By how many degrees fahrenheit can the temperature of one metric ton of water be raised with the addition of 110 thousand btus of heat?

Answers

Answer 1
1 BTU mean 1 degree of F increased for 1 pound mass. Since the mass unit is in kg that means you need to convert it into pound. One metric ton means 1000kg and 1 kg mean 2.20462lb, then the water mass is: 1000kg x 2.20462lb/kg= 2204.62lb.

You are putting 110,000 btus of heat, so the calculation would be

Temperature increase= energy/mass = 110000btus/ 2204.62lb= 49.9 degree
Answer 2

Final answer:

The amount by which the temperature can be raised is calculated by using the specific heat capacity of water and converting BTUs to joules to find the temperature change in Celsius, which is then converted to Fahrenheit.

Explanation:

To determine by how many degrees Fahrenheit one metric ton of water can be raised with the addition of 110 thousand BTUs of heat, we need to use the specific heat capacity of water and convert the energy from BTUs to a compatible unit.

The specific heat capacity of water is 4.184 joules per gram per degree Celsius (J/g°C). Moreover, the conversion from BTUs to joules is necessary for the calculations. Since 1 BTU corresponds to approximately 1055.06 joules, 110,000 BTUs is equivalent to 110,000 × 1055.06 joules.

A metric ton of water is equal to 1,000,000 grams. Assuming the specific heat capacity of water is 4.184 J/g°C, we can set up an equation to find the temperature change in Celsius, which can then be converted to Fahrenheit.

The heat energy equation is heat energy (Q) = mass (m) × specific heat capacity (c) × temperature change (ΔT).

Q = m × c × ΔT

Substituting our values, we have:

110,000 BTUs × 1055.06 J/BTU = 1,000,000 g × 4.184 J/g°C × ΔT

This gives us the temperature change in Celsius, which can then be multiplied by 1.8 to convert to Fahrenheit.


Related Questions

The Doppler Effect means that all observers of a moving wave source detect the same wave frequency.

True or False... Please explain

Answers

False, it depends on where the source is, and where all observers are located. If some are further than other, then they will feel less of the frequency then those closer, and will not be as affected

hope this helps

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.

Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?

hang time



range



height



angle of launch

Answers

its the angle of launch, its one of many variables that will effect the over all out come of this test, angle at the ground and launch and well you get the point.

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.

Find the equivalent capacitance of a 4.20 uf capacitor and an 8.50 uf in parallel and in series

Answers

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

Final answer:

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.

Explanation:

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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Which planet do most known extrasolar planets most resemble?

Answers

The awnser is A>Most known exoplanets resemble gas giants known as "hot Jupiters" as a result of being large objects orbiting close to they're host star.
Most known exoplanets that we have detected are likely or confirmed to be gas giants. This is because one of the most used detection techniques is the Transit technique.

This technique involves looking at a distance star's light curve (the changes in the brightness of star over a period of time). Most stars dim and brighten over time by a small bit. But if the light curve shows a periodic and large dip in brightness, this is a sign that something large is passing (or transiting) in front of it.

The reason most detected exoplanets are gas giants is simply because they're the easiest to detect. They cause a larger dip in the curve than a smaller planet would. A small planet's dip are masked by the star's normal dimming and brightening.

Imagine having a flashlight shining on a wall. If you pass a large object through the light beam, it has a large shadow. If you pass a smaller object through, it has a smaller shadow. The change in the light you see on the wall is similar to what you will see in a star's light curve. Now if you pass an object closer to flashlight, as opposed to closer to the wall, this changes the shadow as well.

This is why most exoplanets are "hot Jupiters". They're large gas giants that are hot because they pass very close to the star, resulting in a larger dip that is easier to see in the curve. They also have shorter orbits so they're more likely to be seen in a few weeks or month's time.


Final answer:

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.

Explanation:

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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An automobile tire is rated to last for 35,000 miles. to an order of magnitude, through how many revolutions will it turn?

Answers

The tire will rotate about 10 million times. An automobile tire is slightly less than 2 and half feet in diameter. It's circumference is that times pi with is a bit over 3. So 2.5 * 3 = 7.5 ft as an estimate for how far the tire rolls per revolution. A mile is a bit over 5000 feet, so call it 700 revolutions per mile. For the 35000 miles, call it 7 times 5000 miles. Now 7 times 7 is a bit under 50, so call 7 * 700 = 5000. And 5000 times 5000 = 25000000. The nearest order of magnitude is 10 million. So as an order of magnitude estimate, a automobile tire will rotate about 10 million times during it's life.

What are some similarities and differences between a volcano and a folded mountain?

