Is this statement true or false? The next generation of nuclear power plants being built in California and South Africa are even safer and more efficient than previous generations of power plants.

Answer:

0.2448 point²

Explanation:

1 gry = 1/10 line

1 line = 1/12 inch

=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch

=> 1 inch = 120 gry

1 point = 1/72 inch

=> 1 inch = 72 points

Therefore,

120 gry = 72 points

=> 1 gry = 3/5 point

Therefore,

1 gry² = (3/5)² point²

1 gry² = 9/25 point²

This means that 0.68 gry² will be:

0.68 gry² = 0.68 * 9/25 point²

=> 0.68 gry² = 0.2448 point²

The area of 0.68 gry2 is 0.2448 point2.

Given that A gry is an old English measure for length, defined as 1/10 of a line. We can write this as given below.

[tex]1 \;\rm gry = \dfrac {1}{10}\;\rm line[/tex]

A line is another old English measure for length, defined as 1/12 inch. We can write this as given below.

[tex]1 \;\rm line = \dfrac {1}{12}\;\rm inch[/tex]

Thus the value of 1 gry in inches will be,

[tex]1 \;\rm gry = \dfrac {1}{10}\times \dfrac {1}{12} \;\rm inch[/tex]

[tex]1 \;\rm gry = \dfrac {1}{120}\;\rm inch[/tex]

[tex]120 \;\rm gry = 1\;\rm inch[/tex]

A common measure for length in the publishing business is a point, defined as 1/72 inch. We can write this as given below.

[tex]1 \;\rm point = \dfrac {1}{72}\;\rm inch[/tex]

[tex]72 \;\rm point = 1 \;\rm inch[/tex]

Now we can equate the values of 1 inch, we get the expression as,

[tex]120 \;\rm gry = 72 \;\rm point[/tex]

[tex]1 \;\rm gry = \dfrac {72}{120}\;\rm point[/tex]

[tex]1 \;\rm gry = 0.6 \;\rm point[/tex]

[tex](1\;\rm gry )^2 = (0.6 \;\rm point)^2[/tex]

[tex]1\;\rm gry ^2 = 0.36 \;\rm point^2[/tex]

We have the value of 1 gry2 in terms of point2. Thus,

[tex]0.68 \;\rm gry^2 = 0.68 \times 0.36 \;\rm point^2[/tex]

[tex]0.68\;\rm gry^2 = 0.2448 \;\rm point^2[/tex]

Hence we can conclude that the area of 0.68 gry2 in points squared is 0.2448.

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Answer:

Fraction of the nickel atoms removed = 1.02 x 10 raised to power -11

Explanation:

The detailed and step by step derivation with appropriate substitution is as shown in the attached files.

To calculate the fraction of nickel atoms' electrons needed to support the weight of a nickel coin when placed 1 meter above it, we use the weight of the coin and Coulomb's law to determine the necessary charge. This charge divided by the elementary charge gives us the number of electrons needed, providing the fraction when compared to the total number of electrons in the coin.

To calculate what fraction of a nickel coin's electrons would need to be removed and placed 1.00 m above it to support its weight, we should use the concept of electrostatic force. We'll begin by determining the weight of the coin in newtons (since weight is a force). Next, we will determine the number of moles of nickel in the coin using its atomic mass, and from that, calculate the number of nickel atoms. With the atomic structure of nickel, we find the total number of electrons in the coin. Using Coulomb's Law, we can equate the electrostatic force due to the separated charges to the weight of the coin to solve for the fraction of electrons needed.

First, we need to convert the mass of the coin into weight using the formula weight (W) = mass (m) * gravity (g). Since the coin has a mass of 5.00 g (which is 0.005 kg) and the gravitational constant is approximately 9.8 m/s², the weight W is 0.049 N. Next, we use the atomic mass of nickel (58.7 u) to find the number of moles in 5.00 g of nickel. This number is given by the mass of the coin divided by the atomic mass of nickel. The number of atoms is the number of moles times Avogadro's number (approximately 6.022 × 10²³ atoms/mole).

