distance in good weather, but this should be increased to at least 6 seconds in heavy rain to accommodate for the increased stopping distances on wet pavement and reduced visibility.

Stopping distances can vary greatly depending on road conditions and driver reaction time, and heavy rain can significantly increase these distances. As braking distance increases with speed and poor weather conditions, it is important to adjust your following distance accordingly to ensure safety.

Referring to the given figures, we can deduce that for a car traveling at 30.0 m/s, the stopping distance will be much longer on wet pavement than on dry. If the driver's reaction time is assumed to be 0.500 s, the total distance traveled before the car comes to a stop will include both the reaction distance and the braking distance. When considering crossing a street, you must take into account that a safe distance is one where you are completely sure that the car can come to a full stop without reaching your crossing point, which can be roughly compared to the stopping distances shown in various figures.

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.

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

[tex]h=ut+\frac{1}{2}gt^2[/tex]

here u=0 and t=time taken

[tex]h=\frac{1}{2}gt^2[/tex]

for [tex]t=1 s[/tex]

[tex]h_1=\frac{1}{2}g[/tex]

for [tex]t=2 s[/tex]

[tex]h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g[/tex]

distance traveled in 2 nd sec[tex]=2g-\frac{1}{2}g=\frac{3}{2}g[/tex]

for [tex]t=3 s[/tex]

[tex]h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g[/tex]

distance traveled in 3 rd sec[tex]=\frac{9}{2}g-2g=\frac{5}{2}g[/tex]

so we can see that distance traveled in each successive second is increasing

The amount of distance travels by free falling object in each succeeding second is greater than the second before.

Speed of the free falling object is the speed, by which the body is falling downward towards the ground level.

At the Earth's surface, the speed of the free falling object will accelerate 9.841 meters per squared second.The odometer is the device, which is used the speed of the body in m/s or km/hrs.

Now we have to find out, whether the distance traveled by the object each succeeding second is constant, deceasing or increasing.

The second equation of the motion for distance can be given as,

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Here,  [tex]u[/tex] is the initial velocity of the body, [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by it.

As for the free falling body the initial velocity is zero. Thus the distance traveled by it for the first second is,

[tex]s=0+\dfrac{1}{2}9.81\times1\\s=4.905\rm m[/tex]

The distance traveled by it for the 2nd second is,

[tex]s=\dfrac{1}{2}9.81\times2^2\\s=19.61\rm m[/tex]

The distance traveled by it for the 3rd second is,

[tex]s=\dfrac{1}{2}9.81\times3^2\\s=44.145\rm m[/tex]

Now the difference between the third and 2nd second of distance traveled by object is greater then the difference between the second and first second.

Thus, the amount of distance travels by free falling object in each succeeding second is greater than the second before.

Learn more about speed of free falling body here;

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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:

nothing, mass is not affected by gravitational force

Explanation:

Weight is the gravitational force a planet exerts on a mass on the surface.

It is the product of the mass of an object with the gravitational acceleration that the planet produces.

The weight is the gravitational force

[tex]W=mg[/tex]

where,

m = Mass of the object

g = Acceleration due to gravity = 9.81 m/s²

Mass is the property that matter has which opposes the force being applied to it. It is intrinsic to the object itself and does not change according to the gravitational force. But, the weight changes.

The correct statement is nothing, mass is not affected by gravitational force.

The gravitational force of attraction of every object in the universe is given by Newton's gravitational law;

[tex]F_1= \frac{GmM_e}{R^2}[/tex]

where;

m is your mass

[tex]M_e[/tex] is mass Earth

R is the radius of the Earth

G is gravitational constant

If the mass of the Earth is cut into half, the gravitational force will be affected as follows;

[tex]F_2= \frac{Gm}{R^2}\times \frac{M_e}{2} =\frac{1}{2} (\frac{GmM_e}{R^2}) = \frac{1}{2} (F_1)[/tex]

The gravitational force will be reduced by 2

Now, let's check how your mass will be affected;

[tex]F_2= \frac{GmM_e}{R^2}\\\\GmM_e = F_2R^2\\\\m = \frac{F_2R^2}{G M_e} \\\\When, M_e \ is \ halved \ (0.5M_e) , \ F_2 = \frac{1}{2} F_1 = 0.5F_1\\\\m = \frac{0.5F_1R^2}{G \times 0.5M_e}\\\\m = \frac{F_1R^2}{G M_e}[/tex]

Your mass is not affected.

Thus, gravitational force is affected by mass but mass is not affected by gravitational force.

The correct statement is nothing, mass is not affected by gravitational force.

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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:

[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.