Temperature°F = (9/5 * °C) + 32°
°C = 5/9 * (°F - 32°)
1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:
1. 5 cm = ________ mm
2. 83 cm = ________ m
3. 459 L = _______ ml
4. .378 Kg = ______ g
5. 45°F = ________ °C
6. 80°C = _________ °F

Answers

Answer 1
Answer:
5cm = 50mm
2.83cm = 0.0283m
3.459l = 3459ml
4.378kg = 4378g
5.45f =  - 47.79c
6.80c = 44.24f

Related Questions

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.
A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?
(a) A woman climbing the Washington Monument metabolizes 6.00×102kJ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?
A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B
A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Calculate the de Broglie wavelength of an electron and a one-ton car, both moving with speed of 100 km/hour. Based on your calculation could you predict which will behave like a "quantum particle" and why. Please explain each step in words and detail.

Answers

Answer :

(a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is 2.38*10^(-38)\ m

Explanation :

Given that,

Speed = 100 km/hr

Mass of car = 1 ton

(a). We need to calculate the wavelength of electron

Using formula of wavelength

\lambda_(e)=(h)/(p)

\lambda_(e)=(h)/(mv)

Put the value into the formula

\lambda_(e)=(6.63*10^(-34))/(9.1*10^(-31)*100*(5)/(18))

\lambda=0.00002622

\lambda=26.22*10^(-6)\ m

\lambda=26.22\ \mu m

(II).  We need to calculate the wavelength of car

Using formula of wavelength again

\lambda_(e)=(6.63*10^(-34))/(1000*100*(5)/(18))

\lambda=2.38*10^(-38)\ m

The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.

Therefore, electron is quantum particle and car is classical.

Hence, (a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is 2.38*10^(-38)\ m.

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a distance of meters.

Answers

So lets fill out what we have first:

Vi or initial velocity = 20 m/s

Acceleration or a = 4 m/s^2

Time for the motion = 10s

Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:

Δx = Vi(t) + 0.5at²

Plug all of our information in:

Δx = (20)(10) + (0.5)(4)(100)

Δx = 400 m

Answer:

400 m

Explanation:

answer on ed

A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.

Answers

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→   θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?

Answers

Answer:

a) x =  5.48 10⁻² m and b)  0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

     Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

     Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

    A = √ (2Em / k)

    A = √ (2 0.12 / 80)

   A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

      Em = K = ½ m v²

Final

     Em = Ke = ½ k x²

     ½ m v² = ½ k x²

     x = √(m/k) v

     x = 2 √(0.50 /800.0)

     x = 0.05 m

Answer:

a) The greatest extension of the spring is 0.055 m

b) The spring compress 0.05 m

Explanation:

Please look at the solution in the attached Word file

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force .

Answers

Complete question:

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.

Answer:

(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N

Explanation:

Given;

initial velocity of the electron, v_0 = +6.18 x 10⁵ m/s

final velocity of the electron, v_f = 2.59 x 10⁶ m/s

the distance traveled by the electron, d = 0.0708 m

The first electric force, F_1 = 8.87*10^(-17) \ N

(a) The net force of the electron is given as;

∑F = F₁ - F₂ = ma

where;

a is the acceleration of the electron

a = (v_f^2 -v_0^2)/(2d) \n\na = ((2.59*10^6)^2 -(6.18*10^5)^2)/(2(0.0708))\n\na = 4.468*10^(13) \ m/s^2

∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)

∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ is given as;

∑F = F₁ - F₂

F₂ = F₁ - ∑F

F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷

F₂ = 4.8 x 10⁻¹⁷ N

Final answer:

The problem involves calculating the acceleration of an electron, then using Newton's second law to find the net force on the electron. This is used to find the magnitude of a second electric force acting on the electron.

Explanation:

First, we can calculate the acceleration of the electron using the formula a = Δv/Δt, where 'a' is acceleration, 'Δv' is the change in velocity, and 'Δt' is the change in time. In this case, Δv = vf - vi = 2.59 x 106 m/s - 6.18 x 105 m/s = 1.972 x 106 m/s. The time taken by the electron to travel 0.0708 m can be found using the equation d = vi t + 0.5 a t₂. We use these values to get Δt which we use to find 'a'.

Next, let's use Newton's second law F = ma to find the net force acting on the electron. The only forces acting on the electron are electric forces, and we know one them is 8.87 x 10-17 N. If we designate this known force as F₁ then the total force F total = F₁ + F₂ where F₂ is the unknown electric force.

Finally, we can find F₂ = F total - F₁. This gives the magnitude of the second electric force.

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An internal explosion breaks an object, initially at rest,intotwo pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?

Answers

Answer:

4500 J and 3000 J

Explanation:

According to conservation of momentum

      0 = m_1 V_1 + m_2 V_2

Given that m_2 = 1.5 m_1 , so

    V_1 = -1.5 V_2

  the kinetic energy of each piece is

    K_2= (1)/(2) m_2v_2^2

    K_1= (1)/(2) m_1v_1^2

substituting the value of V1 in the above equation

    K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2

  Given that

         K_1 + k_2 = 7500 J

       1.5 K_2 + K_2 = 7500

         K_2 = 7500 / 2.5

               = 3000 J

this is the KE of heavier mass

      K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

Final answer:

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.

Explanation:

The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

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