A spring scale is used to measure the weight of various objects. When a 100 g empty glass bottle is placed on the scale, the spring stretches down 1.8 cm. How much water should be poured into the glass bottle to stretch the spring down by 2.7 cm?

A: 44 mL
B: 50 mL
C: 33 mL
D: 150 mL

Answers

Answer 1
The answer is B. 50ml
Answer 2

50 mL of water have to poured to the glass bottle of 100 g to increase the stretching of the spring from 1.8 cm to 2.7 cm.

What is spring stretching?

A spring can hang when it loads a weight. The stretching of spring is the change in length of the spring end from its initial length.  Spring balance works with the principle of Hooks law.

The stretching is made by the elasticity of the spring which make it in a wave like motion.

Given that the glass bottle with 100 g weight when loaded on the spring scale, it will stretch down to 1.8 cm. Then the weight required to make a stretch of 2.7 cm is calculated as follows:

weight required  = (100 g × 1.8 cm) /2.7 cm

                            = 150 g.

A total of 150 g is needed to stretch into 2.7 cm. Thus we have to add 50 g of water to the glass weighing 100 g . 50 g is equivalent to 50 ml, since density of water is 1 g/L.

Hence, 50 ml of water should be poured to the glass.

To find more about spring scale, refer the link below:

https://brainly.com/question/2568071

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Related Questions

In the absence of air resistance, as an object falls freely, its
O velocity increases
O acceleration increases
both A and B
O none of the above

Answers

Answer:

None of the above

Explanation:

None of the above

If 1 light year = 9.46x 1015.
m and 1 mile = 1.6 km, how many miles are in a light year? You

Answers

Answer:

5 912 500 000 000 = 5.91x10^12 miles

Explanation:

The first thing we need to do is convert the light year distance to kilometres, since we have the kilometre to mile conversion. There are 1000m in a kilometre, so we need to divide the light year value that we have by 1000, giving us 9.46x10^12 km for 1 light year.

Now that we have our value in kilometres, we simply need to divide it by 1.6 to get it in miles. Doing so gives us 5 912 500 000 000 miles, or 5.91x10^12 miles.

Hope that made sense, and let me know if you have any further questions!

How does the suns energy contibute to the movement of water in the water cycle

Answers

Explanation:

The sun is what makes the water cycle work. The sun provides what almost everything on Earth needs to go—energy, or heat. Heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds... clouds that move over the globe and drop rain and snow.

Answer:

"The sun is what makes the water cycle work. ... Heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds... clouds that move over the globe and drop rain and snow. This process is a large part of the water cycle."

Explanation:

Please help as soon as possible! I will give Brainliest! I just need to know the answers to this diagram.

Answers

Hope it will help you

6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar
force of 5000N on the ball. What linear impulse is delivered to the ball as a result?

Answers

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

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3)(20pts) Un barco invasor se encuentra a 560m de un puerto de defensa, el puerto tiene cañones que
disparan con una velocidad de 82m/s. a) ¿Qué ángulo debe de tener el barco para poder golpear el
barco invasor? b) ¿Cuál es el tiempo de vuelo de las balas disparadas por el cañón utilizando el ángulo
de la respuesta "a"? c) ¿Qué tan lejos debe alejarse el barco invasor de la costa para lograr estar fuera
del alcance de los cañones?

Answers

a) Las balas de cañon disparadas desde el puerto deben tener una ángulo de 27.39° para que puedan impactar al barco, con una velocidad inicial de 82 m/s.

b) El tiempo de vuelo de las balas de cañon para alcanzar al barco que está a 560 m de distancia es de 7.69 s.

c) Sabiendo que el ángulo calculado en el inciso a) es para una distancia de 560 m, el barco debe estar a una distancia mayor para que las balas no lo alcancen.

a)

Podemos usar las ecuaciones de tiro parabólico para encontrar el ángulo que permita derribar al barco invasor.

[tex]x=\frac{v_{i}^{2}sin(2\alpha)}{g}[/tex] (1)

Donde:

v(i) es la velocidad inicial del cañon (82 m/s)α es el ángulo de tirog es la gravedad (9.81 m/s²)x es el desplazamiento total (560 m)

Lo que debemos hacer es depejar α de la ecuación 1

[tex]sin(2\alpha)=\frac{xg}{v_{i}^{2}}[/tex]

[tex]sin(2\alpha)=\frac{560*9.81}{82^{2}}[/tex]

[tex]sin(2\alpha)=0.82[/tex]

[tex]\alpha=\frac{sin^{-1}(0.82)}{2}[/tex]

[tex]\alpha=27.39^{\circ}[/tex]

Por lo tanto, el ángulo para que el cañón impacte en el barco es de 27.39 °.

b)

Sabemos que la componente de la velocidad en el eje x es constante, así que podemo usar la siguiente ecuación.

