One sphere has a radius of 5.10 cm; another has a radius of 5.00 cm. What is the difference in volume (in cubic centimeters) between two spheres?

We are given that there are two capacitors:

4.20 uf

8.50 uf

A. In parallel

The equivalent capacitance of capacitors is similar to calculating that of a current, they are added when in parallel. Therefore the equivalent capacitance is:

equivalent capacitance = C1 + C2 + C3 + ...

equivalent capacitance = 4.20 uf + 8.50 uf

equivalent capacitance = 12.70 uf

B. In series

When the capacitors are placed in series, the formula for the equivalent capacitance is:

equivalent capacitance = 1 / (1/C1 + 1/C2 + 1/C3 + ...)

equivalent capacitance = 1 / (1/4.20 + 1/8.50)

equivalent capacitance = 2.81 uf

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances, so we get 12.70 µF. For capacitors in series, we calculate using the reciprocal formula and get approximately 2.79 µF.

To solve this problem, we need to know the formulas to calculate capacitance in a series and parallel circuit. For capacitors in parallel, the total equivalent capacitance (Ceq) is the sum of the capacitors, so Ceq = C1 + C2, where C1 and C2 are the capacitances of your individual capacitors. Here, C1 is 4.20 µF and C2 is 8.50 µF. So, Ceq = 4.20 µF + 8.50 µF which equals 12.70 µF.

For capacitors in a series connection, the total equivalent capacitance is calculated using the reciprocal formula: 1/Ceq = 1/C1 + 1/C2. So, 1/Ceq = 1/4.20 µF + 1/8.50 µF. Solving for Ceq gives us an equivalent capacitance of about 2.79 µF.

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Answer:

Entropy

Explanation:

100%

The landing speed of the airplane is 1.5 m/s.

The question is asking for the landing speed of an airplane given its landing time and the distance traveled on the airstrip. To find the landing speed, we can use the formula:

Speed = Distance / Time

Plugging in the values given:

Using the formula:

Landing Speed = 45 m/s / 30 s = 1.5 m/s

Therefore, the landing speed of the airplane is 1.5 m/s.

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The question pertains to the deceleration of an airplane after it lands, specifically calculating the final velocity given an initial velocity and a constant deceleration over a certain time span.

The student's question is related to the deceleration of an airplane after landing. An airplane that lands with an initial velocity of 70.0 m/s, and then decelerates at 1.50 m/s2 for 40.0 s, will have a final velocity that can be calculated using the following formula: final velocity = initial velocity + (acceleration × time). Here, the acceleration is negative because it is opposite the direction of motion (deceleration).

In the example given, if an airplane lands with a velocity of 70.0 m/s and decelerates at 1.50 m/s2, the final velocity after 40 seconds can be found by:

Final Velocity = 70.0 m/s - (1.50 m/s2 × 40.0 s) = 70.0 m/s - 60.0 m/s = 10.0 m/s. The plane would slow down to 10.0 m/s before heading to the terminal.

The acceleration of the car is the ratio of its change in velocity to the change in time. The acceleration of the car is  6 m/s².

Acceleration of a moving body is the rate of change in its velocity. Acceleration is a vector quantity and is characterised by a magnitude and direction. The change in magnitude or direction or both in velocity results in an acceleration of for the body.

Acceleration is the ratio  of the change in a velocity to the change in time. Hence the expression relating the time, velocity and acceleration is given by,

a = U - V/ t0 - T

Where, u be the initial velocity and v is final velocity. T0 is initial time and T be the final time .

Given that the car is starts from rest. Hence initial time and velocity is zero. The final velocity is 42 m/s and time is 7 seconds.

Then, acceleration = velocity / time

                          =  42 m/s / 7 s = 6 m/s².

Therefore, the acceleration of the car is  6 m/s².

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Answer:

False

Explanation:

The observers will detect different frequencies based whether the source is approaching them or receding from them. Based on the location of observer, the source may either approach the observer or recede from the observer.

If the source is receding

[tex]f_{observed}=f_{source}(\frac{v}{v+v_{source}} )[/tex]

if the source is approaching

[tex]f_{observed}=f_{source}(\frac{v}{v- v_{source}} )[/tex]

For example, consider a scenario where an ambulance moves from west to east.

There are two observers one located west of the ambulance and other located east of the ambulance The ambulance is receding from the west side observer and approaching the east side observer.

The observer west of the ambulance will hear lower pitch and the observer eat of the ambulance will hear higher pitch.

One student uses a flask held up at eye level to measure liquids for the experiment.

The second student measuring the liquids for the experiment while the flask is on the table will probably cause the largest amount of inaccuracy, hence option C is the right response.

Scientific claims are statements made in science based on an experiment.

As given in the problem statement One student uses a flask held up at eye level to measure liquids for the experiment.

The other student measuring the experiment's liquids while the flask is on the table is most likely to introduce the highest amount of mistakes.

Thus, therefore the correct answer is option C.

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Answer:

Heat is most closely related to THERMAL ENERGY.