Answers

Folded Mountains - The term fold is used in geology when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of plastic (that is, permanent) deformation. 
Volcanic Mountains - are mountains that form when molten rock erupts onto the Earth's surface. They can either form on land or in the ocean 

What are the units of ppb mass mass/mass?

Answers

Mass concentration is typically represented as mass units of chemical per million units total mass or parts per million (ppmm), where the subscript 'm' indicates mass). The conversion factors are 109 and 1,012 for parts per billion (ppbm) and parts per trillion (pptm), respectively. hope this helps 

Elena: Y ____________________ es Patricia López, mi mejor amiga.

Answers

a.eres(b.es)c.somosd.son

Answer:

Ella es.

Explanation:

In spanish you have to conjugate every verb for each different person that you are talking about, in this example you are using the third person, in singular, since you are refering to Patricia Lopez, and you have to refer to her as She or "Ella" so you would have to say "Ella es Patricia López, mi mejor amiga"

The relationship among mass, force, and acceleration

Answers

Force is mass times acceleration, or F= m x a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass. This is Newton's Second Law of Motion.

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.


A car traveling at speed v takes distance d to stop after the brakes are applied. 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.

Answers

49d

Further explanation

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.

[tex]\boxed{u \ or \ v_i = initial \ velocity}[/tex][tex]\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}[/tex][tex]\boxed{a = acceleration \ (constant)}[/tex][tex]\boxed{d = distance \ travelled}[/tex]

The formula we follow for this problem are as follows:

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

a = acceleration (in m/s²)u = initial velocity  v = final velocityd = distance travelled

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.

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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.

Final answer:

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

Explanation:

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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Two planets have the same surface gravity, but planet b has twice the radius of planet
a. if planet a has mass m, what is the mass of planet b?

Answers

Planet A;
m = the mass 
Let r =  the radius 

Planet B:
Let M =  the mass
The radius is 2r (twice the radius of planet A)

The surface gravitational acceleration of planets A and B (they have the same surface gravity) are
[tex]g= \frac{Gm}{r^{2}} \, and \, g= \frac{GM}{(2r)^{2}} \\\\ m= \frac{M}{4} \\\\ M=4m[/tex]

Answer: The mass of planet B is 4m.

The mass of planet B is 4m

[tex]\texttt{ }[/tex]

Further explanation

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]

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[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

According to newton, the gravity of the ________ is needed to explain planetary orbits.

Answers

According to newton, the gravity of the Sun is needed to explain planetary orbits.
mass is the correct answer.

How much energy is produced when the sun converts 1kg of mass into energy?

Answers

E=mc^2 = (1kg)(299792458^2)= 89875517873681764J 

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

What is Energy?

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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Heat is most closely related to blank energy

Answers

the correct answer is thermal energy

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.

Whenever a body is in motion, there is always____ to opposite the motion

Answers

when in motion it stays in motion unless acted on by an outside force

an object travels 10 meters in the first second of travel, 10 meters again during the second second of travel, and 10 meters again during the third second of travel. what is the objects accelaration of the object?

A) 0 m/s2
B) 20m/s2
C) 10m/s2
D) 30m/s2

Answers

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.

A car starts from rest and after 7 seconds it is moving at 42 m/s. what is the carâs average acceleration?

Answers

a= (v2 - v1) / (t2 - t1)
a = (42-0)/ (7-0)

42/7 = 6

a(avrg.)= 6 m/s(sqaured)
i hope this is right, i tried if its wrong, i truly did try!!

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.

What is acceleration?

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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The vestibulo-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head. Please select the best answer from the choices provided T F

Answers

Hi Andijwiltbank

The vestibule-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head.

Answer: A) True or the first option.

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.

Calculate the amount of energy released per gram of peanut burned

Answers

Use this formula
4.2 Is the specific heat capacity of water
Mass of peanut and water should be in (g)
Increase in temperature : (final - initial)
Your answer unit is in (kJ/g)
If answer wants calorie (cal), 1 cal =4.2 J .

The approximate amount of energy released per gram of peanut burned is around 106,712 J/g.

Given that the volume of water is 200.0 mL and the density of water is approximately 1 g/mL, we have:

Mass of water = 200.0 g

The heat transfer equation:

Mass of water = Volume of water × Density of water

Change in temperature = Final temperature - Initial temperature

Change in temperature = 14.2 °C

The energy released = Heat capacity of water × Mass of water × Change in temperature / Mass of peanut burned

The mass of the peanut burned is the difference between its initial and final masses:

Mass of peanut burned = Initial mass - Final mass

Mass of peanut burned = 0.609 - 0.053

Mass of peanut burned = 0.556 g

Energy released = 4.184× 200.0 × 14.2 / 0.556

Energy released = 106,712 J/g

Rounded to a more practical number, the approximate amount of energy released per gram of peanut burned is around 106,712 J/g.