Therefore, the total number of electrons in the 5.00 g of nickel is the number of atoms times the number of electrons per atom, which for nickel is 28. To support the coin at a distance of 1.00 m, the electrostatic force (Felectrostatic) should be equal to the weight of the coin, or 0.049 N. Using Coulomb's law, Felectrostatic = k * |q₁ * q₂| / r², where k is Coulomb's constant (approximately 8.988 × 10⁹ N·m²/C²), q₁ and q₂ represent the charges, and r is 1.00 m in our case. By solving this equation for q (assuming both q₁ and q₂ would be the same since it's the same electrons that have been separated), we can find the necessary charge to create a force equal to the weight of the coin.

After finding this charge, we divide by the elementary charge (approximately 1.602 × 10⁻¹⁹ C) to get the number of electrons needed. Dividing this number by the total number of electrons in the coin gives us the fraction of the nickel atoms' electrons needed to support the weight of the coin.

The weight of the book is calculated using the formula for weight, which is the mass of the object times the acceleration due to gravity. Given the mass of the book and the standard acceleration due to gravity, the weight of the book is 12.74 Newton.

To calculate the weight of the book, we need to remember from our physics knowledge that weight is the force exerted by an object under the influence of gravity. The formula for weight (W) is W = m*g, where m is the mass of the object, and g is the acceleration due to gravity. In most parts of the Earth, the acceleration due to gravity is approximately 9.8 m/s^2.

So in this case, the weight of the book is calculated as follows: W = 1.3 kg * 9.8 m/s^2 = 12.74 N. Therefore, the weight of the book is 12.74 Newton. Note that the coefficient of static friction is not needed to calculate the weight of the book.

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The weight of a book with a mass of 1.3 kg is calculated using the formula W = m x g, resulting in 12.74 N.

The weight of the book in Newtons can be calculated using the formula for weight, which is the mass of the object (m) multiplied by the acceleration due to gravity (g). The acceleration due to gravity is approximately 9.80 m/s² on Earth. Therefore, to find the weight (W) of the book, you use the equation W = m  imes g.

For a book with a mass of 1.3 kg, the calculation is W = 1.3 kg times 9.80 m/s² = 12.74 N.

This is the weight of the book that the woman must counteract with the normal force exerted by her hands, which is equal in magnitude to the weight of the book but in the opposite direction, as explained by Newton's third law of motion.

Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps.Which statement is TRUE?

answer choices

Marble 1 has a faster speed than Marble 2.

Marble 2 has a faster speed than Marble 1.

Both the marbles travel at the same speed.

There is not enough data to compare the speeds of marbles.

Marble 2 has a faster speed than Marble 1.

Option B.

The speed is defined as the distance covered per unit time. Here in the question, 2 balls cover equal distances in same time.

Time taken by the ball = 10 seconds.

Distance covered by 1st ball = 20 cm.

Distance covered by 2nd ball = 3cm.

So speed of the 1st ball = 2cm/sec.

Speed of the 2nd ball = 3 cm /sec.

So,it's very much evident that speed of 2nd Marble is much higher than the speed of the 1st marble.

Answer:

54.67 N

Explanation:

The total energy produced is the product of power and time duration:

E = Pt = 82 * 1 = 82 J

Which is converted from work, product of forced extended over a displacement

W = E = Fs = F*1.5 = 82

F = 82 / 1.5 = 54.67 N

So the magnitude of the force exerting on the handle is 54.67 N

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

The relationship between the amount of stretch or compression  in an elastic object and the amount of force that is applied to that object is directly proportional to each other.

Answer:

Option B

Explanation:

Ideal elastic objects are those which will retain its original form after the removal of external force on it. So this specialty of retaining their original form after deforming due to external force is termed as elasticity. And the objects exhibiting this property is termed as elastic object. The most important law in the field of elasticity which is obeyed by all elastic object is the Hooke's law. As per Hooke's law, the external force applied on any object for deformation will be directly proportional to the changes or deformation exhibited by the object due to force. In other or simple words, the strain produced due to the external force is directly proportional to the stress acting on that object.

