If a car stops suddenly, you feel "thrown forward." We'd like to understand what happens to the passengers as a car stops. Imagine yourself sitting on a very slippery bench inside a car. This bench has no friction, no seat back, and there's nothing for you to hold on to.
A. Draw a picture and identify all of the forces action on you as thecar travels at a perfectly steady speed on level ground.
B. Draw your free-body diagram. Is there a net force on you?
C. Describe what happens to you as the car slowsdown?
D. Suppose now that the bench is not slippery. As the carslows down, you stay on the bench and dont slide off. What force is responsible for you deceleration?
E. In which directin does this froce point?

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

           [tex]F_{e} = F_{m}[/tex]

           q E = qv B

           v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

Waves transfer energy, but not matter. (C)

Answer:

C. Waves transfer energy not matter

Explanation:

wave is a disturbance

Answeri believe it is the first one but not sure

Explanation:

Answer:

The answer is sensorimotor.

Explanation:

Complete Question

C2B.7

Suppose I drop a 60-kg anvil from rest and from such a height that the anvil reaches a speed of 10 m/s just before hitting the ground. Assume the earth was at rest before I dropped the anvil.

(a) What is the earth's speed just before the anvil hits?

b)     How long would it take the earth to travel [tex]1.0 \mu m[/tex] (about a bacterium's width) at this speed?

Answer:

a

  [tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]

b

  [tex]t = 9.95 *10^{15} \approx 10 *10^{15} \ s[/tex]

Explanation:

From the question we are told that

     The mass of the anvil is [tex]m_a = 60\ kg[/tex]

     The speed at which it hits the ground is  [tex]v = 10 \ m/s[/tex]

Generally the mass of the earth  has a value  [tex]m_e = 5972*10^{24} \ kg[/tex]

Now according to the principle  of momentum conservation

   [tex]P_i = P_f[/tex]

 Where [tex]P_i[/tex] is the initial momentum which is zero given that both the anvil and the earth are at rest

   Now  [tex]P_f[/tex] is the final momentum which is mathematically represented as

     [tex]P_f = m_a * v + m_e * v_1[/tex]

So  

      [tex]0 = m_a * v + m_e * v_1[/tex]

substituting values

     [tex]0 = 60 * 10 + 5.972 *10^{24} * v_1[/tex]

=>    [tex]v_1 = -1.0*10^{-22} \ m/s[/tex]

Here the negative sign show that it is moving in the opposite direction to the anvil

  The magnitude of the earths speed is

      [tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]

The time it would take the earth is  mathematically represented as

        [tex]t = \frac{d}{|v_1|}[/tex]

substituting values

        [tex]t = \frac{1.0*10^{-6}}{1.0 *10^{-22}}[/tex]

        [tex]t = 10 *10^{15} \ s[/tex]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

Where:

[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse

[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.

The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]

To determine the fraction:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]

ln (MR) = 2.17

MR = [tex]e^{2.17}[/tex]

MR = 8.76

Part 2: Isp = 2000s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]

ln (MR) = 0.49

MR = [tex]e^{0.49}[/tex]

MR = 1.63

Answer:

2790 J/C

Explanation:

charge on sphere Q = 62 nC = [tex]62*10^{-9} C[/tex]

radius of the sphere r = 5.0 cm = 0.05 m

distance away from reference point d = 15.0 cm = 0.15 m

total distance of charge relative reference point R = r + d = 0.05 + 0.15 = 0.2 m

electric potential V is given as

[tex]V = \frac{kQ}{R}[/tex]

where k = Coulumb's constant = [tex]9*10^{9}[/tex] kg⋅m³⋅s⁻⁴⋅A⁻²

[tex]V = \frac{9*10^{9} * 62*10^{-9} }{0.2}[/tex] = [tex]\frac{9*62}{0.2}[/tex]

V = 2790 J/C

Answer:

About [tex]1.795 \times 10^{-3}[/tex] seconds

Explanation:

[tex]\Delta p=F \Delta t[/tex], where delta p represents the change in momentum, F represents the average force, and t represents the change in time.

The change of velocity is:

[tex]37-(-27.1)=64.1m/s[/tex]

Meanwhile, the mass stays the same, meaning that the change in momentum is:

[tex]64.1\cdot 0.14kg=8.974[/tex]

Plugging this into the equation for impulse, you get:

[tex]8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s[/tex]

Hope this helps!

Answer:

6

Explanation:

We are given that

[tex]\theta=2.12^{\circ}[/tex]

Slid width,a=0.110 mm=[tex]0.11\times 10^{-3} m[/tex]

[tex]1mm=10^{-3} m[/tex]

Wavelength,[tex]\lambda=582 nm=582\times 10^{-9}[/tex] m

[tex]1nm=10^{-9} m[/tex]

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.

