Answer:
Eris
Explanation:
The object, temporarily designated 2003UB313 and later named Eris, orbits the sun about once every 560 years, its distance varying from about 38 to 98 AU.
Answer:
me
Explanation:
1 point
1)A 1 kg brick is carried to up the Eiffel Tower to the 275 m platform. How
much gravitational potential energy does it acquire?
Answer:
The gravitational potential energy is 2,750 Joules
Explanation:
Given the following data;
Mass of brick = 1kg
Height of Eiffel Tower = 275m
Potential energy can be defined as the energy possessed by an object due to its position.
Mathematically, potential energy is;
[tex] P = mgh[/tex]
Also, we know acceleration due to gravity = 10m/s2
[tex] P = mgh[/tex]
[tex] P = 1 * 275 * 10 [/tex]
P = 2750 Joules.
Hence, the gravitational potential energy the brick acquired is 2750 Joules.
. A car traveling at 30m/s is accelerated uniformly at the rate of 2.5m/s2for 6 sec. What is its final velocity?
Answer:
45m/s^1
Explanation:
v=u+at
v=(30)+(2.5)(6)
v=45m/s^1
Which of the following describes the relationship between work and power
Answer:
Power is the rate which work is done.
Explanation:
Power is the rate which work is done. Power is measured in watts.
Work is the use of force to move an object. Work is measured in joules
You mix 1 glass of water at 20 C and 1 glass of water at 80 C. What will be temperature of mixture
30
40
50
60
Explanation:
When Two Samples of Water are Mixed, what Final Temperature Results?
Go to Mixing Two Amounts of Water: Problems 1 - 10
Go to calculating the final temperature when mixing water and a piece of metal
Worksheet #2
Back to Thermochemistry Menu
Example #1: Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C.
This is problem 8a from Worksheet #2.
First some discussion, then the solution. Forgive me if the points seem obvious:
1) The colder water will warm up (heat energy "flows" into it). The warmer water will cool down (heat energy "flows" out of it).
2) The whole mixture will wind up at the SAME temperature. This is very, very important.
3) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water)
This problem type becomes slightly harder if a phase change is involved. For this example, no phase change. What that means is that only the specific heat equation will be involved
Solution Key Number One: We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.
The warmer water goes down from to 46.8 to x, so this means its Δt equals 46.8 − x. The colder water goes up in temperature, so its Δt equals x − 14.9.
That last paragraph may be a bit confusing, so let's compare it to a number line:

To compute the absolute distance, it's the larger value minus the smaller value, so 46.8 to x is 46.8 − x and the distance from x to 14.9 is x − 14.9.
These two distances on the number line represent our two Δt values:
a) the Δt of the warmer water is 46.8 minus x
b) the Δt of the cooler water is x minus 14.9
Solution Key Number Two: the energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:
qlost = qgain
However:
q = (mass) (Δt) (Cp)
So:
(mass) (Δt) (Cp) = (mass) (Δt) (Cp)
With qlost on the left side and qgain on the right side.
Substituting values into the above, we then have:
(32.2) (46.8 − x)(4.184) = (32.2) (x − 14.9) (4.184)
Solve for x
What is the depth of the crater if the time is 6.3 s? An astronaut stands by the rim of a crater on the moon, where the acceleration of gravity is 1.62 m/s2. To determine the depth of the crater, she drops a rock and measures the time it takes for it to HIT the bottom
Complete Question
An astronaut stands by the rim of a crater on the moon, where the acceleration of gravity is 1.62 m/s2. To determine the depth of the crater, she drops a rock and measures the time it takes for it
to hit the bottom. If the time is 6.3 s, what is the depth of the crater?
Answer:
The depth is 32 m
Explanation:
From the question we are told that
The time is t = 6.3 s
The acceleration due to gravity is [tex]g = 1.62 \ m/s^2[/tex]
Generally from kinematic equation
[tex]s = ut + \frac{1}{2} - at^2[/tex]
Here the u is the initial velocity and the value is 0 m/s
[tex]s = 0 + \frac{1}{2} - (1.62) * (6.3)^2[/tex]
[tex]s = 32 \ m [/tex]
A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerated at 4m/s2 until it reaches an altitude of 1200 m. At that point the engines fail and the rocket goes into free-fall. Disregard air resistance.