Stress ∝ Strain

Stress is the measure of force applied on that object and strain is the measure of amount of stretch or compression due to stress or applied force.

So option B is the correct option.

Thus, the relationship between the amount of stretch or compression  in an elastic object and the amount of force that is applied to that object is directly proportional to each other.

The amount of stretch or compression in an elastic object is directly proportional to the force applied, based on the principle known as Hooke's Law. This relationship is linear, meaning that changes in force result in proportional changes in deformation.

The amount of stretch or compression in an elastic object is directly proportional to the amount of force that is applied. This principle is captured by Hooke's Law, which states when you apply force to stretch or compress a spring, the resulting spring force is proportional, and in the opposite direction, to the displacement or deformation from its position of equilibrium. Essentially, if you double the force, the amount of stretch or compression doubles, and if you halve the force, the stretch or compression is halved as well. This relationship can be represented by the equation F = -kx, where F is the force, k is the spring's force constant, and x is the displacement from the equilibrium position.

Tension and compression play a role here, where tension forces stretch the object while compression forces push the object together. Hooke's Law applies to ideal springs and other elastic objects provided the deformation does not exceed the material's elastic limit.

Answer:

Toxicity is a quantitative property

Explanation:

.

Answer:

the launched cannonball

Explanation:

In a cannon firing a cannonball, while the total momentum is conserved, the cannonball has greater kinetic energy than the recoiling cannon due to its higher velocity and the relationship between kinetic energy and velocity.

The subject of this question is Physics, specifically dealing with momentum conservation and kinetic energy in the context of a cannon firing a cannonball. The law of conservation of momentum states that the total momentum before and after an event is constant if no external forces are acting on the system. When a cannon fires a cannonball, the momentum is conserved, meaning the cannon and the cannonball together possess the same total momentum post-explosion as the stationary cannon had before firing (which is zero). Though both the cannon and the cannonball have momentum of equal magnitude but opposite direction, the distribution of kinetic energy between them is not equal.

Kinetic energy, which depends on both mass and the square of velocity, will not be conserved in the same way as momentum. The kinetic energy is given by the relation KE = (1/2)mv2. Considering the masses of the two objects, a lighter cannonball and a heavier cannon, if both the cannon and cannonball have the same magnitude of momentum, the cannonball will have a much higher velocity due to its smaller mass. Since kinetic energy depends on the square of velocity, the cannonball, with a higher velocity, will have greater kinetic energy than the heavier, slower-moving cannon. Thus, even though momentum is conserved, the kinetic energies will differ, with the cannonball gaining more kinetic energy than the recoiling cannon.

In summary, the law of conservation of energy applies, but kinetic energy is not conserved like momentum. The chemical energy stored in the gunpowder is converted into heat and the kinetic energy of both the cannon and the cannonball. Although the cannon has a larger mass, it will recoil with less velocity than the fired cannonball, which means the cannonball will have a greater kinetic energy.

Answer:

An unbalanced force is defined as a force that is not canceled out by one or more other forces acting in other directions.

Explanation:

According to Newton's first law of motion,  “An object in a constant state of motion remains in that constant state of motion unless acted upon by an unbalanced force.” An unbalanced force is a force that changes the speed, position, or direction of an object. It is not canceled out by other forces acting in other directions.

An unbalanced force is defined as a force that is not canceled out by other forces acting in different directions. This results in a net force that causes an object to accelerate, changing its state of motion.

An unbalanced force is defined as a force that is not canceled out by other forces acting in different directions. allows an object to remain at rest cause an object to accelerate in a direction opposite to the unbalanced force is not canceled out by one or more other forces acting in other directions allows an object to move at constant speed.