[tex]asin\theta=\frac{2N+1}{2}\lambda[/tex]

Using the formula

[tex]0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})[/tex]

[tex]2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}[/tex]

[tex]2N+1=13.98[/tex]

[tex]2N=13.98-1=12.98[/tex]

[tex]N=\frac{12.98}{2}\approx 6[/tex]

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

Answer:

a) A positive current will be induced in the coil

Explanation:

Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.

According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.

Answer: 9.312 m/s

Explanation:

The friction force (opposite to the motion) is Fa = μ*m*g*cos(α) with μ = kinetic friction. The force that makes the motion is

F = m*g*sin(α).

The Newton's law gives:

F - Fa = m*a

m*g*sin(α) - μ*m*g*cos(α) = m*a

g*sin(α) - μ*g*cos(α) = a so a = 4.335 m/s²

It's a uniformly accelerated motion:

Space

S = 0.5*a*t²

10 = 0.5*a*t²

=> t = 2.148 s

Velocity

V = a*t = 9.312 m/s.

We have that the speed at the bottom of the plane is

[tex]v-9.3m/s[/tex]

From the question we are told that:

Angle of slide [tex]\theta =3.7 \textdegree[/tex]

Coefficient of kinetic friction [tex]\mu= 0.20[/tex]

Length [tex]L=10m[/tex]

Generally, the equation for acceleration along the slide is mathematically given by

[tex]a=gsin \theta-\mu cos\theta[/tex]

[tex]a=(9.8sin37-0.20*9.8*cos37[/tex]

[tex]a=4.33m/s^2[/tex]

Therefore

Velocity v is  is mathematically given by

[tex]v=\sqrt{2as}[/tex]

[tex]v=\sqrt{2*4.33*10}[/tex]

[tex]v-9.3m/s[/tex]

In conclusion

The speed at the bottom of the plane is

[tex]v-9.3m/s[/tex]

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Answer:

21.48 km 2.92° north of east

Explanation:

To find the resultant direction, we need to calculate a sum of vectors.

The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)

The second vector has module 16 and angle = 40°

Now we need to decompose both vectors in their horizontal and vertical component:

horizontal component of first vector: 13 * cos(315) = 9.1924

vertical component of first vector: 13 * sin(315) = -9.1924

horizontal component of second vector: 16 * cos(40) = 12.2567

vertical component of second vector: 16 * sin(40) = 10.2846

Now we need to sum the horizontal components and the vertical components:

horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491

vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922

Going back to the polar form, we have:

[tex]module = \sqrt{horizontal^2 + vertical^2}[/tex]

[tex]module = \sqrt{460.0639 + 1.1929}[/tex]

[tex]module = 21.4769[/tex]

[tex]angle = arc\ tangent(vertical/horizontal)[/tex]

[tex]angle = arc\ tangent(1.0922/21.4491)[/tex]

[tex]angle = 2.915\°[/tex]

So the resultant direction is 21.48 km 2.92° north of east.

Answer:

blue  θ₂ = 22.26º

red    θ₂ = 22.79º

Explanation:

When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media

let's apply this equation to each wavelength

λ = blue

in this case n₁ = 1, n₂ = 1,645

       sin θ₂ = n₁/ n₂ sin₂ θ₁

let's calculate

       sin θ₂ = 1 / 1,645 sint 38.55

       sin θ₂ = 0.37884

       θ₂ = sin⁻¹ 0.37884

       θ₂ = 22.26º

λ = red

n₂ = 1,609

         sin θ₂ = 1 / 1,609 sin 38.55

         sin θ₂ = 0.3873

         θ₂ = sim⁻¹ 0.3873

         θ₂ = 22.79º

the refracted rays are between these two angles

Answer:

5.95m/s to 2 decimal places

Explanation:

In physics speed is measured in metres per second so convert 8mins to seconds

8x60=420 seconds

The formula needed:

Speed (m/s)= Distance (m)/Time (s)

2500/420=5.95m/s

Answer:

2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)

This is a combination reaction.

Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)

This is a replacement reaction.

Explanation:

A combination reaction is a reaction in which two reagents are combined into one product. The reaction has the following general form:

A + B → AB

where A and B represent any two chemical substances.

2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)

This is a combination reaction because a single compound forms from two or more reacting species.

Double Substitution, Double Displacement or Metastasis Reactions are those in which two elements found in different compounds exchange their positions forming two new compounds. These chemical reactions do not present changes in the number of oxidation or relative load of the elements. So they are not considered redox reactions.

The solvent of the double displacement reactions usually is water and the reagents and products are usually ionic compounds (cations or anions are exchanged), although they can also be acids or bases.

