Answer:
second lag
Explanation:
If in a cup game, a specified time limit is assigned to both teams to score high. If both teams are unable to score or if score of both the teams is equal then there is another second lag played where each team tries to score high. Even if in second lag both teams fail to score higher than other the last third lag is played or else game is declared draw.
The force that the left team pulls with is 1000 N. If the right team's total mass is 300 kg and they accelerate by 1.2 m/s2, what is the force of resistance on the right team
Answer:
the force of resistance on the right team is 360 N
Explanation:
Given;
force of the left team, = 1000 N
total mass of the right team, m = 300 kg
acceleration of the right team, a = 1.2 m/s²
The force of resistance of the right team is calculated as;
Force = mass x acceleration
Force, F = 300 x 1.2
Force = 360 N
Therefore, the force of resistance on the right team is 360 N
An inquisitive physics student and mountian climber climbs a 43.6 m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.35 s apart and observes that they cause a single splash. The first stone has an initial velocity of 1.8 m/s. How long after release of the first stone do the two stones hit the water
Answer:
Explanation:
What we are basically looking for here is how long it takes the first stone to hit the water. We have everything we need to figure that out. We will use the equation
Δx = . Filling in, we will solve for t, the time is takes the first stone to hit the water (which is the same for both since they both hit the water at the same time):
which is a quadratic that we will have to factor. Get it into standard form, setting it equal to 0:
and factor to get that
t = 3.2 s and t = -2.8 s
Since time can't ever be negative, it takes 3.2 s for the stones to hit the water.
what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s
Answer:
[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]
Eli states that sodium phosphate is a mixture because it is composed of both sodium ions and phosphate ions.
Which is the best analysis of Eli’s statement?
The question is incomplete, the complete question is;
Eli states that sodium phosphate is a mixture because it is composed of both sodium ions and phosphate ions. Which is the best analysis of Eli’s statement? It is correct because each ion is a pure substance, so sodium phosphate is made up of two pure substances. It is correct because the composition of sodium phosphate changes depending on the sample. It is incorrect because sodium phosphate is a compound that has a single composition. It is incorrect because the two types of ions in sodium phosphate cannot be seen.
Answer:
It is incorrect because sodium phosphate is a compound that has a single composition
Explanation:
A compound is a neutral substance made up of two or more atoms which are chemically combined together.
Ionic substances are made up of ions. These ions are not separate entities, they are part of the compound.
Hence, Eli's statement is incorrect because sodium phosphate is a compound that has a single composition.
a motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20km/h-¹ find the average speed ?
PLEASE DO IT AS FAST YOU CAN AND NO SCAM ANSWERS OR ELSE I WILL REPORT
Answer:
Average speed = 24 km/h
Explanation:
Let the distance be x.Given the following data;
Uniform speed A = 30 km/h
Uniform speed B = 20 km/h
To find the average speed;
Mathematically, the average speed of an object is given by the formula;
[tex] Average \; speed = \frac {total \; distance}{total \; time} [/tex] ..... equation 1
Total time = TA + TB
[tex] T_{A} = \frac {x}{30} [/tex]
[tex] T_{B} = \frac {x}{20} [/tex]
[tex] Total \; time = \frac {x}{30} + \frac {x}{20} [/tex]
[tex] Total \; time = \frac {5x}{60} [/tex]
Total distance = x + x
Total distance = 2x
Substituting the values into equation 1;
[tex] Average \; speed = \frac {2x}{\frac {5x}{60}} [/tex]
[tex] Average \; speed = \frac {2x*60}{5x} [/tex]
[tex] Average \; speed = \frac {120}{5} [/tex]
Average speed = 24 km/h
A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throws a 0.50 kg piece of equipment so that it moves at a speed of 4 m/ s directly away from the spaceship towards the left . How long will it take him to reach the ship? *hint find his speed after the collision and consider it a constant speed all the way back to his spaceship*
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
[tex][m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a[/tex] which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
[tex][(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a[/tex]
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
In the hydraulic system depicted, the cylinder on the left has a diameter of 2 inches and the cylinder on the right has a diameter of 6 inches. If 100 lbs of force was applied to the cylinder on the left, what force would be exerted on the cylinder on the right
Answer:
F2 = 900 lbs
Explanation:
From pascal principle;
F1/A1 = F2/A2
Force on cylinder at left; F1 = 100 lbs
Diameter of cylinder at left; d1 = 2 inches
Diameter of cylinder at right; d2 = 6 inches
Formula for area of top of cylinder = πr²
Thus;
Area of top of left cylinder; A = π × 2² = 4π
Area of top of right cylinder; A = π × 6² = 36π
Thus;
100/4π = F2/36π
F2 = (36π × 100)/4π
F2 = 900 lbs
An object 2cm high is placed 3cm in front of a concave lens of focal length 2cm, find the magnification?