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The complete question is:

A student followed the procedure of this experiment to determine the Caloric content of a Planter's cocktail peanut. The peanut studied had a mass of 0.609 g before combustion and 0.053 g after combustion. The energy released during combustion caused a 14.2 degrees C increase in the temperature of 200.0 mL of water in the calorimeter.

Calculate the amount of energy released per gram of peanut burned.

Two students are working together on an experiment that measures the effect of different liquid fertilizers on the thickness of plants’ stems. Which is most likely to result in the greatest amount of error in their scientific experiment?

One student measures the plant stem widths by using calipers. The other student measures the widths of the same plant stems again by using a different pair of calipers.

One student measures the plant stem widths by using calipers. The other student types the measurements as they are taken into a computer spreadsheet.

One student measures liquids for the experiment by holding the flask up at eye level. The other student measures liquids for the experiment while the flask sits on the table.

One student weighs liquids for the experiment by using a calibrated scale. The other student weighs liquids for the experiment by using a different calibrated scale.

Answers

I'm pretty sure the answer is C.

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.

What is the scientific claim?

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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Is the center of each galaxy a black hole containing new galaxy?

Answers

there are black holes in the middle of big galaxies

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:

Have a good day

Viteza de aterizare a unui avion este de 45 m/s iar timpul deplasarii pe pista aerodromului este egal cu 30 s

Answers

Final answer:

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

Explanation:

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:



Speed = 45 m/sTime = 30 s



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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Final answer:

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.

Explanation:

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 scientific measure of disorder is called _____. thermodynamics matter entropy conservation

Answers

The answer is entropy

Answer:

Entropy

Explanation:

100%

a man stands on a flat surface and shoots an arrow vertically into the sky at avelocity of 60 meters per second. calculate the maximum height the arrow reached. what was the velocity of the car when it hit the ground?

Answers

a) Vi= 60 m/s
a= -9.81 m/s^2
Vf= 0m/s^2
d=?

Vf^2=Vi^2+2ad
(0m/s)^2=(60m/s)^2+2(-9.81m/s^2)d
0=3600+(-19.62)d
-3600=(-19.62)d
(-3600)/(-19.62)=(-19.62)d/(-19.62)
d=183.5m

c) Vi= 60 m/s
a= -9.81 m/s^2
Vf= 0m/s^2
t=10s
d=?

d=Vi*t+(1/2)a*t^2
d=60m/s(10s)+(1/2)(-9.81m/s^2)(10s)^2
d=600+(-4.905)(100)
d=600+(-490.5)
d=108.5m
Final answer:

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.

Explanation:

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 outer planets are mostly formed of the lighter elements, hydrogen and helium, because these elements condense at a __________ temperature, further from the Sun.
1. Lower
2.Equal
3.Higher

Answers

lower is the correct answer

the correct answer is A  *LOWER*

which amount of power is the smallest

Answers

I think this question is intended to compare units of power and choose the smallest one.
Since no choices are given, I will just tell you what each prefix equals and you can compare that to your choices:
gigawatt = 10^9 watt
megawatt = 10^6 watt
kilowatt = 10^3 watt
watt = 1 watt
So, if you have these units in your choices, the smallest amount of power will be the one having "watt" as a unit.

What s the density of a gasoline sample that has a volume of 13.3 ml and a mass of 8.86g?

Answers

In this question, you are given the volume( 13.3ml) and the mass(8.86) of the gasoline. You are asked to determine the density. Density is calculated by dividing mass by volume. Then the calculation would be:

Density= mass/ volume
Density= 8.86g/13.3ml= 0.666g/ml

Which statement(s) correctly compare the masses of protons, neutrons, and electrons?

Answers

electron. the central part of an atom containing protons and neutrons ... which of the following is necessary to calculate the atomic mass of an element? ... which of the statements correctly comparesthe relative size of an ion to its neutral atom?

A 2.1W iPod is used for 30 minutes. How much energy does it use? (3

Answers

it us 200 energy if it the right awnser


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.

Calculate the period of a satellite orbiting the moon, 94 km above the moon's surface. ignore effects of the earth. the radius of the moon is 1740 km.

Answers

The period of a satellite orbiting the moon is 7048 s

Further explanation

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]

Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441

Answer details

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.

To learn more about the Period, here:

https://brainly.com/question/28103637

#SPJ6

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