If you are driving a car with a velocity of -25 m/s and you have an acceleration of -2 m/s^2, are you speeding up or slowing down? Why?

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

[tex]2gh = v_f^2 - v_i^2[/tex]

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]

(c)

Now we will consider the downward motion and use the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

Answer:

3.034 m

Explanation:

From the law of conservation of energy, the energy at the top of the incline equals the energy at the bottom of the incline since at the top of the incline, the horizontal surface is frictionless and along the incline there is no friction.

The work done in moving the crate a distance, d = 2.7 m with a force of F = 121 N to the top of the incline is W = Fd = 121 N × 2.7 m = 326.7 J.

From work-kinetic energy principles, this work W = kinetic energy of the crate at the top of the incline, K₁.

Now, the total mechanical energy at the top of the incline, E equals the total mechanical energy at the bottom of the incline E' since there is no friction along the incline.

So, E = E'

U₁ + K₁ = U₂ + K₂ where U₁ = potential energy of crate at top of incline = mgh where m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s², h = height of incline = 1.8 m, K₁ = kinetic energy of crate at top of incline = 326.7 J, U₂ = potential energy of crate at bottom of incline = 0 J(since it is at an elevation h = 0) and K₂ = kinetic energy of crate at bottom of incline

So, substituting the values of the variables into the equation, we have

U₁ + K₁ = U₂ + K₂

mgh + K₁ = U₂ + K₂

28.9 kg × 9.8 m/s² × 1.8 m + 326.7 J = 0 J + K₂

509.796 J + 326.7 J = K₂

K₂ = 836.496 J

K₂ ≅ 836.5 J

Now since the vertical wall is a distance d2 away and the long ideal spring has a length dy = 3.4 m, let x be the compression of the spring. So, the distance moved by the crate is thus D = d2 - dy - x.

Now, the change in kinetic energy of the crate ΔK equals the work done by friction and that done by the spring W.

So ΔK = -W (from work-kinetic energy principles)

Let W' = work done by friction = μmgD where  μ = coefficient of kinetic friction between surface and crate = 0.41, m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s² and D = distance moved by crate = D = d2 - dy - x = 5.2 m - 3.4 m - x = 1.8 - x

So, W' = μmgD

W' = 0.41 × 28.9 kg × 9.8 m/s² (1.8 - x)

W' = 116.12(1.8 - x)

W' = 2090.16 - 116.12x

The work done by the spring W" = 1/2k(x₀² - x²) where k = spring constant = 154 N/m, x₀ = initial spring length = dy = 3.4 m and x = final spring compression.

So,  W" = 1/2k(x₀² - x²)

W" = 1/2 × 154 N/m[(3.4 m)² - x²]

W" = 77 N/m[11.56 m² - x²]

W" = 890.12  - 77x²

So, W = W' + W"

W = 2090.16 - 116.12x + 890.12  - 77x²

W = 2980.28 - 116.12x - 77x²

Since the crate stops, final kinetic energy K₃ = 0. So, ΔK = K₃ - K₂ = 0 - 836.5 J = -836.5 J

Also, ΔK = -W

-836.5 = -(2980.28 - 116.12x - 77x²)

836.5 = 2980.28 - 116.12x - 77x²

77x² + 116.12 -2980.28 + 836.5 = 0

77x² + 116.12x -2143.78 = 0

dividing through by 77, we have

x² + 1.508x -27.841 = 0

Using the quadratic formula to find x, we have

[tex]x = \frac{-1.508 +/-\sqrt{1.508^{2} - 4 X 1 X (-27.841)} }{2 X 1.508} \\x = \frac{-1.508 +/-\sqrt{2.274064 + 111.364} }{3.016} \\x = \frac{-1.508 +/-\sqrt{113.638064} }{3.016} \\x = \frac{-1.508 +/- 10.66}{3.016} \\x = \frac{-1.508 - 10.66}{3.016} or x = \frac{-1.508 + 10.66}{3.016} \\x = \frac{-12.168}{3.016} or x = \frac{9.152}{3.016} \\x = -4.03 or 3.034[/tex]

x = -4.03 or 3.034

Since the compression of the spring is positive, we choose x = 3.034

So, the crate compresses the spring 3.034 m

Answer:

The mass decreases

Explanation:

Just smart

Answer:

The location of helicopter is behind the packet.

Explanation:

As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.

The horizontal velocity remains same as there is no force in the horizontal direction. The vertical  velocity goes on increasing as acceleration due to gravity acts.

So, the helicopter is behind the packet.