When forces on an object are unbalanced, it means that the net force acting on the object is not zero, leading to a change in the object's motion. This unbalanced force results in acceleration according to Newton's second law of motion, which states that [tex]\( F = ma \)[/tex] force equals mass times acceleration.

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

[tex]x=Acos\alpha \\[/tex]

we can express the velocity by the derivative of the displacement,

Hence

[tex]V=-Awsin\alpha \\[/tex]

at equilibrium, the velocity becomes

[tex]V=wA\\w=2\pi f[/tex]

Hence if we substitute values we arrive at

[tex]V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s[/tex]

Answer:

haha lol

Explanation:

sorry but i dont know this

Answer:

[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]

Explanation:

We are given that

[tex]y=(0.05x^2)[/tex]

Where x (in m)

x component of velocity, [tex]v_x=-4m/s[/tex]

x- component of acceleration,[tex]a_x=-1.2 m/s^2[/tex]

x=3 m

We have to find the y-component of velocity and acceleration at this instant.

Differentiate w.r.t. time

[tex]\frac{dy}{dt}=0.05\times 2x\frac{dx}{dt}[/tex]

[tex]\frac{dy}{dt}=0.1x\frac{dx}{dt}=0.1xv_x[/tex]

By using the formula

[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]

[tex]v_x=\frac{dx}{dt}[/tex]

Substitute the values

[tex]v_y=\frac{dy}{dt}=0.1(3)\times (-4)=-1.2m/s[/tex]

Again differentiate w.r.t. t

[tex]a_y=\frac{dv_y}{dt}=0.1v_x+0.1x\frac{d(v_x)}{dt}[/tex]

Substitute the values

[tex]a_y=0.1(-4)+0.1(3)(-1.2)=-0.4-0.36=-0.76m/s^2[/tex]

Where [tex]a_x=\frac{dv_x}{dt}[/tex]

Hence, the y- component of velocity and acceleration

[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]

To find the y components of velocity and acceleration, differentiate the equation y = 0.05x^2 with respect to time. Plug in the given values to find vy and ay at x = 3m.

To find the corresponding y components of velocity and acceleration at x = 3m, we need to differentiate the equation y = 0.05x^2 with respect to time. Differentiating y with respect to time gives vy = dy/dt = d/dt(0.05x^2) = 0.1x(dx/dt). Plugging in the given values dx/dt = -4m/s and x = 3m, we can find vy. Similarly, to find the y-component of acceleration, we differentiate vy with respect to time, which gives ay = dvy/dt = d/dt(0.1x(dx/dt)) = 0.1(dx/dt)(dx/dt) + 0.1x(d^2x/dt^2). Plugging in the given values dx/dt = -4m/s and d^2x/dt^2 = -1.2m/s^2, we can find ay.

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Final answer:

To find the displacement, subtract the distance traveled in one direction from the distance traveled in the opposite direction.

Explanation:

To solve this problem, we need to consider the concept of displacement. Displacement is a vector quantity that represents the change in position of an object. To find the displacement, we can subtract the distance traveled in one direction from the distance traveled in the opposite direction. In this case, the girl rides 5.4 km due east and then turns around and rides back to her starting point. So her displacement is 5.4 km - 5.4 km = 0 km.

The time period of the mass attached to a spring that oscillates and completes 51 full cycles in 35 seconds is approximately 0.686 seconds, and the frequency is about 1.458 Hz.

To determine the time period (T) and frequency (f) of the oscillating mass attached to a spring, we can use the relationship between the number of cycles (n), total time (t), and the time period (T = t/n). If the mass completes 51 full cycles in 35 seconds, the time period (T) is calculated as:

In conclusion, the time period of the system is approximately 0.686 seconds, and the frequency is about 1.458 Hz.