In general, this type of reaction can be expressed as:

AB + CD ⇒ AD + CD

In the reaction:

Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)

This is a replacement reaction because it is a double replacement reaction in which the ions are exchanged to form new compounds.

Answer:

  c.  the net work a spring does on a body is zero when the body returns to its initial position

Explanation:

A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.

An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

Final velocity,

Initial velocity,

By using Newton's 2nd law of motion, we get

→ Impulse, [tex]I = m(v-u)[/tex]

By substituting the values, we get

                     [tex]= 2.4[2.5-(1.5)][/tex]

                     [tex]= 2.4(2.5+1.5)[/tex]

                     [tex]= 2.4\times 4[/tex]

                     [tex]= 9.6 \ Ns[/tex]

Thus the above answer is right.    

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Answer:

0.00185 °C

Explanation:

From the question,

The potential energy of the bird = heat gained by the water in the fish tank.

mgh = cm'(Δt)................... Equation 1

Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.

make Δt the subject of the equation

Δt = mgh/cm'............... Equation 2

Given: m = 1 kg, h = 40 m, m' = 50.5 kg

constant: g = 9.8 m/s², c = 4200 J/kg.K

Substitute into equation 2

Δt = 1(40)(9.8)/(50.5×4200)

Δt = 392/212100

Δt = 0.00185 °C

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

[tex]\Sigma F = F - F' = M\cdot a[/tex]

Box with mass 2M

[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]

Box with mass 3M

[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]

On the third equation, acceleration can be modelled in terms of F'':

[tex]a = \frac{F''}{3\cdot M}[/tex]

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

[tex]F' = 2\cdot M \cdot a + F''[/tex]

[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]

[tex]F' = \frac{5}{3}\cdot F''[/tex]

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]

[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]

[tex]F = 2\cdot F''[/tex]

[tex]F'' = \frac{1}{2}\cdot F[/tex]

Afterwards, F' as function of the external force can be obtained by direct substitution:

[tex]F' = \frac{5}{6}\cdot F[/tex]

The net forces of each block are now calculated:

Box with mass M

[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]

[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]

Box with mass 2M

[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]

[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]

Box with mass 3M

[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

Answer:

The auroras C.

Explanation:

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex]       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

[tex]v_x=v_o[/tex]

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

[tex]v_y^2=v_{oy}^2+2gy[/tex]         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

[tex]v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}[/tex]

Finally, you use the equation (1) to find the angle:

[tex]\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°[/tex]

The angle below the horizontal is 12.60°

The angle below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.

Given the following data:

To find the angle below the horizontal of the velocity of the puck just before it hits the ground:

First of all, we would determine the horizontal and vertical components of the hockey puck.

For horizontal component:

[tex]V_y^2 = U_y^2 + 2aS\\\\V_y^2 = 0^2 + 2(9.81)(2)\\\\V_y^2 = 39.24\\\\V_y = \sqrt{39.24} \\\\V_y = 6.26 \; m/s[/tex]

For vertical component:

[tex]V_x = U_x\\\\V_x = 28.0 \;m/s[/tex]

Now, we can find the angle by using the formula:

[tex]\Theta = tan^{-1} (\frac{V_y}{V_x} )[/tex]

Substituting the values, we have:

[tex]\Theta = tan^{-1} (\frac{6.26}{28.0} )\\\\\Theta = tan^{-1} (0.2236)\\\\\Theta = 12.60[/tex]

Angle = 12.60 degrees.

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Answer:

3.08*10^-6 m

Explanation:

Given that

Total shearing force, F = 640 N

Shear modulus, S = 1*10^9 N/m²

Height of the cylinder, L = 0.7 cm

Diameter of the cylinder, d = 4.3 cm

The solution is attached below.

We have our shear deformation to be 3.08*10^-6 m

Answer:

The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]

Explanation:

From the question we are told that  

    The mass of the bird  is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]

    The initial speed of the bird is  [tex]u_1 = 6.2 \ m/s[/tex]

     The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]

       The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]

The negative sign is because it is moving in opposite direction  to the bird

According to the principle of linear momentum conservation

       [tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]

substituting values

        [tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]

    [tex]1.51 = 0.31 v_f[/tex]

     [tex]v_f = 4.87 \ m/s[/tex]

The Final velocity of Bird =  4.87 m/s

Mass of the bird = 300 g = 0.3 kg

Velocity of bird = 6.2 m/s

Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 =  1.86 kgm/s

Mass of the insect = 10 g = 0.01 kg

Velocity of insect =   - 35 m/s

Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35  kgm/s

According to the law of conservation of momentum We can write that

In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.