Answer:

O The environment did work on an object

Explanation:

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:

x-axis:

[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]

[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]

y-axis:

[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]

Use Eqn 1 to solve for T,

[tex]T = m_1(g \sin \theta - a)[/tex]

Substitute this expression for T into Eqn 2,

[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]

Collecting all similar terms, we get

[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]

or

[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]

Answer:

the moment of inertia of this system of masses about the y-axis is 99 kgm²

Explanation:

Given the data in the question;

mass m₁ = 5.0 kg at point ( 3.0, 4.0 )

mass m₂ = 6.0 kg at point ( 3.0, -4.0 )

Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;

Moment of inertia [tex]I[/tex]ₓ = mixi²

Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²

we substitute

Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ]  + [ 6.0 × ( 3 )² ]

Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ]  + [ 6.0 × 9 ]

Moment of inertia [tex]I[/tex] = 45 + 54

Moment of inertia [tex]I[/tex] = 99 kgm²

Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²

A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.

A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.

A positron is an electron with a positive charge.

During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.

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Answer:

Dragging a Chair, Climbing, Rubbing Hands

Answer:

Total cost = 56.16 cents

Explanation:

Given the following data;

Power = 45 Watts

Time = 4 hours

Number of days = 30 days

Cost = 10.4 cents

To find how much does it cost her to watch TV for one month;

First of all, we would determine the energy consumption of the TV;

Energy = power * time

Energy = 45 * 4

Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).

Energy consumption = 0.18 Kwh

Next, we find the total cost;

Total cost = energy * number of days * cost

Total cost = 0.18 * 30 * 10.4

Total cost = 56.16 cents

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]

Height of helicopter [tex]h=105\ m[/tex]

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]

So, package will take 5.21 s to reach the ground

Answer:

d. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

Explanation:

The current through a capacitor is given by i = CdV/dt where C = capacitance of capacitor and V = voltage across capacitor. At steady state dV/dt = 0 and V = constant. So, i = CdV/dt = C × 0 = 0.

So, in dc steady state, the voltage across a capacitor is constant and the current zero.

The voltage across an inductor is given by V = Ldi/dt where L = inductance of inductor and i = current through inductor. At steady state di/dt = 0 and V = constant. So, V = Ldi/dt = L × 0 = 0.

So, in dc steady state, the voltage across an inductor is zero and the current constant.

So, In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

The answer is d.

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Answer:

coz teaching is their profession.

Answer:

The centripetal force is 0.54 N.

Explanation:

mass, m = 0.56 kg

radius, r = 0.72 m

angular speed, w = 1.155 rad/s

The centripetal force is given by

[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]

Answer:

O Distant objects are blurry. describes farsightedness.

Explanation:

Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.

When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

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Answer:

8 N

Explanation:

Applying,

(F'+F) = ma............... Equation 1

Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.

From the question,

Given: F = 12 N, m = 4 kg, a = 5 m/s²

Substitute these values into equation 1

(F'+12) = 4×5

(F'+12) = 20

F' = 20-12

F' = 8 N.

Hence Ahmed's force is 8 N

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

Answer: C.

Explanation:  plato users

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

Answer:

Sultan Kösen

here is a pic

Answer:

[tex]t=1.9 sec[/tex]

Explanation:

From the question we are told that:

Height [tex]h=28m[/tex]

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

Generally the Newton's equation for Initial velocity upward is mathematically given by

 [tex]s=ut+\frtac{1}{2}at^2[/tex]

 [tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]

 [tex]u=24.03m/s[/tex]

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 [tex]v=\sqrt{24.03-2*9.8*28}[/tex]

 [tex]v=5.35m/s and -5.35m/s[/tex]

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 [tex]v=u+at[/tex]

 [tex]5.35=24.03-9.8t[/tex]

 [tex]t=\frac{28.03-5.35}{9.8}[/tex]

 [tex]t=1.9 sec[/tex]

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Answer:

The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization

Explanation:

Answer:

the radio can tune wavelengths between 1.88 and 5.97 m

Explanation:

The signal that can be received is the one that is in resonance as the impedance of the LC circuit.

         X = X_c - X_L

         X = 1 / wC - w L

at the point of resonance the two impedance are equal so their sum is zero

         X_c = X_L

         1 / wC = w L

         w² = 1 / CL

         w = [tex]\sqrt{\frac{1}{CL} }[/tex]

let's look for the extreme values

C = 1  10⁻¹² F

         w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]

         w = [tex]\sqrt{1 \ 10^{18}}[/tex]

         w = 10⁹ rad / s

C = 10 10⁻¹² F

         w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6

         w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018

         w = 0.316 10⁹ rad / s

Now the angular velocity and the frequency are related

           w = 2π f

           f = w / 2π

the light velocity  is

           c = λ f

           λ = c / f

we substitute

          λ = c 2π/w

we calculate the two values

 C = 1 pF

          λ₁ = 3 10⁸ 2π / 10⁹

          λ₁= 18.849 10⁻¹ m

          λ₁ = 1.88 m

C = 10 pF

           λ₂ = 3 10⁸ 2π / 0.316 10⁹

           λ₂ = 59.65 10⁻¹ m

           λ₂ = 5.97 m

so the radio can tune wavelengths between 1.88 and 5.97 m

Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Answer:

Ray A - incident ray

Ray B - reflected ray