Answer:
0.4
Explanation:
A concave lens is a diverging lens, so it will always have a negative focal length. Image distance is always negative for a concave lens because it forms virtual images.
From the lens formula;
1/f = 1/u+ 1/v
- 1/2 = 1/3 - 1/v
1/v = 1/3 + 1/2
v= 6/5
v= 1.2 cm
Magnification = image distance/object distance
Magnification = 1.2cm/3cm
Magnification = 0.4
Question 23 of 23
Suppose a current flows through a copper wire. Which two things occur?
O A. The field is parallel to the direction of flow of the current.
B. An electric field forms around the wire.
OC. A magnetic field forms around the wire.
U
D. The field is perpendicular to the direction of flow of the current.
SUBM
Answer:
The field is parallel to the direction of flow of the current.
When the E string of a guitar (frequency 330 Hz) is plucked, the sound intensity decreases by a factor of 2 after 4 s. Determine
Answer:
[tex]Q=50.3[/tex]
Explanation:
From the question we are told that:
Frequency [tex]F=330Hz[/tex]
Sound intensity drop [tex]I_d=2[/tex]
Time [tex]T=4s[/tex]
Therefore
Sound intensity Ratio
[tex]\frac{I}{I_x}=\frac{1}{2}[/tex]
Generally the equation for Sound intensity is mathematically given by
[tex]\frac{I}{I_x}=e^{-4\ \=t}[/tex]
[tex]\frac{1}{2}=e^{-4\ \=t}[/tex]
[tex]\=t =5.8s[/tex]
Generally the equation for Quality Factor is mathematically given by
[tex]Q=2 \pi\frac{E}{\triangle E}[/tex]
[tex]Q=2 \pi\frac{E}{\frac{E}{2*4}}[/tex]
[tex]Q=50.3[/tex]
After landing the aeroplane's momentum becomes zero. Explain how
the law of conservation holds here.
Answer:
The law of momentum conservation can be stated as follows. For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
a ball is thrown straight up into the air while the ball is traveleling upwards what are the magnitue and direction
Answer: hi your question is incomplete attached below is the complete question
answer :
magnitude of acceleration : | a | = g = 9.81 m/s^2
direction : a = - g j
Explanation:
Neglecting Air resistance
magnitude of acceleration :
| a | = g = 9.81 m/s^2
Direction of acceleration
a = - g j ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )
5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura
Answer:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
Explanation:
A continuación, veremos que representa cada caso:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
What is the dependent variable in this
experiment?
DONE
Biologists designed an experiment to test
the effect of compost on the development
of root crops. They tested several different
crops, including carrots, potatoes, beets,
and onions. They grew most of the plants
in the greenhouse, but due to space issues,
they had to grow some outdoors. They gave
all the plants the same amount of compost.
They obtained the compost from a local
farmer and from the local hardware store.
They ran out of the farmer's compost, so
some of the plants received that compost
when the seeds were planted and other
plants got hardware store compost after
the plants had already started growing.
What is the independent variable in this
experiment?
DONE
Answer:
"the plants had already started growing."
Explanation:
I think this is the answer because the definition of a dependent variable is the variable that is being affected by the change. Since the plants had already started growing BECAUSE of "They ran out of the farmer's compost, so
some of the plants received that compost
when the seeds were planted and other
plants got hardware store compost after
the plants had already started growing."
Sorry if I am wrong, I am just a 4th grader, pls don't hate on me, I am just trying to help :)
Answer:
It's compost
Explanation:
In case you needed the dependent variable, its the amount of plant growth
The large reservoir of comet nuclei far beyond Pluto, from which we believe new long-period comets come into the inner solar system, is called:
Answer:
Oort cloud
Explanation:
The large reservoir of comet nuclei far beyond Pluto, from which we believe new long-period comets come into the inner solar system, is called Oort cloud.
The comets hich have the large periods that means more than 200 years to orbit the Sun generaly comes from Oort cloud hich is also knon as the cometary cloud.
While traveling north on an expressway, a car traveling 60 mph (miles per hour) slows down to 30 mph in 12 minutes due to traffic conditions
Answer:
acceleration = - 150 m/s^2
distance = 9 miles.
Explanation:
initial speed, u = 60 mph
time, t = 12minutes = 0.2 hour
final speed, v = 30 mph
Let the acceleration is a and the distance is s.
By the first equation of motion
v = u + at
30 = 60 + a x 0.2
a = - 150 m/s^2
Let the distance is s.
Use third equation of motion is
[tex]v^2 = u^2 + 2 a s \\\\30^2 = 60^2 + 2 \times 150\times s\\\\s = 9 miles[/tex]
Which statement or question is a good hypothesis
Please answer will mark brainleist
[tex]option \: (a) \: is \: correct.[/tex]
Explanation:
Yes, I was wrong. Pressure increases as the area decreases.