Explanation:

Time taken to rotate 2π radians = 60 seconds

Angular displacement = 2π radians

Time taken = 60 seconds

Angular displacement = Angular velocity x Time

2π = Angular velocity x 60

Angular velocity = 0.105 rad/s

Angular velocity of the second hand on a clock is 0.105 rad/s

Answer:

0.0016 cm

Explanation:

[tex]\alpha_b[/tex] = Thermal coefficient of expansion of brass = [tex]19\times 10^{-6}\ /^{\circ}C[/tex]

[tex]\alpha_g[/tex] = Thermal coefficient of expansion of glass = [tex]9\times 10^{-6}\ /^{\circ}C[/tex]

[tex]\Delta T[/tex] = Change in temperature = [tex](60-20)^{\circ}C[/tex]

[tex]R_0[/tex] = Initial radius = 4 cm

Change in radius of material is given by

[tex]R=R_0(1+\alpha\Delta T)[/tex]

Difference in radii of the lid and jar

[tex]\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm[/tex]

The size of the gap is 0.0016 cm or 0.000016 m

Final answer:

Heating the brass lid and glass jar causes the lid to expand more than the jar, creating a gap between them. To find the size of the gap, we can calculate the change in radius of the lid. Using the given information, we can find that the size of the gap is approximately 19.2 x [tex]10^-^3[/tex] cm.

Explanation:

When the brass lid and glass jar are heated in hot water, their temperatures increase. The lid, being made of metal, expands more than the glass jar due to the higher coefficient of expansion for metals. This causes the lid to loosen its grip on the jar, creating a gap between them.

To find the size of the gap, we need to calculate the difference in radius of the lid before and after heating. The inside diameter of the lid is given as 8.0 cm at 20 degrees C. We can calculate the radius using the formula r = ½ d, where d is the diameter. The radius at 20 degrees C is therefore 4.0 cm.

Now, let's calculate the radius at 60 degrees C. To do this, we need to know the coefficient of linear expansion for brass. Assuming it is 19x10^-6 /°C, we can use the formula: ∇r = αr0∇T, where ∇r is the change in radius, α is the coefficient of linear expansion, r0 is the initial radius at a reference temperature, and ∇T is the change in temperature. Plugging in the values, we get: ∇r = (19x[tex]10^-^6[/tex] /°C) x (4.0 cm) x (60 °C - 20 °C) = 19.2 x [tex]10^-^3[/tex] cm.

Therefore, the size of the gap that develops by heating the lid and jar is approximately 19.2 x [tex]10^-^3[/tex] cm.

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time  of computer A is [tex]CT_{A} =200 ps[/tex]

Clock time  of computer B is [tex]CT_{B} =250 ps[/tex]

Effective CPI of computer A is [tex]CPI_{A} =1.5[/tex]

Effective CPI of computer B is[tex]CPI_{B} =1.7[/tex]

CPU time of A is

[tex]CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec[/tex]

CPU time of B is

[tex]CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec[/tex]

Hence Computer A is Faster by [tex]\frac{425}{300} =1.41[/tex]

Computer A is 1.41 times faster than the Computer B

Answer:

While the Applied Force (force exerted) is decreasing, the velocity is decreasing

Explanation:

From Newton's second law of motion, which states that the rate of change of linear momentum is directly proportional to the applied force, and takes place in the direction of the applied force.

Momentum (P) = MV

Thus, F ∝ MV/t

where;

F is the applied force

M is the mass of the object

V is the velocity of the object

From the equation above, force is directly proportional to the velocity of the refrigerator (F∝V). That is, as the applied force is decreasing, the velocity is decreasing and vice versa.

Therefore, while the Applied Force (force exerted) is decreasing, the velocity is decreasing.

Answer:

on mastering physics its b.increasing

Explanation:

Answer: c*m on now

Explanation:

u dumb b***h

Answer:

2.5m

Explanation:

PE = MGH

50 = 2*10*H

50 = 20H

H = 2.5m

Answer:

As magnification increases, the area of the field of view increases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.

As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.

Area of the field of vision: As magnification is increased, the field of view narrows, and you can see less of the object or scene that you are watching. The area you can see reduces as a result of the field of view is smaller.

