16. An object has a gravitational potential energy 41,772.5 Jof and has a mass of 1550 kg. How high is it
above the ground?
Plz help

Answer:

a=∆v/∆t

Explanation:

The definition of Acceleration is the change in velocity in a given time. So this means you first calculate ∆v (Change in velocity), and you calculate ∆t which is the time taken to apply that change in velocity. Then you find a= ∆v/∆t. This gives us the equation of Acceleration.

Answer:

Explanation:

Answer:

Heat pump needs 1.965 megajoules each hour to run.

Explanation:

The Coefficient of Performance ([tex]COP[/tex]), no unit, of a Carnot's heat pump is:

[tex]COP = \frac{Q_{H}}{W}[/tex] (1)

Where:

[tex]Q_{H}[/tex] - Heat received by the building, measured in megajoules.

[tex]W[/tex] - Work needed to run the heat pump, measured in megajoules.

If heat pump is a Carnot engine working in reverse, then the amount of work needed to run the heat pump is the least possible work. If we know that [tex]Q_{H} = 7.27\,MJ[/tex] and [tex]COP = 3.70[/tex], then the amount needed by the heat pump each hour is:

[tex]W = \frac{Q_{H}}{COP}[/tex]

[tex]W = \frac{7.27\,MJ}{3.70}[/tex]

[tex]W = 1.965\,MJ[/tex]

Heat pump needs 1.965 megajoules each hour to run.

Answer:

Acceleration due to gravity, g = 2.68m/s²

Explanation:

Given the following data;

Period = 5.14s

Length = 0.25m

To find acceleration due to gravity, g;

[tex] Period, T = 2 \pi \sqrt {lg} [/tex]

Substituting into the equation, we have;

[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]

[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]

[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]

[tex] 0.8180 = \sqrt {0.25g} [/tex]

Taking the square of both sides

[tex] 0.8180^{2} = 0.25g [/tex]

[tex] 0.6691 = 0.25*g[/tex]

[tex] g = \frac {0.6691}{0.25} [/tex]

Acceleration due to gravity, g = 2.68m/s²

Answer:

Option D

2Na + Cl₂ —> 2NaCl

Explanation:

We'll begin by stating the law of conservation of matter.

The law of conservation of matter states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.

For an equation to comply with the law of conservation of matter, the number of atoms of each element must be the same on both side of the equation. This simply means that the equation must be balanced!

NOTE: An unbalanced equation simply means matter has been created or destroyed.

Now, we shall determine which equation is balanced. This can be obtained as follow:

For Option A

Na + Cl₂ —> 2NaCl

Reactant:

Na = 1

Cl = 2

Product:

Na = 2

Cl = 2

Thus, the equation is not balanced!

For Option B

2Na + 2Cl₂ —> 2NaCl

Reactant:

Na = 2

Cl = 4

Product:

Na = 2

Cl = 2

Thus, the equation is not balanced!

For Option C

2Na + Cl₂ —> NaCl

Reactant:

Na = 2

Cl = 2

Product:

Na = 1

Cl = 1

Thus, the equation is not balanced!

For Option D

2Na + Cl₂ —> 2NaCl

Reactant:

Na = 2

Cl = 2

Product:

Na = 2

Cl = 2

Thus, the equation is balanced!

From the above illustrations, only option D has a balanced equation. Thus, option D illustrate the law of conservation of matter.

Answer:

3.2N

Explanation:

Given parameters:

Coefficient of kinetic friction  = 0.43

Weight of box  = 7.4N

Unknown:

Force of kinetic friction  = ?

Solution:

The force of kinetic friction is given as:

        Force of kinetic friction  = UN

U is the coefficient of friction

N is the weight

  Force of kinetic friction  = 0.43 x 7.4  = 3.2N

Answer:

[tex]0.3\ \text{m}^3/\text{kg}[/tex]

[tex]36\ \text{kJ}[/tex]

[tex]18\ \text{kJ/kg}[/tex]

Explanation:

V = Volume of air = [tex]0.6\ \text{m}^3[/tex]

P = Power = 10 W

t = Time = 1 hour

m = Mass of air = 2 kg

Specific volume is given by

[tex]v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}[/tex]

The specific volume at the final state is [tex]0.3\ \text{m}^3/\text{kg}[/tex]

Work done is given by

[tex]W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}[/tex]

The energy transfer by work, is [tex]36\ \text{kJ}[/tex]

Change in specific internal energy is given by

[tex]\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}[/tex]

The change in specific internal energy of the air is [tex]18\ \text{kJ/kg}[/tex]

Answer:

Explanation:

60 mph = 60 x 1760 x 3 / (60 x 60) ft /s

speed of car , v = 88 ft /s

kinetic energy of car = 1/2 m v²

= .5 x 3000 x 88²

= 11616 x 10³ poundal - foot

Potential energy = mgh

= 3000 x 32 x 100

=  9600 x 10³ poundal - foot

Total energy = potential energy + kinetic energy

= ( 11616 + 9600 )x 10³

= 21216 x 10³ poundal - foot .