The bird opens the mouth and enjoys the free lunch  hence

Let the final velocity of bird is [tex]v_f[/tex]

Initial momentum of the system = Final momentum of the system

1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )

1.51 =  [tex]v_f[/tex] 0.31

[tex]v_f[/tex] = 4.87 m/s

The Final velocity of Bird =  4.87 m/s

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Answer:

 Torque = 1191.68 N-m

Explanation:

Given data

mass m = 76 kg

standingdistance r  = 1.6 m

Solution

we get here torque  that si express as

torque  = force × distance ................1

torque  = r × F sin(theta)

and we know that

F = mg   .........2

and g = 9.8 m/s²

put here value in equation 1 we get

Torque = 76 × 1.6 × 9.8 × sin(90)

 Torque = 1191.68 N-m

"1 watt" means 1 joule of energy per second.

75 W means 75 joules/sec .

Energy = (75 Joule/sec) x (12 min) x (60 sec/min)

Energy = (75 x 12 x 60) (Joule-min-sec / sec-min)

Energy = 54,000 Joules

Answer:

M = 3.66 kg

Explanation:

Here is the complete question

A block of mass M rests on a block of mass M1 = 5.00 kg which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction  μ

k  at both surfaces equals 0.330. A force of F = 56.0 N pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed. Determine the mass of the upper block. (Express your answer to three significant figures.)

Solution

The forces on mass M are

F - μMg = Ma  (1)

The forces on mass M₁ are

F - μ(M + M₁)g = M₁a   (2) (since both weights act downwards on M)

From (1) a = (F - μMg)/M

Substituting a into (2), we have

F - μ(M + M₁)g = M₁((F - μMg)/M)  

Cross-multiplying M we have

MF - μ(M + M₁)Mg = M₁F - μMM₁g

Expanding the bracket, we have

MF - μM²g + μM₁Mg = M₁F - μMM₁g

We now collect like terms

MF - μM²g + μM₁Mg + μMM₁g = M₁F

MF - μM²g + 2μM₁Mg - M₁F = 0

- μM²g + 2μM₁Mg + MF - M₁F = 0

Dividing through by -1, we have

- μM²g + (2μM₁g + F)M - M₁F = 0

μM²g - (2μM₁g + F)M + M₁F = 0

M² - (2M₁ + F/μg)M + M₁F/μg = 0

We now have a quadratic equation in M. We now substitute the values of the variables int o the quadratic equation to get

M² - (2(5 kg) + 56 N/(0.33 × 9.8 m/s²))M + (5 kg × 56 N)/(0.33 × 9.8 m/s²) = 0

M² - (10 kg) + 56 N/3.234 m/s²)M + (5 kg × 56 N)/(3.234 m/s²) = 0

M² - (10 kg + 17.32 kg) M + 86.58 kg = 0

M² - 27.32 kg M + 86.58 kg = 0

Using the quadratic formula

with a = 1, b = -27.32 and c = 86.58,

[tex]M = \frac{-(-27.32) +/- \sqrt{(-27.32)^{2} - 4 X 1 X 86.58} }{2 X 1} \\= \frac{27.32 +/- \sqrt{746.38 - 346.32} }{2}\\= \frac{27.32 +/- \sqrt{400.06} }{2}\\= \frac{27.32 +/- 20.001 }{2}\\= \frac{27.32 + 20.001 }{2} or \frac{27.32 - 20.001 }{2}\\= \frac{47.32 }{2} or \frac{7.32 }{2}\\\\=23.66 or 3.66[/tex]

Since M cannot be greater than M₁ for M to move over M₁, we take the smaller number.

So, M = 3.66 kg

Answer:

C = $5.08

it costs $5.08 to run the LED bulb for one year if it runs for four hours a day

Explanation:

Given;

Power of Led bulb P = 12 W

Rate r = $0.29 per kilowatt-hour

Time = 4 hours per day

The number of hours used in a year is;

time t = 4 hours per day × 365 days per year

t = 1460 hours

The energy consumption of Led bulb in a year is;

E = Pt

E = 12 W × 1460 hours

E = 17520 watts hour

E = 17.52 kilowatt-hour

The cost of the energy consumption is;

C = E × rate = Er

C = 17.52 × $0.29

C = $5.08

it costs $5.08 to run the LED bulb for one year if it runs for four hours a day

Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

Given that,

Height =1.5 m

Angle = 45°

We need to find the greater speed of the ball

Using conservation of energy

[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]

[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]

Here, initial velocity and final potential energy is zero.

[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]

Put the value into the formula

[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]

[tex]v_{f}^2=2\times9.8\times1.5[/tex]

[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]

[tex]v_{f}=5.42\ m/s[/tex]

Hence, the greater speed of the ball is 5.42 m/s.