Depth of field: The range of distances inside an object or scene that simultaneously seem in focus is referred to as depth of field. The depth of field tends to get shallower as magnification gets bigger. This makes it more difficult to simultaneously examine objects at various depths because only a limited range of distances will ever be in focus.

The working distance is the separation between the thing being observed and the objective lens (or magnifying device). The working distance often reduces as magnification rises. To keep the object in focus, you must move it closer to the lens or microscope.

Higher magnification frequently necessitates additional light in order to preserve image quality and visibility. The features become more noticeable when you zoom in on a smaller region, although they might need more light to be clearly visible. At greater magnifications, insufficient light can produce a hazy or dim image, therefore more light would be required.

Hence, As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.

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The phenomenon in question describes charging by induction. The presence of a negative charge induced a redistribution of charges in two uncharged metal balls, and grounding allowed excess charges to move away. The two uncharged balls ended up positively charged after the negatively charged ball was removed and the balls separated.

The subject of this question relates to the concept of charging by induction, a common concept in Physics. When the third ball, which carries a negative charge, was brought near Ball Y and X, it caused the redistribution of charges in both balls due to the phenomenon of induction. As Ball Y was grounded, it provided a path for the electric charges to move away, thus becoming neutral again. However, Ball X retained the overall net positive charge due to the negative charge being near earlier.

After the ground wire is disconnected and the negatively charged ball removed, Ball Y and X still have net positive charges. This happens because the negative charge in the vicinity induced a redistribution of charges earlier, and when they were removed, there weren't any free charges nearby to neutralize this charge.

The separation of Ball Y and X made their charges apparent. Both balls are left with a positive charge as the excess electrons moved to the earth during grounding.

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Ball Y ends up with a positive charge, and Ball X with a negative charge due to induction caused by a nearby negatively charged ball. Grounding Ball Y neutralizes any negative charge, leaving it positively charged. The scenario exemplifies charging by induction.

Charging by Induction

This question deals with the concept of charging by induction in physics. Let's break down the scenario step-by-step:

To address the specific parts of the question:

a. Charges on Each Ball

Ball Y will have a positive charge, and Ball X will have a negative charge. This distribution occurs due to the transfer of electrons from Y to X when the negatively charged third ball is near.

b. How the Balls Get These Charges

The negatively charged third ball induces a redistribution of charges in Balls Y and X. Electrons are repelled from Ball Y towards Ball X, leaving Ball Y positively charged and Ball X negatively charged.

c. Effect of Grounding Ball Y

When Ball Y is grounded while the negatively charged ball is close, the excess electrons on Ball Y will flow into the ground, neutralizing any negative charge and leaving it positively charged. When the ground connection is removed, Ball Y remains positively charged.

This process illustrates induction as the primary method of charging without direct contact.

Answer:

Explanation:when a tape is moved past a small coil of wire due to a non uniform field,the voltage is induced in the coil. A practical example of this is a tape recorder in which a moving tape produces sound energy..

Answer:

Explanation:

Maximum height of a projectile

[tex]H_{max}=\frac{u^2\sin ^2\theta }{2g}[/tex]

(1)[tex]u=30.7\ m/s[/tex]

[tex]\theta =35.7^{\circ}[/tex]

[tex]H_1=\frac{30.7^2\times \sin ^2(35.7)}{2\times 9.8}[/tex]

[tex]H_1=16.37\ m[/tex]

(2)[tex]u=47.9\ m/s[/tex]

[tex]\theta =33.2^{\circ}[/tex]

[tex]H_2=\frac{47.9^2\times \sin ^2(33.2)}{2\times 9.8}[/tex]

[tex]H_2=35.09\ m[/tex]

(3)[tex]u=21.8\ m/s[/tex]

[tex]\theta =67.7^{\circ}[/tex]

[tex]H_3=\frac{21.8^2\times \sin ^2(67.7)}{2\times 9.8}[/tex]

[tex]H_3=20.75\ m[/tex]

(4)[tex]u=7.7\ m/s[/tex]

[tex]\theta =62.1^{\circ}[/tex]

[tex]H_4=\frac{7.7^2\times \sin ^2(62.1)}{2\times 9.8}[/tex]

[tex]H_4=2.36\ m[/tex]      

Therefore height reached by pumpkin can be arranged as

[tex]H_4<H_1<H_3<H_2[/tex]

Final answer:

The maximum height reached by the pumpkin depends on the vertical component of the initial velocity, which is highest for the trial with a speed of 21.8m/s and angle of 67.7 degrees and lowest for the speed of 7.7m/s at an angle of 62.1 degrees.

Explanation:

To rank the maximum height reached by the pumpkin in each trial, we can rely on the principle of projectile motion. The maximum height depends primarily on the vertical component of the initial velocity and can be compared by isolating this component using the initial speeds and angles given.

Using the formula for the vertical component of velocity: [tex]Vy = V * sin(\(\theta\)),[/tex] where V is the initial speed and [tex]\(\theta\)[/tex] is the launch angle. Then, the maximum height H can be roughly compared using [tex]H \(\propto Vy^2[/tex], assuming gravitational acceleration is constant and air resistance is negligible.

Answer:

[tex]\displaystyle V_l=V\ \frac{\rho_o -\rho_w }{\rho_l-\rho_w}[/tex]

Explanation:

Density

It's a physical magnitude that relates the mass of an object with the volume it occupies. The formula is

[tex]\displaystyle \rho = \frac{m}{V}[/tex]

Equivalently

[tex]m=\rho V[/tex]

The density of water is 1 gr/ml, 1 Kg/lt, 1000 kg/m^3 or any equivalent unit.  

Assume the wood sphere has a volume V and a density [tex]\rho_w[/tex], thus

[tex]m_w=\rho_w V[/tex]

A hole is to be drilled inside the wood so it means part of its volume will be filled with lead, which mass will be

[tex]m_l=\rho_l V_l[/tex]

The volume of wood is the total volume V minus the (unknown) volume of lead, thus

[tex]V_w=V-V_l[/tex]

The total mass of the modified sphere is

[tex]m_s=m_w+m_l=\rho_w V_w+\rho_l V_l[/tex]

Substituting Vw

[tex]m_s=m_w+m_l=\rho_w (V-V_l)+\rho_l V_l[/tex]

Operating

[tex]m_s=m_w+m_l=\rho_w V-\rho_w V_l+\rho_l V_l[/tex]

[tex]m_s=m_w+m_l=\rho_w V+ V_l(\rho_l -\rho_w)[/tex]

The total volume of the sphere doesn't change, which means  

[tex]V_s=V[/tex]

The new density of the modified sphere is

[tex]\displaystyle \rho_s = \frac{m_s}{V}[/tex]

[tex]\displaystyle \rho_s = \frac{\rho_w V+ V_l(\rho_l -\rho_w)}{V}[/tex]

This density must be equal to the density of water [tex]\rho_o[/tex]

[tex]\displaystyle \rho_o = \frac{\rho_w V+ V_l(\rho_l -\rho_w)}{V}[/tex]

Operating and solving for vl

[tex]\displaystyle V_l=\frac{\rho_o V-\rho_w V}{\rho_l-\rho_w}[/tex]

Or equivalently

[tex]\displaystyle \boxed{V_l=V\ \frac{\rho_o -\rho_w }{\rho_l-\rho_w}}[/tex]

The equation would be enough for the volume of wood to keep the sphere neutrally buoyant would be [tex]Vt= Vp0-Pw/Pt-Pw[/tex]

The formula for the physical magnitude is p= m/v

Given that the density of water is 1 gr/ml, 1 Kg/lt, 1000 kg/m^3 or any equivalent unit.  