Answer:
The rocket above the ground is in 44 sec.
Explanation:
Given that,
Initial velocity = 92 m/s
Acceleration = 4 m/s²
Altitude = 1200 m
Suppose, How long was the rocket above the ground?
We need to calculate the time
Using equation of motion
[tex]s=ut-\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]1200=92t+\dfrac{1}{2}\times4t^2[/tex]
[tex]2t^2+92-1200=0[/tex]
[tex]t=10\ sec[/tex]
We need to calculate the velocity
Using equation of motion
[tex]v=u+at[/tex]
Put the value into the formula
[tex]v=92+4\times10[/tex]
[tex]v=132\ m/s[/tex]
When the rocket hits the ground,
Then, h'=0
We need to calculate the time
Using equation of motion
[tex]h'=h+ut-\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]0=1200+132t-\dfrac{1}{2}\times9.8t^2[/tex]
[tex]4.9t^2-132t-1200=0[/tex]
[tex]t=34\ sec[/tex]
When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground
So, the total time will be
[tex]t'=34+10[/tex]
[tex]t'=44\ sec[/tex]
Hence, The rocket above the ground is in 44 sec.
A 250 kg cannon containing a cannonball is initially at rest. If the cannon shoots the 20 kg cannonball at 15 m/s, what is the magnitude of the recoil velocity of the cannon?
Answer:
hhgyujhkhgjghhjhhghjkhlhjkghghgh
Explanation:
What amount of force would be required to lift a 10 kg dog?
Answer:
5 Kg force I think try
Explanation:
the force is equal to the weight divided by two. It will require a 5 Kg force to lift a 10Kg weight.
PLEASE HELP!!
A person is traveling at 5m/s for around 20 minutes. how much distance did he cover? hint: change minutes to seconds.
Choose the scientific theories that have been disproved through scientific advancement. Select all that apply.
a) steady-state theory
b) theory of evolution
c) big bang theory
d) geocentric universe
e) phrenology
One particle has a charge of 4.2 x 10-°C, while another particle has a charge
of 1.10 10-9 C. If the two particles are separated by 0.005 m, what is the
electromagnetic force between them? The equation for Coulomb's law is
F = kongs, and the constant, k, equals 9.00 x 10°Nm2/C2
Answer:
1.66*10^-3N
Explanation:
Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. The child pushing the sled is exerting a horizontal a force of 8.00 N. There is a force of friction of 5.00 N.
What is the weight of the sled?
(blank) N
What is the normal force exerted on the sled? Round the answer to the nearest whole number.
(blank) N
What is the acceleration of the sled? Round the answer to the nearest hundredth.
(blank) m/s2
Answer:
Explanation:
Find the diagram of the scenario in the attached file.
From the diagram, the weight = mass * acceleration due to gravity
mass of sled = 30.0kg
acc. due to gravity = 9.81m/s
weight of the sled = 30.0*9.81
weight of the sled = 294.3N
The normal reaction force R in the diagram will be gotten by resolving the 12N force along the vertical
Ry = 12sin45°
R = 12 * 1/√2
R = 12√2/√2
R = 6√2 N
R = 8.49N ≈ 9.0N
The normal force exerted on the sled is approximately is 9.0N
Get the acceleration
Using the formula [tex]\sum Fx = ma_x[/tex] where;
m is the mass of the sled = 30.0kg
[tex]a_x[/tex] is the acceleration of the sled
[tex]\sum Fx = 8 + 12cos45^0\\\sum Fx = 8 + 12(0.7071)\\\sum Fx = 8 + 8.49\\\sum Fx = 16.49N[/tex]
Substitute into the formula;
[tex]a_x = \frac{\sum Fx }{m} \\ a_x = \frac{16.49}{30}\\ a_x = 0.55 m/s^2[/tex]
Hence the acceleration of the sled rounded to the nearest hundredth is 0.55m/s²
An space probe built on Earth has a mass of 750 kg. Calculate the weight of the space probe on Earth.
