The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?

Answer:

A - 5

B - 1

C - 4

D -2

Explanation:

I don't have one i just know...

The fire blanket is a tightly woven fabric. The spill containment kit consists of absorbent material. Fire extinguishers control small fires and the safety data sheet provides chemical, health, and safety information.

(a) The fire blanket is a blanket, which may be quickly thrown over a fire to snuff out the flames, and comprises fire-resistant materials.

Hence, option (a) matches with option (5)

(b) In order to contain a chemical spill, absorbent items like pads, socks, or booms are frequently included in spill containment kits.

Hence, option (b) correctly matches with option (1).

(c) A fire extinguisher is a tool used to put out small fires during emergencies.

Hence, option (c) correctly matches with option (4).

(d) A Safety Data Sheet (SDS) gives in-depth details regarding a specific chemical or chemical mixture. It provides information about the physical characteristics of the chemical, any potential risks, safe handling and storage practices, emergency response strategies, and more.

Hence, option (d) correctly matches option (2).

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Answer:

[tex]R1/R2=\frac{2}{3}[/tex]

Explanation:

From the question we are told that:

1st's Length [tex]l=L[/tex]

1st's radius [tex]r=r[/tex]

2nd's Length [tex]l=6L[/tex]

2nd's radius [tex]r=2r[/tex]

Generally the equation for Resistance R is mathematically given by

 [tex]R=\frac{\rho L}{\pi r^2}[/tex]

Therefore

 [tex]R_1=\frac{\rho L}{\pi r^2}[/tex]

And

 [tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]

Therefore

 [tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]

 [tex]R1/R2=\frac{2}{3}[/tex]

Answer:

X = 0.6

Explanation:

The resistance of the unknown resistor can be found by using the formula of the Wheatstone bridge:

[tex]\frac{M}{N}=\frac{P}{X}\\\\\frac{850}{15} = \frac{33.48}{X}\\\\X = \frac{(33.48)(15)}{850}[/tex]

X = 0.6

Hence, the unknown value of resistance is found to be 0.6 units.

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

Answer:

Gravity between you and the sun

Answer:

Graph B express the magnetic relationship of magnetic flux and electronic flow

"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.

The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)

Using Newton's second law, we set up the following equations.

• p = 100 N

∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0

∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0

• P = 350 N

∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0

∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0

(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)

Solve for n and N :

n = p sin(θ) + mg cos(θ)

N = P sin(θ) - mg cos(θ)

Substitute these into the corresponding equations containing µ, and solve for µ :

µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))

µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

Next, you would set these equal and solve for m :

(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

...

Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with

m ≈ 36.5 kg

µ ≈ 0.256

The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.

From the given,

The force that opposes the crate by sliding is P = 100N

In X-axis, the sum of forces is zero.

ΣF = 0

Pcosθ - mgsinθ-Ff = 0

Ff = Pcosθ - mgsinθ

In Y-axis

Psinθ - mgcosθ - N = 0

N = Psinθ-mgcosθ

Frictional force, Ff = μN, μ is the coefficient of friction

Ff = μN

Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0

μ = mgsin30-Pcos30/Psin30+mgcos30 ------1

The block is sliding with the horizontal force, F = 350N

X-axis

P₂cosθ - mgsinθ-Ff = 0

Y-axis

P₂sinθ - mgcosθ - N = 0

N = P₂sinθ-mgcosθ

μ = P₂cos30-mgsin30/P₂sin30-mgcos30   -----2

Equate equations 1 and 2

mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30

4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49

41.7m² + 123m - 1.516×10⁴ = 0

-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)

83.4m² - 2207m -3.03×10⁴ = 0

m= 36.4 kg

Hence, the mass of the crate is 36.4 Kg.

Substitute the value of m in equation 1,

μ = 4.905(36.4) - 86.6 / 50 + 8.49

μ  = 0.256

Thus, the coefficient of static friction is 0.256.

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Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]

Answer:

7.84 m/s

Explanation:

Height, h = 2 m

Initial velocity, u = 10 m/s

Angle, A = 80°

(a) Let the time taken to go to the net is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]

Time cannot be negative.

So, t = 0.2 s

The vertical velocity at t = 0.2 s is

v = u + at

v = 10 sin 80 - 9.8 x0.2

v = 9.8 - 1.96 = 7.84 m/s

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

The question is incomplete. The complete question is :

A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?

Solution :

Given :

Speed of the bullet train, v = 90 mi/h

                                            = [tex]$90 \times \frac{5280}{3600}$[/tex]

                                            = 132 ft/s

Time = 15 minutes

        = 15 x 60

        = 900 s

Acceleration from rest,

[tex]$a(t) = 4 \ ft/s^2$[/tex]

[tex]$v(t) = 4t + C$[/tex]

Since, v(0) = 0, then C = 0, so velocity is

v(t) = 4t ft/s

Then find the position function,

[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]

      [tex]$=2t^2+C$[/tex]

It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :

[tex]$s(t) = 2t^2$[/tex]

Time to get maximum cruising speed is :

4t = 132

t = 33 s

Distance travelled (at cruising speed) by speed to get the remaining distance travelled.

[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]

Total distance travelled, converting back to miles,

[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]

                      = 22.9125 mi

Therefore, the distance travelled is 22.9125 miles

Answer:

Reflective

Explanation:

The radiation pressure of the wave that totally absorbed is given by;

[tex]P_{abs}= \frac{I}{C}[/tex]

and While the radiation pressure of the wave totally reflected is given by;

[tex]P_{ref}= \frac{2I}{C}[/tex]

Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.