Explanation:

As we know that heat is a dynamic nature of energy in which heat will flow from high temperature to low temperature.

As we know that thermal energy is the energy due to the kinetic energy of all molecules of the given system

This thermal energy is exchanged from one system to other system only due to the temperature gradient.

So here heat always flows from high temperature system to low temperature system.

So here correct answer would be

Heat is most closely related to THERMAL ENERGY.

Answer:

The statement is true. The vestibulo-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head.

Explanation:

The vestibulo-ocular reflex is a reflection of ocular movement that stabilizes the image in the retina during the movement of the head, producing an eye movement in the opposite direction to the movement of the head, conserving the image in the center of the visual field. For example, when the head moves to the right, the eyes move to the left, and vice versa. As there are slight movements of the head at all times, the RVO is very important to stabilize the vision: patients who have damaged RVO find it difficult to read printed media, because they can not stabilize the eyes during small tremors of the head.

Final answer:

The object's velocity remains constant as it covers the same distance each second, implying that the acceleration is zero (0 m/s^2). So the correct option is A.

Explanation:

The question relates to constant velocity and acceleration and to identifying the acceleration of an object moving 10 meters each second for three seconds. Since the object travels the same distance in each second, it means that its velocity remains constant. Therefore, the rate of change of velocity, which is acceleration, is zero. The correct answer to the question is A) 0 m/s2. To understand this, consider the definition of acceleration: a change in velocity over some time. Since there is no change in velocity here (the object keeps moving at a constant rate), there is no acceleration.

Answer:

Energy used, E = 3780 Joules

Explanation:

It is given that,

Power, P = 2.1 W

Time, t = 30 minutes = 1800 seconds

We need to find the energy used by the iPod. The product of power and time is called energy used i.e.

E = P × t

[tex]E=2.1\ W\times 1800\ s[/tex]

E = 3780 Joules.

So, the energy used by the iPod is 3780 Joules. Hence, this is the required solution.

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

The energy produced when the sun converts one kg of mass into energy is equal to 89875517873681764 J .

Energy is the capacity to do work in physics. Potential, kinetic, thermal, electrical, chemical, radioactive, and other kinds are just a few of the many possible manifestations. Additionally, heat and effort are two other ways that energy is transported from one body to another. Following a transfer, energy is always classified according to its nature. Thus, the generation of thermal energy may result from the transmission of heat, whereas the generation of mechanical energy may result from the performance of work.

Motion is a property of all types of energy. Anybody that is in motion, for instance, has kinetic energy. Even while at rest, a tensioned object like a bow or spring has the capacity to move; this is because of the way it is designed.

Now, let's find out the energy,

E=mc²

= (1 kg) × (299792458)²

= 89875517873681764 J

Therefore, the energy produced by the sun is 89875517873681764 J.

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The maximum height reached by the arrow shot vertically at 60 meters per second is 183.67 meters. Upon returning to the ground, the arrow's velocity will be -60 meters per second.

The student is dealing with a problem that involves projectile motion, a topic within physics. To calculate the maximum height the arrow reached when shot vertically into the sky at a velocity of 60 meters per second, we use the kinematic equation that ignores air resistance:

Maximum height (h) = (v^2) / (2g)

where v is the initial velocity and g is the acceleration due to gravity (9.8 m/s^2). Substituting the given values:

h = (60^2) / (2*9.8) = 3600 / 19.6 = 183.67 meters

The maximum height reached by the arrow is 183.67 meters. When the arrow hits the ground, the velocity will be the same as the initial velocity in magnitude but in the opposite direction, so the velocity of the arrow when it hits the ground is -60 meters per second (the negative sign indicates the direction is downward).

The period of a satellite orbiting the moon is 7048 s

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

To find the period of the satellite can be carried out in the following way:

[tex]F = G \frac{m_{moon} \times m_{satellite}}{R^2}[/tex]

[tex]m_{satellite} \times \omega^2 \times R = G \frac{m_{moon} \times m_{satellite}}{R^2}[/tex]

[tex]\omega^2 \times R = G \frac{m_{moon}}{R^2}[/tex]

[tex]\omega^2 = G \frac{m_{moon}}{R^3}[/tex]

[tex]\omega = \sqrt{G \frac{m_{moon}}{R^3}}[/tex]

[tex]\omega = \sqrt{6.67 \times 10^{-11} \frac{7.35 \times 10^{22}}{(1.834 \times 10^6)^3}}[/tex]

[tex]\omega = 8.91 \times 10^{-4}[/tex]

[tex]\frac{2 \pi}{T} = 8.91 \times 10^{-4}[/tex]

[tex]T = \frac{2 \pi}{8.91 \times 10^{-4}}[/tex]

[tex]\boxed {T \approx 7048 ~ seconds}[/tex]

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

The period of a satellite orbiting the moon, 94 km above the moon's surface is 71903 seconds.