Answer:
7357.5N
Explanation:
Weight is the force used to pull a mass of object towards the centre of the Earth. The force is 9.81 N/kg.
so, Weight = mass × 9.81
the weight of the space probe should be: 750 × 9.81 = 7357.5 N
Answer:
7350N
Explanation:
Data
m-750kg
g-9.8
W=mg
=750 x 9.8
=7350N
what are two different examples of positive exeleration
Answer:Accelerating your car to get up to speed on a freeway
An airplane accelerating to take off
Accelerating out of the starting blocks at a track meet
Explanation:
Between 1990 and 2010, more than 8.6 million acres of forest were destroyed in Indonesia, Malaysia, and Papua New Guinea to build palm oil plantations. These forests are homes to orangutans, elephants, tigers, and other animals. How will this activity affect the forest ecosystem?
Answer:
1.More number of natural disasters will occur.
2. Disease will spread.
Explanation:
An ecosystem is a geographic area where plants, animals, and other organisms live in and interact with each other in an environment.
Forests are homes to orangutans, elephants, tigers, and other animals.
As more than 8.6 million acres of forest were destroyed in Indonesia, Malaysia, and Papua New Guinea to build palm oil plantations, this may lead to the following consequences:
1.More number of natural disasters will occur.
2. Disease will spread.
y u y u bully me121212121
Answer:
omg, dude you're giving out this many free points??
TYSM!!!!
God bless!
4
Select the correct answer.
A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?
ОА.
100 kilometers/hour south
OB.
200 kilometers/hour
O C.
200 kilometers/hour north
OD
100 kilometers/hour
Reset
Next
Use what you learned in the passage to
determine which pitcher's fastball has
the average speed.
A. Dennis, 60 feet in 0.5 seconds
B. Johnny, 60 feet in 0.45 seconds
C. Travis, 60 feet in 0.52 seconds
D. Michael, 60 feet in 0.48 seconds
Dennis, 60 feet in 0.5 seconds is the average speed of a pitcher's fastball. So, the correct option is (A).
What is the Average speed?Average speed is defined as the total distance covered by the object in a particular time interval. Average speed is a scalar quantity which is represented by the magnitude and not the direction.
The formula for average speed is found by calculating the ratio of the total distance traveled by the body to the time taken to cover that distance which is expressed as:
Average speed= Total distance travelled/ Time taken
The unit of the average speed is meter/second.
In the above given information, the pitcher's fastball average speed of Dennis is 120
Thus, Dennis, 60 feet in 0.5 seconds is the average speed of a pitcher's fastball. So, the correct option is (A).
Learn more about Average speed, here:
https://brainly.com/question/12322912
#SPJ5
if 6.75g of oxide of lead was reduced to 6.12g of metal, calculate the empirical formula of the oxide
What was not a positive benefit of Spanish exploration in Oklahoma?
A.
Future interactions between the Spanish and French would have a buffer zone.
B.
Historians gained a written account of Oklahoma in the 15th and 16th centuries.
C.
Geographers gained new knowledge about the continent of North America.
D.
The Spanish and American Indian tribes enjoyed exclusively friendly relations.
the answer is:
d. the spanish and american indian tribes enjoyed exclusively friendly relations.
Answer:
b. the spanish and american indian tribes enjoyed exclusively friendly relations.
Explanation:
If a racer maintained a constant speed, during which sections of the race would the racer's velocity be the same?
Sections 1 and 2
Sections 2 and 3
Sections 2 and 4
Sections 3 and 4
pls help!!
Answer:
Sections 3 and 4
Explanation:
The correct answer would be sections 3 and 4.