This energy is dissipated as heat when brakes are applied on the car to stop the car .

Answer:

L = 1.11 x [tex]10^{6}[/tex] m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x [tex]10^{5}[/tex] m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x [tex]10^{-2}[/tex] m

Length = ?

So,

we know that,

U = 1/2 C Δ[tex]v^{2}[/tex]

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δ[tex]v^{2}[/tex]  

52000 J = (0.5) x (C) x ([tex]610^{2}[/tex])

C = 0.28 F

And we also know that,

C = [tex]\frac{K*E*A}{d}[/tex]

E = 8.85 x [tex]10 ^{-12}[/tex]

K = 3.7

A = 0.20 x L

d = 2.6 x [tex]10^{5}[/tex] m

Plugging in the values into the formula, we get:

0.28 = [tex]\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }[/tex]

Solving for L, we get:

L = 1.11 x [tex]10^{6}[/tex] m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.  

Answer:

Endothermic, because the reactants are lower in energy (C)

Explanation:

From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that  when the enthalpy change  Eproducts is gretater than  Ereactants, the reaction is said to be endothermic.

Answer:

NE DIYON INGILIZ MISIN SEN

Answer:

v = 4 m/s

Explanation:

Given that,

Initial speed of hare, u = 0

The hare accelerates uniformly at a rate of 1 m/s² for 4 seconds.

We need to find how fast is the hare going 1.6 seconds after it starts. Let the speed be v. So,

v = u+at

Substitute all the values,

v = 0+1×4

v = 4 m/s

So, the required speed of the hare is 4 m/s after it starts.

Answer:

Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.

Answer:

Charge at rest only produces electric field. Moving charge produces both electric field and magnetic field.

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Solution :

The volume rate of flow is given by : R = 5.0 L/s

                                                                 [tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]

The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]

∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]

             = 0.637 meter per second

Then the speed of the water at wider section,

[tex]$R=A_1v_1$[/tex]

Similarly, the speed of water at narrow pipe.

The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m

[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]

             = 2.55 meter per sec

Now from Bernoulli's theorem,

[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]

[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]

    [tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]

    = 50 kPa + 3.05 kPa

    = 53.05 kPa

or 53000 Pa

This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.

The pressure reading in the wide section is "53.05 KPa".

First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.

[tex]V = A_1v_1[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m

A₁ = Area of narrow section = πr₁² = π(0.025 m)²

v₁ = velocity at narrow section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]

Similarly,

[tex]V = A_2v_2[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m

A₂ = Area of wide section = πr₁² = π(0.05 m)²

v₂ = velocity at wide section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]

Now, we will use Bernoulli's Theorem to find out the pressure wide section.

[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]

where,

[tex]\rho[/tex] = density of water = 1000 kg/m³

P₁ = pressure in narrow section = 50 KPa = 50000 Pa

P₂ = pressure in wide section = ?

Therefore,

[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]

P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa

P₂ = 53046.45 Pa = 53.05 KPa

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Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

We have that the speed  must be at the speed below if they want to just hit their target

From the Question we are told that

Distance [tex]d=50m[/tex]

Height [tex]h=70m[/tex]

Constant Velocity [tex]v= 7.0 m/s[/tex]

Generally the equation for the time  is mathematically given as

[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]

Therefore

The velocity required to make horizontal movement is

[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]

Given that

Velocity on the Vertical axis is

[tex]v_y=7m/s[/tex]

Velocity on the  horizontal axis is

[tex]v_x=5m/s[/tex]

Therefore resultant speed

[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]

[tex]v_r=8.6023m/s[/tex]

In conclusion

[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target

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Answer:

Large; small.

Explanation:

A telescope can be defined as an optical instrument or device which comprises of a curved mirror and lenses used for viewing distant objects i.e objects that are very far away such as stars and other planetary bodies. The first telescope was invented by Sir Isaac Newton.

To have the highest magnification in a telescope, the focal length of the objective lens should be large and the focal length of the eyepiece lens should be small.

This ultimately implies that, the eyepiece lens has a small focal length while the objective lens has a large focal length.

Answer:

Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion.

Explanation:

Hello!

[tex]\large\boxed{4000 N}[/tex]

Use the following equation to solve for the net force (N):

∑F = m × a

Plug in the given mass (kg) and speed (m)

∑F = 2000 * 2

Simplify:

∑F = 4000 N

Answer:2000 cm³

Here, pressure remains constant.

