A 2.80 kg mass is dropped from a height of 4.50 m. Find its potential energy when it reaches the ground.

it would be the 3rd one. so C

Answer:

The apparent depth d = 19.8495 cm

Explanation:

The equation for apparent depth can be expressed as:

[tex]d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}[/tex]

here;

[tex]d_1 = d_2 = 15 \ cm[/tex]

[tex]\mu_1[/tex] = refractive index in the first liquid = 1.75

[tex]\mu_2[/tex] = refractive index in the second liqquid= 1.33

∴

[tex]d = \dfrac{15}{1.75}+\dfrac{15}{1.33}[/tex]

[tex]d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})[/tex]

[tex]d = 15( 0.5714 +0.7519)[/tex]

d = 15(1.3233 ) cm

d = 19.8495 cm

Answer: The correct answer is C

Explanation:

Answer:

I think it's C

Explanation:

Molar mass (M) is equal to the mass of one mole of a particular element or compound; as such, molar masses are expressed in units of grams per mole (g mol–1) and are often referred to as molecular weights. The molar mass of a particular gas is therefore equal to the mass of a single particle of that gas multiplied by Avogadro’s number (6.02 x 1023 ). To find the molar mass of a mixture of gases, you need to take into account the molar mass of each gas in the mixture, as well as their relative proportion.

Answer:

Its pulling down on the book

Explanation: if it was pushing up the book would be floating and the other choices don't make sense

Answer:

b.  The objects velocity is decreasing.

Explanation:

I know this because if you read a graph left to right it will tell u whether it is increasing or decreasing.

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)

Answer:

Explanation:

This is a common question on Khan Academy's "Calculating change in kinetic energy from a force" practice exercises. (AP Physics 1)

Here's the real question without all the formatting:

A truck pushes a pile of dirt in the horizontal direction with a force of 20 N to a distance of 15 m then the kinetic energy of the dirt is 300 J.

In physics, the term "work" refers to the measurement of energy transfer that takes place when an item is moved over a distance by an externally applied, at least a portion of which is applied inside the direction of the displacement.

The duration of the path is multiplied by the component of a force acting along the path to calculate work if the forces are constant. The work W is theoretically equivalent to the pressure f times the range d, or W = fd, to represent this idea. Work is done when a force is applied at an angle of to a displacement, or W = fd cos.

The work done, [tex]W = Force * Displacement[/tex]

W = 15× 20

W = 300 J

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[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]

[tex]\\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]

[tex]\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]

[tex]\\[/tex]

☯ For left penetration (s₂)

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]

[tex]\\[/tex]

[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]

Answer:

[tex]\theta = 16.70 ^{\circ}[/tex]

Explanation:

The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.  

Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.

Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.

We are given the mass of the block (kg) and we know that g = 9.8 m/s².

Now we can use this force in the equation:

Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.

Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].

Evaluate this equation by taking the inverse tangent of both sides of the equation.

The minimum angle at which the block will begin to slide is about 16.70 degrees.

Answer:

The maximum angular velocity is 20 rad/s

Explanation:

Given;

torque, τ = 10 N

maximum mechanical power, P = 200 J/s

The output power of the pmdc is given as;

P = τω

where;

P is the maximum mechanical power

ω is the maximum angular velocity

ω = P / τ

ω = (200) / (10)

ω = 20 rad/s

Therefore, the maximum angular velocity is 20 rad/s

Answer:

the period of the pendulum will be twice as long.

Explanation:

because i looked it up

Answer: 2

Explanation:

:}

Answer:

3.33 Hz

Explanation:

The first step is to calculate the wavelength

= speed × period

= 13.3 × 0.3

= 3.99

Therefore the frequency of the wave can be calculated as follows

= speed/wavelength

= 13.3/3.99

= 3.33 Hz

Answer:

Knowing that these metals are infact good conductors of electricity we can infer that metals are able to hold and conduct certain temperatures. Another thing we can infer is that these good conductors can be used in connection to transferring energy or electricity.