The velocity of an object is the speed of the object in a particular direction. In this case, the speed of the racer is constant irrespective of the distance or the direction. Hence, the sections of the race during which the racer's velocity would be the same would be the sections in which the racer faces the same direction - which is sections 3 and 4.
4. Which of the following statements is correct?
A Mass and weight are different names for the same thing
B The mass of an object is different if the object is taken to the Moon
C The weight of a cer is one of the forces acting on the car.
D The weight of a chocolate beris measured in kilograms
Answer:
Explanation:
A: wrong. Mass and weight are different.
B: Wrong. The mass here and the mass on the moon are the same. The weight, which is Mass * the acceleration is equal to weight.
C: Correct.
D: Wrong. Weight is not measured in Kg. Mass is.
find fk and f in the picture shown
Two objects gravitationally attract with a force of 18.0 N. If the mass of one of these objects is doubled, the new force of attraction is ____ N.
Two objects gravitationally attract with a force of 18.0 N. If the mass of both of these objects is doubled, the new force of attraction is _____ N.
Two objects gravitationally attract with a force of 18.0 N. If the mass of one of these objects is tripled, the new force of attraction is _____ N.
Two objects gravitationally attract with a force of 18.0 N. If the mass of both of these objects is tripled, the new force of attraction is ______ N.
Two objects gravitationally attract with a force of 18.0 N. If the mass of one of these objects is quadrupled, the new force of attraction is ____N.
Two objects gravitationally attract with a force of 18.0 N. If the mass of both of these objects is quadrupled, the new force of attraction is ____N
2. A police car drives from a police station to a distance of 170 km in 4 hours. He then stops at a gas station for 1 hour to fill gas and rest. Then the driver goes for 100 Km in 1 hour 30 minutes to his final destination.
a. What is the speed of the police car between the police station and the gas station?
b. What is the speed of the police car between the gas station and the final destination?
c. What is the average speed of the police car?
Answer:
a.42.5 km/h or .71 km/m
b. 66.7 km/h or 1.1repeating km/m
c.41.5 km/h or .69 km/m
Explanation:
These are all average speed problems, meaning you divide the distance traveled by the time [tex]s=\frac{d}{t}[/tex]
A. 170km/4hrs or 170km/240 mins
B. 100km/1.5hrs or 100km/90mins
C. The only tricky thing here is the question seems like it wants the speed for all of the information given, so you do include the one hour that he stops
270km/6.5 hrs or 270km/390 mins
How could you categorize or classify the five major types of
forces?
Answer:
Contact forces can be classified according to six types: tensional, spring, normal reaction, friction, air friction, and weight. Noncontact forces: Forces that take place when two objects do not touch. These forces can be classified according to three types: gravitational, electrical, and magnetic.
Explanation:
field lines point out from all sides on a object. Some of the lines point into a second object nearby. Which statement best describes the objects? A. both objects are permanent magnets B. both objects are electromagnets C. the first object is negatively charged, and the second object is positively charged D. the first object is positively charged, and the second object is negatively charged
Answer:
The first object is positively charged, and the second object is negatively charged. (D)
Explanation:
If your speed is 3 meters in 1 second (3m/s), how long would it take you to go 15
meters?
Answer:
5 seconds
Explanation:
Answer:
5 seconds
Explanation:3=1 6=2 9=3 12=4 15=5 15 equal 5 mean 15 meters in 5 sec
i really need help please just answer at least one
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
[tex]v_{f}^{2} = v_{i}^{2}-(2*a*x)\\[/tex]
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
[tex]v_{f} =v_{i} - (a*t)[/tex]
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
[tex]v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})[/tex]
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
[tex]v_{f}= v_{i}+a*t\\[/tex]
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
[tex]v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})[/tex]
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
[tex]v_{f}= v_{i}+a*t\\[/tex]
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have at the top of the circle? 3.74 m/s 9.67 m/s 5.86 m/s None of the above
Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :
[tex]F=\dfrac{mv^2}{r}[/tex]
v is the maximum speed
[tex]v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s[/tex]
Hence, the maximum speed of the ball is 5.86 m/s.