prove mathematically :
1. v = u + at
2. s = ut+1*2 at ​

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

Explanation:

someone to check if the answer is correct

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]

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:

(d)

Explanation:

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

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:

(I)

[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]

(ii)

[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]

Answer:

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

Answer:

Gravity between you and the sun

Answer:

m = 0.057 kg = 57 g

Explanation:

Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water

[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]

where,

E = Energy added to water = 145 KJ

m = mass of water = ?

C = specific heat capacity of water = 4.2 KJ/kg.°C

ΔT = change in temperature = 100°C - 35°C = 65°C

H = Latent heat of vaporization of water = 2260 KJ/kg

Therefore,

[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]

m = 0.057 kg = 57 g

The mass of water that can be heated is equal to 0.527 kilograms.

Given the following data:

Scientific data:

To calculate the mass of water that can be heated:

Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.

Mathematically, this is given by this expression:

[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]

Making m the subject of formula, we have:

[tex]m=\frac{E}{c\theta + H}[/tex]

Substituting the parameters into the formula, we have;

[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]

Mass, m = 0.527 kilograms.

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Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3

Answer:

True

Explanation:

A magnetohydrodynamic drive or MHD accelerator is a method which is used for propelling the vehicles using only by applying the electric and magnetic fields. It has no moving parts. It accelerates an electrically conductive propellant (liquid or gas) with magnetohydrodynamics.

Its working principle is same as an electric motor except that in an MHD drive, the moving rotor is replaced by the fluid acting directly as the propellant.

An MHD accelerator is reversible.

So,  the statement is true.

Answer:

Option C. 0.73 g/cm³

Explanation:

From the question given above, the following data were obtained:

Mass = 80 g

Area (A) = 10000 cm²

Thickness = 0.11 mm

Density =?

Next, we shall convert 0.11 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

0.11 mm = 0.11 mm × 1 cm / 10 mm

0.11 mm = 0.011 cm

Thus, 0.11 mm is equivalent to 0.011 cm.

Next, we shall determine the volume of the paper. This can be obtained as follow:

Area (A) = 10000 cm²

Thickness = 0.011 cm

Volume =?

Volume = Area × Thickness

Volume = 10000 × 0.011

Volume = 110 cm³

Finally, we shall determine the density of the paper. This can be obtained as follow:

Mass = 80 g

Volume = 110 cm³

Density =?

Density = mass / volume

Density = 80 / 110

Density = 0.73 g/cm³

Therefore the density of the paper is 0.73 g/cm³

We have that  the distance to Betelgeuse, and the uncertainty in that measurement is

From the Question we are told that

Betelgeuse (in Orion)  has a parallax of 0.00451 + 0.00080

Generally

[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]

Where

Parallax [tex]P =0.00451[/tex]

Uncertainty [tex]U = 0.00080[/tex]

Generally the equation for the distance  is mathematically given as

[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]

Therefore

[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]

[tex]d=(221.7\pm39.33)pc[/tex]

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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

Answer:

because both petrol and diesel are oil

Explanation:

oil floats on water that's why if we will try to extinguish with water so the fire will float on water

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please mark methe brainest

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:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

Answer:

a)   p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the  constructor equation

         [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distances to the object and the image, respectively

the power of the lens is

         P = 1 / f

         f = 1 / P

         f = 1 / 2.25

         f = 0.4444 m

the distance to the object is

         [tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]

the distance to the image is

          q = 85 -2

           q = 83 cm

we must have all the magnitudes in the same units

           f = 0.4444 m = 44.44 cm

we calculate

           [tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]

           1 / p = 0.010454

            p = 95.66 cm

b) if they were contact lenses

            q = 85 cm

            [tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]

             1 / p = 0.107375

             p = 93.13 cm

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

Let m be the mass of the second car, so the first car's mass is 2m.

Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.

Let u and v be the speeds of the first car and the second car, respectively. At the start,

• the first car has kinetic energy

K/2 = 1/2 (2m) u ² = mu ²   ==>   K = 2mu ²

• the second car starts with kinetic energy

K = 1/2 mv ²

It follows that

2mu ² = 1/2 mv ²

==>   4u ² = v ²

When their speeds are both increased by 2.76 m/s,

• the first car now has kinetic energy

1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²

• the second car now has kinetic energy

1/2 m (v + 2.76 m/s)²

These two kinetic energies are equal, so

m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²

==>   2 (u + 2.76 m/s)² = (v + 2.76 m/s)²

Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.

Energy = (power) x (time)

Energy = (0.69 W) x (0.36 sec)

Energy = 0.25 Joule

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.

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]

Explanation:

someone to check if the answer is correct

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 - 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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