Answer:

The value is   [tex]E_t = 17958.2 \ J[/tex]

Explanation:

From the question we are told that

     The atmospheric temperature is [tex]T_a = 720 \ K[/tex]

       The molar mass of carbon dioxide is  [tex]Z = 44 \ g/mol[/tex]

        The pressure is [tex]P = 92 \ atm =[/tex]

      The number of moles is [tex]n = 3 \ moles[/tex]

Generally the translational kinetic energy is mathematically represented as

        [tex]E_t = \frac{f}{2} * n * R T[/tex]

       Here  R is the gas constant with value  [tex]R = 8.314 J\cdot K^{-1}\cdot mol^{-1}[/tex]

Generally the degree of freedom of carbon dioxide in terms of  translational motion is  f =  3

     So

           [tex]E_t = \frac{ 3}{2} * 2 * 8.314 * 720[/tex]

=>         [tex]E_t = 17958.2 \ J[/tex]

Answer:

Δx = 39.1 m

Explanation:

        [tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

        [tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]        

The car will travel a distance of 39.1 m before its stops.

To solve the problem above, use the equations of motion below.

Equation:

Where:

From the question,

Given:

Substitute these values into equation 1

Solve for s

Hence, The car will travel a distance of 39.1 m before its stops.

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Answer:

D. 2.5 x 10^4

Explanation:

10 to the power of 4 is 10,000.

10,000 x 2.5 = 25,000

D. 2.5 x 10⁴ is the scientific notation of Ellie records.

Scientific Notation is the expression of a number n in the form a∗10ᵇ. where a is an integer such that 1≤|a|<10. and b is an integer too. It is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.

According to the question,

The given number is 25,000 m

i.e.

25 x 10³

Since we have to put the decimal  between 2 and 5

It becomes,

2.5  x 10⁴

or

10 to the power of 4 is 10,000.

10,000 x 2.5  = 25,000

Therefore,

The answer is 2.25 x 10⁻⁶ M

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Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg  , where m = mass of body , g= acceleration due to gravity .

Given: Weight  = 475 N

[tex]g= 9.8\ m/s^2[/tex]

Substitute all values in formula ,  we get

[tex]475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg[/tex]

Hence, his mass = 48.47 kg.

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]

Answer:

c. because A will land first becuase its heavier

Explanation:

Answer: 54.88 meters/s

Explanation:

The final velocity will be calculated by using the formula:

v = u + at

where,

v = final velocity

u = initial velocity = 0

a = 9.8

t = 5.6

Therefore, we slot the value back into the formula. This will be:

v = u + at

v = 0 + (9.8 × 5.6)

v = 0 + 54.88

v = 54.88 meters per second

Therefore, the final velocity is 54.88m/s

Answer:

600 KPa.

Explanation:

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

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

Answer:

Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.

Explanation:

Hope this helps

Answer:

The answer is "1557 meters".

Explanation:

speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]

[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]

Answer:

The value is [tex]T_2 =416.9 \ K[/tex]

Explanation:

From the question we are told that

     The mass of the aluminum baking sheet is  [tex]m = 225 \ g = 0.225 \ kg[/tex]

      The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]

       The initial  temperature is  [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]

Generally the heat absorbed is mathematically represented as

         [tex]Q = m * c_a * [T_2 - T_1][/tex]

Here  [tex]c_a[/tex] is the specific heat capacity of  aluminum with value  [tex]c_a = 897 \ J / kg \cdot K[/tex]

So

           [tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]

=>         [tex]T_2 - 298 = 118.915[/tex]

=>          [tex]T_2 =416.9 \ K[/tex]

Answer:

2.57 seconds  (rounded to 2.6 Seconds)

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

Before we can solve this question we need to create a formula that calculates the final speed. The formula will be the following,

Where:

Vf is the final Velocity

Vi is the initial velocity

A is the acceleration

t is the time in seconds

Now that we have the formula we can plug in the values given to us in the question and solve for the amount of time (t).

Finally, we can see that it would take the cyclist 2.57 seconds (approximately) to reach a speed of 18 m/s

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:

the placement of the ammeter in the circuit

Explanation: