Which statement would most likely be found in an advertisement from a
cell phone provider

Answers

Answer 1
Answer:

Answer:B

Explanation:


Related Questions

A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?
Hello I would like to know if there is a course or a book in which to learn quantum mechanics is easier, I am currently studying quantum mechanics, but it is a little complicated because I have some mathematical doubts, but nevertheless I would like to know if there is a course or book that explains and develops the problems so that they are understandable. and if there is no book or course I would appreciate an honest and complete answer, thank you very much and greetings.Sorry I know that this is not a problem but what better to ask the experts do not believe.
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. What is the ratio of the radius of A to that of B
How can socialism impact populations?
After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.0 m. Determine the magnitude of the impulse delivered to the ball by the floor.

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, m_1 = 4.5 kg

Mass of the raven, m_2 = 1.2 kg

Initial speed of raven, v_1 = 4.5 m/s

Final speed of raven, v_2 = - 1.5 m/s

Moment of Inertia of the gate about the axis passing through one end:

I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2

Angular momentum of the gate, L = I \omega

L = 5.33 \omega

Using the law of conservation of angular momentum:

m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n

In the calorimetry experiment which energy will be calculated during the heat exchange if water is used?

Answers

From the coffee cup to the thermometer

The assumption behind the science of calorimetry is that the energy gained or lost by the water is equal to the energy lost or gained by the object under study. So if an attempt is being made to determine the specific heat of fusion of ice using a coffee cup calorimeter, then the assumption is that the energy gained by the ice when melting is equal to the energy lost by the surrounding water. It is assumed that there is a heat exchange between the iceand the water in the cup and that no other objects are involved in the heat exchanged. This statement could be placed in equation form as

Qice = - Qsurroundings = -Qcalorimeter

The role of the Styrofoam in a coffee cup calorimeter is that it reduces the amount of heat exchange between the water in the coffee cup and the surrounding air. The value of a lid on the coffee cup is that it also reduces the amount of heat exchange between the water and the surrounding air. The more that these other heat exchanges are reduced, the more true that the above mathematical equation will be. Any error analysis of a calorimetry experiment must take into consideration the flow of heat from system to calorimeter to other parts of the surroundings. And any design of a calorimeter experiment must give attention to reducing the exchanges of heat between the calorimeter contents and the surroundings.

The energy calculated while dealing with the calorimeter experiment are the latent heat of vaporization, latent heat of fusion and the heat required to change the temperature of the substances.

Further Explanation:

The calorimeter works on the principle of conservation of energy. The amount of heat given by one part of the system is equal to the amount of heat gained by another part provided that the calorimeter does not loss any heat to the environment.

Consider that ice is mixed with water at some temperature. Then the water being at higher temperature losses heat to the ice at lower temperature. The ice gains the heat from the water and the system reaches an equilibrium at which the solution of ice and water has the same amount of energy at a particular temperature.

The different types of energies dealt with in the calorimetry experiment are as follows:

Latent heat of fusion:

The amount of energy required by a body when it is melted from its frozen state or freezes from its melted state is termed as the latent heat of fusion.

For example:  

The small amount of ice is mixed with water in a calorimeter. Here, the ice requires the latent heat of fusion that leads to the melting of ice and converts it into water.

Latent heat of vaporization:

The amount of heat required to convert one gram of liquid to vapor without raising its temperature is known as latent heat of vaporization.

For example:

The water is boiling at   in a calorimeter. Here, the water requires latent heat of vaporization which leads to the vaporization of water and convert it into vapors.

Thus, the latent heat of fusion, latent heat of vaporization and the heat required to change the temperature of the substance are the energies measured with the calorimeter.

Learn more:

1.  Transnational kinetic energy brainly.com/question/9078768.

2.  Expansion of gas brainly.com/question/9979757.

3. Conservation of momentum brainly.com/question/9484203.

Answer Details:

Grade: College

Subject: Physics

Chapter: Heat and Energy

Keywords:

Heat, energy, calorimeter, latent heat, vaporization, fusion, experiment, temperature, melting, boiling, liquid, vapor, evaporation, condensation, freeze.

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

=  Tensile \ strength* Area

On substituting the values, we get

=  (3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)

=  (3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)

=  21195 \ N

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

Learn more about Tensile Strength here:

brainly.com/question/14293634

#SPJ3

An electron experiences a magnetic force of magnitude 4.60 x 10^-15 N when moving at an angle of 60.0° with respect to a magnetic field of magnitude 3.50 x 10^-3 T. Find the speed of the electron.

Answers

Answer:

9.49 × 10⁶ m/s

Explanation:

Data provided in the question:

Magnitude of Magnetic force, F = 4.60 × 10⁻¹⁵ N

Angle, θ = 60°

Magnitude of magnetic field, B = 3.50 × 10⁻³ T

Now,

we know

F = qVBsin(θ)

here ,

q is the charge of electron = 1.6 × 10⁻¹⁹ V

V is the speed of electron

Therefore,

4.60 × 10⁻¹⁵ = ( 1.6 × 10⁻¹⁹ ) × V × ( 3.50 × 10⁻³ ) × sin(60°)

or

V = 9.49 × 10⁶ m/s

Answer:

9.49*10^6(m/s)

Explanation:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the

Answers

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)

k = 1.4

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K

Work done is given as;

W = (1)/(2) *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

A celestial body moving in an ellipical orbit around a star

Answers

A celestial body moving in an elliptical orbit around a star is a planet.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

Other Questions