We would assume the wood sphere has a volume V and a density Pw,

[tex]Mw= PwV[/tex]

Since a hole is to be drilled inside the wood so it means part of its volume will be filled with lead, the mass will be [tex]Mt= PtVt[/tex]

The volume of wood is the total volume V minus the (unknown) volume of lead, thus

[tex]Vw= V- Vt[/tex]

Hence, after further evaluation, the density would be [tex]Vt= V Po- Pw/Pt- Pw[/tex]

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Answer:

The solution is in the attached files below

Explanation:

Answer:

a) Average velocity = 156.7m/s

b) Speed of blast wave is 0.457 times the speed of sound wave

Explanation:

The step by step calculations is as shown in the attached file.

Answer:

Velocity,c = 156.67m/s

Explanation:  Altitude, d = 23.5km or 23500m

Time duration, T = 2 minutes + 30 seconds = (120 + 30) seconds = 150 seconds

a) Frequency, F = 1/T= 1/150 Hz

Hence, Velocity, c = d X F = 23500 X 1/150

∴ c = 156.67 m/s

b)  The speed of sound increases as the wave moves from a lighter medium to denser medium. While the of the superbolide meteor decreases from a lighter medium to denser medium.

Answer:

B. When the racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket

Explanation:

Newton's third law of motion states that for every action there is an equal and opposite reaction. Therefore, when Krista's racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket.

When the racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket. Therefore option B is correct.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction.

In this scenario, when Krista hits the tennis ball with her racket (the action), the tennis ball exerts an equal but opposite force back onto the racket (the reaction).

This is depicted by option B, which correctly states that the tennis ball applies an equal but opposite force to the racket. This law underscores the idea that forces always come in pairs, and the forces are exerted on two different objects.

The force Krista applies to the ball is met with an equal force from the ball back onto her racket, causing the ball to fly in the opposite direction.

Therefore option B is correct.

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Answer

given,

initial speed = 18 m/s

distance, d = 130 m

reaction time, t_r = 0.50 s

a) distance traveled in the reaction time

   d= s x t_r

   d=18 x 0.5

  d = 9 m

distance of the traffic light, d = 130 - 9 = 121 m

b) deceleration

  [tex]a = \dfrac{v^2-u^2}{2s}[/tex]

  [tex]a = \dfrac{0^2-18^2}{2\times 121}[/tex]

         a = -1.34 m/s²

c) Using equation of motion

   v = u + a t

   0 = 18 - 1.34 t

    t = 13.43 s

Total time, T = 13.43 + 0.5 = 13.93 s

Part A. The distance from the intersection is 121 m.

Part B. The acceleration required is 1.34 m/s2

Part C. The total time to stop at the signal is 13.93 s.

Given that the initial speed is 18 m/s. The distance is 130 m from an intersection when the traffic light turns red. The rejection time is 0.50 s.

Part A

The distance traveled in rejection time is given below.

distance traveled in rejection time =  speed [tex]\times[/tex] rejection time

[tex]d_r = s\times t_r[/tex]

[tex]d_r = 18\times 0.50[/tex]

[tex]d_r = 9\;\rm m[/tex]

Hence the distance from the traffic light is given below.

[tex]d = 130 -d_r[/tex]

[tex]d = 130 -9[/tex]

[tex]d = 121\;\rm m[/tex]

The distance from the intersection is 121 m.

Part B

The acceleration can be calculated as given below.

[tex]a = \dfrac {v^2 - u^2}{2s}[/tex]

Where u is the initial speed, v is the final speed and s is the distance from the intersection.

[tex]a = \dfrac {0^2 - (18)^2}{2\times 121}[/tex]

[tex]a = - 1.34 \;\rm m/s^2[/tex]

The acceleration required is 1.34 m/s2. Here negative sign shows that this is deacceleration.

Part C

The time to stop at the signal can be calculated as given below.

[tex]v = u +at[/tex]

[tex]0 = 18 - 1.34 \times t[/tex]

[tex]t = \dfrac {18}{1.34}[/tex]

[tex]t = 13.43 \;\rm s[/tex]

Total Time = [tex]13.43 + 0.5[/tex]

Total Time =  13.93 s

Hence the total time to stop at the signal is 13.93 s.

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