Answer:

A. (b)

B. (a)

Explanation:

The electric dipole moment is the product of charge and the length of the dipole.

The torque on the dipole placed in the external electric field is given by

torque = p E sin A

where, p is the electric dipole moment, E is the electric field, A is the angle between the field and dipole moment.

When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium.

When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.

So, the option (b) is correct.

Teh energy is given by

U = - p  E cos A

When the angle A is zero , the potential energy is negative and it is minimum.

In this exercise we have to use the knowledge about dipole to be able to mark the correct alternative for each question, in this way we find that:

A) Letter b

B) Letter a

So knowing that the electric dipole moment is the product of charge and the length of the dipole and the torque on the dipole placed in the external electric field is given by:

[tex]torque = p E sin (A)[/tex]

where:

When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium. When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.

Now the energy is given by:

[tex]U = - p E cos (A)[/tex]

We can say that when the angle A is zero , the potential energy is negative and it is minimum.

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Answer:

Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N

Answer:

10 m/s2 or 70 newtons.

Explanation:

............................

............

Answer:

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water the weight of the two beakers is the same

Explanation:

The beaker weight is

 beaker A

          W_total = W_ empty + W_water

Beaker B

            W_total = W_ empty + W_water + W_roca

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same

Answer:

20 km/hr

Explanation:

Distance = 10km

Time = 30 minutes = 1/2 hour

Average Speed = Total distance / Total Time Taken

                           = 10 ÷  1/2

                           = 10 x 2

                           = 20 km/hr

Average speed = (distance covered) / (time to cover the distance)

Average speed = (10 km) / (30 minutes)

Average speed =  1/3 km/min

Most people would probably want to see it in a more convenient, more familiar unit, such as km/hour or m/second.

(10 km / 30 min) x (60 min / hour) = (10 x 60 / 30) (km-min / min-hour)

Average speed = 20 km/hour

AvgSpd = (10 km / 30 min) x (1,000 m / km) x (min / 60 sec)

AvgSpd = (10x1,000 / 30x60) (km-m-min / min-km-sec)

Averge Speed =  5.56 m/s

Answer:

(d)

Explanation:

because dU = Q -W so ,that the option d(D) is correct

Answer:

0.000064 cubic meters.

Explanation:

Given the following data;

Length of side = 4 centimeters

Conversion:

100 centimeters = 1 meters

4 cm = 4/100 = 0.04 meters

To find the volume of cuboid;

Mathematically, the volume of a cuboid is given by the formula;

Volume of cuboid = length * width * height

However, when all the sides are equal the formula is;

Volume of cuboid = L³

Volume of cuboid = 0.04³

Volume of cuboid = 0.000064 cubic meters.

Explanation:

First, we convert the energy from eV to Joules:

[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]

[tex]= 4.64×10^{-19}\:\text{eV}[/tex]

We know from definition that

[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]

so the wavelength of the photon is

[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

Answer:

Explanation:

The energy for an isothermal expansion can be computed as:

[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)

However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:

[tex]V_b = 2V_a[/tex]

Equation (1) can be written as:

[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]

Also, in a Carnot engine, the efficiency can be computed as:

[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]

[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]

In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]

relating the above two equations together, we have:

[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]

Making the work done (W) the subject:

[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]

From equation (1):

[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]

[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

If we consider the adiabatic expansion as well:

[tex]PV^y[/tex] = constant

i.e.

[tex]P_bV_b^y = P_cV_c^y[/tex]

From ideal gas PV = nRT

we can have:

[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]

[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]

From the question, let us recall  aw we are being informed that:

If the volumes changes by a factor = 5.7

Then, it implies that:

[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]

∴

[tex]T_H = T_L (5.7)^{y-1}[/tex]

In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]

As such:

[tex]T_H = T_L (5.7)^{1.6-1}[/tex]

[tex]T_H = T_L (5.7)^{0.67}[/tex]

Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]

From in the question:

W = 930 J and the moles = 1.90

using 8.314 as constant

Then:

[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]

[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]

[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]

[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]

From [tex]T_H = T_L (5.7)^{0.67}[/tex]

[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]

[tex]\mathbf{T_H \simeq 125K}[/tex]

Answer: 19.15 meters on the waves away from the survivor.

Explanation:

Answer:

[tex]M_b=6kg[/tex]

Explanation:

From the question we are told that:

Coefficient of restitution [tex]\mu=1.00[/tex]

Mass A [tex]M_a=6kg[/tex]

Initial Velocity of A [tex]U_a=6m/s[/tex]

Initial Velocity of B [tex]U_b=0m/s[/tex]

Generally the equation for Coefficient of restitution is mathematically given by

 [tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]

 [tex]1=\frac{v_B}{6}[/tex]

 [tex]V_b=6*1[/tex]

 [tex]V_b=6m/s[/tex]

Generally the equation for conservation of linear momentum  is mathematically given by

 [tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]

 [tex]6*6+=M_b*6[/tex]

 [tex]M_b=6kg[/tex]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

Explanation:

someone to check if the answer is correct

5 9 . 4

- 3 7 . 2

2 2 . 2

Explanation:

Use the algorithm method.

5 9 . 4

- 3 7 . 2

2 2 . 2

2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.

22.2

22.2

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.