So, b the gas law

V/V' = T/ T'

1000 / V' = 300 / 600

V' = 2000 cm³

Explanation:also pls mark brainliest

Answer:

the Potential Energy is 2.304 × 10⁻¹⁶ J

Explanation:  

Given the data in the data in the question;

The expression for the electric potential energy between the charges can be expressed as follows;

PE = qV ------equ 1

where q is the charge and V is the electric potential

Also the formula for electric potential due to point a point in a field is;

V = kq /  r -------equ 2

where k is the electrostatic constant and r is the distance form the charged particle

input equation 2 into 1

PE = q × kq /  r

PE = kq²/r ------- equ 3

so we substitute into equation 3; 1.00×10⁻¹² for r, 9.00×10⁹ for k( constant ) and 1.60×10⁻¹⁹ for q( charge )

PE = ((9.00×10⁹) (1.60×10⁻¹⁹)²) / 1.00×10⁻¹²

PE = 2.304 × 10⁻²⁸ / 1.00×10⁻¹²

PE = 2.304 × 10⁻¹⁶ J

Therefore, the Potential Energy is 2.304 × 10⁻¹⁶ J

Answer:

Explanation:

Pressure due to water column as deep as 10994 meters can be given by the following expression

Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .

Pressure = 10994 x 10³ x 9.8

= 10.77 x 10⁷ N / m²

Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .

Effective surface = 3.14 x 0.1²

= .0314 m²

Total force = pressure due to water column x effective surface

= 10.77 x 10⁷  x .0314 N.

= 33.82 x 10⁵ N .

Answer:

There is net loss of gravitational energy .

Explanation:

When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .

Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .

Answer:

A. oceanic crust

Explanation:

I remember that the ocean is said to cover 71% of the Earth's surface. If you look at a globe, notice that most all the surface is blue like the ocean.

One other note: the surface is the Earth's outermost layer. Think of it this way: surface implies the top of something, something exposed to the outside.

Therefore, the answer is A. Hope this helps you understand the question more! Have a great day, 'kay?

Answer:45 m/s north

Explanation:

Answer:

6.29 μC/m³

Explanation:

Volume charge density is the quantity of charge per unit volume.

The direction of the electric field was not specified, therefore the volume charge density (ρ) is given by:

2πRLE = ρπR²L/ε₀

ρ = 2Eε₀ / R

Where E = electric field strength = 16 kN/C = 16 * 10³ N/C, R = radius of rod = 4.5 cm = 0.045 m, ε₀ = relative permittivity of free space = 8.85 * 10⁻¹² C² / Nm²

Therefore:

ρ = 2(16 * 10³ N/C)(8.85 * 10⁻¹² C²/Nm²) / 0.045 m = 6.29 * 10⁻⁶ C/m³

ρ = 6.29 μC/m³

Answer:

A. the average acceleration was the same

Explanation:

Acceleration is calculated by finding the difference of the initial velocity from the final velocity (on impact, usually 0) and then dividing by the amount of time that took place. If we assume that both balls were thrown at the same initial force, and ended up hitting the wall at the same time then we can say that the average acceleration was the same. If the initial velocity was not the same then we would need the initial velocity of each ball in order to calculate the acceleration of each object and determine which had a greater acceleration.

Answer:

the work done by the field is positive and the potential energy of the electron field system decreases

Explanation:

This exercise asks to find the work and the potential energy of an electron in an electric field.

Work is defined by

         W = F .d = F d cos θ

the electric force is

          F_e = q E

         W = q E d cos θ

since the charge of the electron is negative the force is in the opposite direction to the electric field

          W = - e E d

we select the direction to the right is positive, point i is to the left of point f,

therefore the work moving from point i to point F has two possibilities

* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated

* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too.  The electron moves away from this point, this is in accordance with the indicated movement

In the latter case the electric field lines go from f to i point, therefore the Work is positive

Now let's examine the potential energy

            ΔU = - q E .d

so we see that this definition is related to work,

            ΔU = -W

Therefore, as the work is positive, the power energy must decrease

When reviewing the different answers, the correct ones are:

the work done by the field is positive and the potential energy of the electron field system decreases

The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.

An electron moves from point i to point f, in the direction of a uniform electric field, then  the potential energy of the electron can be calculated s given below.

[tex]\Delta V=-qEd[/tex]

Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.

The work done by the electron in the uniform electric field can be calculated as,

[tex]W = F\times d \times cos\theta[/tex]

Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and  for uniform electric field, the value of [tex]\theta[/tex] is zero.

Hence  [tex]W=F\times d\times 1\\W=F \times d[/tex]

Electric force  [tex]F = q E[/tex]

By substituting the value of electric force on the above formula,

[tex]W = qEd[/tex]

Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.

[tex]W = -\Delta V[/tex]

The work done by the electron is negative and the potential energy of the electron increases.

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Answer:

D

Explanation:

A and C are balanced, B has a resultant force of 5N right, and D has a resultant force of 20N right.