Given:
Distance, d = 94 km

Radius, R = 1740 km

Average distance = R + D

Average distance = 1740 km + 94 km = 1834 km

Convert average distance into meters:

Average distance = 1834 km × 1000 = 1,834,000 meters

The period is:

Period² = (4π² / G)  (Average distance)³

Substituting the values, we get:

Period² = (4π² / (6.67430 × 10⁻¹¹ N·m²/kg²)) ×  (1,834,000 meters)³

Period² = 5180781218 seconds²

Period = √(5180781218 seconds²) = 71903 seconds

Hence, the period of a satellite orbiting the moon, 94 km above the moon's surface is 71903 seconds.

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Most known extrasolar planets resemble Earth, with a significant number also resembling what we call "super Earths", or planets with two to ten times the mass of Earth. Our solar system may be unusual, with many systems potentially hosting Earth-like planets closer to their respective stars.

The student asks which planet most known extrasolar planets most resemble. Based on the available data from missions like Kepler, it is clear that the majority of these extrasolar planets or exoplanets most closely resemble Earth. Analyses of the data show that small planets, like the terrestrial ones in our system, are much more common than giant ones. Also relatively common are the so-called "super Earths", which are planets with two to ten times the mass of our planet. In this respect, it is important to note that our solar system may actually be unusual in the organization and types of its planets, and that a large number of planetary systems in our galaxy could potentially host Earth-like planets closer to their star.

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The mass of planet B is 4m

[tex]\texttt{ }[/tex]

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

gravitational acceleration of Planet A = g₁ = g

gravitational acceleration of Planet B = g₂ = g

radius of Planet A = R₁ = R

radius of Planet B = R₂ = 2R

mass of Planet A = M₁ = m

Asked:

mass of Planet B = M₂ = ?

Solution:

We will compare the gravitational acceleration of the two planets as follows:

[tex]g_1 : g_2 = G\frac{M_1}{(R_1)^2} : G\frac{M_2}{(R_2)^2}[/tex]

[tex]g_1 : g_2 = \frac{M_1}{(R_1)^2} : \frac{M_2}{(R_2)^2}[/tex]

[tex]g : g = \frac{m}{(R)^2} : \frac{M_2}{(2R)^2}[/tex]

[tex]1 : 1 = m : \frac{1}{4}M_2[/tex]

[tex]\frac{1}{4}M_2 = m[/tex]

[tex]\boxed{M_2 = 4m}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

the correct answer is A  *LOWER*

49d

This case is about uniformly accelerated motion.

Given:

The initial speed was v takes distance d to stop after the brakes are applied.

Question:

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

The Process:

The list of variables to be considered is as follows.

The formula we follow for this problem are as follows:

[tex]\boxed{ \ v^2 = u^2 + 2ad \ }[/tex]

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

[tex]\boxed{ \ 0 = v^2 + 2ad \ }[/tex]

[tex]\boxed{ \ v^2 = -2ad \ }[/tex]

Both sides are divided by -2d, we get [tex]\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }[/tex]

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

[tex]\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }[/tex]

Here d' is the stopping distance that we want to look for.

[tex]\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }[/tex]

We crossed out 2 in above and below.

[tex]\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }[/tex]

We multiply both sides by d.

[tex]\boxed{ \ v^2 d' = 49.0v^2 d \ }[/tex]

We crossed out v^2 on both sides.

[tex]\boxed{\boxed{ \ d' = 49.0d \ }}[/tex]

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

The stopping distance when a car is initially traveling at speed 7.0v is (7.0v^2)/d.

To determine the stopping distance when a car is initially traveling at speed 7.0v, we can use the fact that the acceleration due to braking is the same in both cases. Since the distance it takes to stop at speed v is d, we can set up a proportion: v/d = 7.0v/x, where x represents the stopping distance when the car is initially traveling at speed 7.0v. We can solve for x by cross multiplying and then dividing: x = (7.0v^2)/d. Therefore, the stopping distance when the car is initially traveling at speed 7.0v is (7.0v^2)/d.

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Answer:

range and angle of launch

Explanation:

Let the horizontal component of velocity be VCosθ = Vx and the vertical component of velocity is VSinθ = Vy.

Here , θ be the angle of projection.

So, if we divide the VSinθ by VCosθ, we get the value of Tanθ, i.e., we get the value of angle of launch.

Now we know that the formula fr the range is given by

[tex]R = \frac{V^{2}\times Sin2\Theta }{g}[/tex]

We can write it as

[tex]R = \frac{V^{2}\times 2\times Sin\Theta \times Cos\Theta }{g}[/tex]

Again we rewrite as

[tex]R = \frac{2\times Vx\times Vy}{g}[/tex]

It shows that the range also depends on the horizontal and vertical component of velocity.

Answer:

8.5 m/s

Explanation:

Answer:

Black holes are a class of astronomical objects that have undergone gravitational collapse, leaving behind spheroidal regions of space from which nothing can escape, not even light. Observational evidence indicates that almost every large galaxy has a supermassive black hole at its center.

Explanation:

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