As per the Figure A, the truck is so heavy and having a greater mass. So, the area of contact between the tyres and the road increases and decreases it's pressure on the road due to the increase in area.
As per the Figure B, You know the nature of Tomato that it is soft and smooth to touch and also lighter mass when it's compared with the truck. As it having a lighter mass, it can be cut through the sharp knife. The area of contact decreases and pressure increases on the vegetables.
Answer:
Option number B
Explanation:
The pressure is indirectly proportional to the area therefore pressure decreases when the area increases, and pressure increases when the area decreases
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz to have a wavelength of 0.790 mm
Answer:
the tension of the rope is 34.95 N
Explanation:
Given;
length of the rope, L = 3 m
mass of the rope, m = 0.105 kg
frequency of the wave, f = 40 Hz
wavelength of the wave, λ = 0.79 m
Let the tension of the rope = T
The speed of the wave is given as;
[tex]v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N[/tex]
Therefore, the tension of the rope is 34.95 N
[tex]what \: is \: mirror \: {?}[/tex]
To overcome the problems that blur images and don't provide the best resolution from Earth, astronomers have started using flexible mirrors that change shape many times each second. This technique is called:
Answer:
adaptive optics
Explanation:
simple
convert the following quantities
[tex]25m {}^{2} \: into \: cm {}^{2} [/tex]
Answer:
[tex] \rm 250000 \; cm^2 [/tex]
Explanation:
Refer to the attachment.
When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre with the strength
Solution :
When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.
The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.
How much energy has 4×10^10m^3 of water collected in a reservoir at a hight of 100 m from the power house ?What kind of energy is that?
Answer:
PE = 3.92x10^16J
potential energy
Explanation:
PE = m*g*h
mass of water = 1000kg/m³
(4*10^10m³)*1000kg = 4*10^13kg
PE = (4*10^13kg)*(9.81m/s²)*(100m)
PE = 3.92x10^16J
Calculate surface tension of an enlarged radius of 4cm to 6cm and amount of work necessary for enlargement was 1.5×10^-4 joules
Answer:
[tex]T=7.5*10^{-5}[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=2cm[/tex]
Work done [tex]W=1.5×10^-4 joules[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=T.dA[/tex]
Therefore
[tex]T=\frac{W}{dr}[/tex]
[tex]T=\frac{1.5*10^{-4}}{2}[/tex]
[tex]T=7.5*10^{-5}[/tex]
A box has a mass of 4kg and surface area 4m². Calculate the
pressure exerted by the box on the floor.
Answer:
10 pa
Explanation:
4kg* 10 (or 9.8m/s2) = 40
40N /4m2 =10
Juanita ran one mile around her school track in six minutes. What is
her average speed, and what is the magnitude of her average velocity?
10 mph, 0 mph
6 mph, 0 mph
6 mph, 6 mph
10 mph, 10 mph
Answer:
The correct option is a) 10 mph, 0 mph.
Explanation:
1. The average speed (S) is a magnitude given by:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance = 1 mi
T: is the total time = 6 min
[tex] S = \frac{D}{T} = \frac{1 mi}{6 min}*\frac{60 min}{1 h} = 10 mph [/tex]
Hence, the average speed is 10 mph.
2. The average velocity is a vector:
[tex] V = \frac{\Delta d}{\Delta t} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex]d_{f}[/tex]: is the final distance
[tex]d_{i}[/tex]: is the initial distance
[tex]t_{f}[/tex]: is the final time
[tex]t_{i}[/tex]: is the initial time
Since Juanita ran one mile around her school track, the final position is the same that the initial position, so the magnitude of the average velocity is zero.
Therefore, the correct option is a) 10 mph, 0 mph.
I hope it helps you!
Look at the distance-time graph below. It shows Angela's journey as she walks to the end of the road and back. The end of the road is 40 m away. After how many seconds did she arrive back where she started?
Answer:
Explanation:
If the distance to the end of the road was 40 m, and she traveled back to where she started, she traveled a total distance of 80 m. It took her 100 seconds to travel the 80 m.
Based on the given distance-time graph, the number of seconds that Angela used to go to the end of the road and arrive back is 100 seconds.
How long did Angela take?The end of the road is 40m and coming back is also 40 m. The total distance is:
= 40 + 40
= 80m
The graph shows that to walk 80 meters, Angela took a 100 seconds because the line on the graph ends at 80 m and 100 seconds.
Find out more uses of distance-time graphs at https://brainly.com/question/13877898.
#SPJ6
6. An object is fired from the gound at 275 m/s at an angle of 55° N of E.
a. How far away did the object first hit the ground?
b. what is the maximum height that the object reaches?
there u go fella hope u understood
The 75.0 kg hero of a movie is pulled upward at a constant velocity by a rope. What is the tension on the rope?
Answer:
750 N
Explanation:
the tension on the rope is the weight of the hero