[tex]v_{x}=\frac{x}{t}[/tex]

La componente x de la velocidad es V(x) = V(i)cos(α) y sabiendo la distancia total de 560 m, el tiempo será:

[tex]v_{i}cos(\alpha)=\frac{x}{t}[/tex]

[tex]t=\frac{x}{v_{i}cos(\alpha)}[/tex]

[tex]t=\frac{560}{82cos(27.39)}[/tex]

[tex]t=7.69\: s[/tex]

El tiempo total de vuelo de las balas de cañon es de 7.69 s.

c)

Sabemos que el ángulo calculado en el inciso a) es de 27.39 °, y ese valor fue calcualdo para una distancia de 560 m, por lo tanto el barco debe estar a una distancia mayor que esa para que las balas no lo alcancen.

Puedes encontrar más información sobre tiro parabólico aquí:

https://brainly.lat/tarea/3605927

Espero te haya sido de ayuda!

A model of a plane in different parts of the atmosphere uses a scale of
1:2,000. If a plane flies at an altitude of 4,000 feet, how high above ground
level is the plane in the model?
A. 8 ft
B. 1/4,000 ft
C. 4ft
0
D. 2 ft

Answers

its D, so 2ft because the 4000 into 2000 is 2, so for every 1 in the model there is 2000. so 2 in the model would be 4000.
its 2.

the force of gravity on Mars is 3.7 m/s^2. what is the force of gravity on Earth?

Answers

The force of gravity on earth is 9.807 m/s^2


A particular car can go from rest to 90 kmh in 10 seconds. What is its
acceleration. (Show your calculation

Answers

Answer:

Im sorry i dont know But thank you for the points

Why is mass not included in the calculated height equation ?

Answers

Answer:

You can't use energy conservation when mass is removed. This has required energy that you are not taking into account.

Answer:

because some time it dose not need to be i equation becasue it is not need

Explanation:

If the tension in the string is equal to 2 W when the stone is at its lowest point, then the tension when the stone is at the highest point will be

Answers

This question involves the concept of tension, centripetal force, centrifugal force, and weight.

The tension of the stone at the highest point is "0 N".

The centrifugal force always acts away from the circle. The tension always acts toward the support that is the center of the circle in this case. When the stone is at its lowest point, the centrifugal force, and the weight of the stone, both act downward. But the tension is pointing in the upward direction. Hence, the equilibrium equation will be:

[tex]Tension = Weight\ + \ Cenrifugal\ Force\\T = W + F_c\\T = 2\ W\ (given)\\2\ W = W+F_c\\F_c = 2\ W - W\\F_c = W[/tex]

Now, we consider the stone at the top position. Here, the centrifugal force will act in the upward direction (away from the center of the circle), while the weight and tension will be acting in the downward direction. Thus, the equilibrium equation, in this case, will be:

[tex]F_c = W+T\\T=F_c-W\\T = W-W\\[/tex]

T = 0 N

Learn more about tension here:

https://brainly.com/question/2287912?referrer=searchResults

The attached picture shows the free body diagram of the stone at different locations.

Two buildings near each other stand 40m and 60 m tall as measured from the ground to the rooftop. You are standing on the rooftop of the shorter building and observe that the rooftop of the taller building is 55o above the horizontal, as shown in the diagram below. Determine the horizontal distance (in meters) between the two buildings.

Answers

Answer:

Explanation:

tanθ = opp/adj

adj = opp/tanθ

adj = (60 - 40)/tan55 = 14 m

Compared to a 1-kg block of solid iron, a 2-kg block of solid iron has the same:
A. Density
B. Mass
C. Volume
D. Inertia

Answers

The Correct choice is :

Density

giải giúp em bài này với ạ?

Answers

vdjs vs jsbdjskshskjsjdjfodjshsvsjsbdbdn

A plane is traveling with an air velocity of 720 kilometers/hour due east. It experiences a headwind of 16 kilometers/hour. Find the resultant velocity and direction of the plane with respect to the ground.

Answers

The resultant velocity and direction of the plane is 45 kilometers per hour.

Taking care of yourself can sometimes be hard work and can even lead to feelings of

Answers

Answer:

deception

Explanation:

a ball thrown horizontally with the velocity of 20 m\s from a 20 m building​ .what is the time

Answers

Answer:2.02s

Explanation:

At the y component, the tract of the ball is free falling body. The initial velocity at y component is 0 m/s.

Height = 1/2 * g * t^2

20 = 1/2 * 9.8 * t^2

t = (20 * 2 / 9.8)^1/2, t = 2.02s

A 55-kg person lands on firm ground after jumping from a height of 2.8 m.

Answers

Explanation:

What are you searching for?

Potential energy?

Answer:

3018.4

Explanation:

So this velocity just before reaching the ground, the final equals square root of two times G times H. So gs 9.8 m per second squared H is 2.8 m. So it gives the velocity of 7.41 m per second squared. That's the velocity right before touching the ground. So to find the impulse required to come to rest here, we just do mass times the velocity minus zero. So zero is the final momentum, final velocity zero. So that makes the final momentum zero. So we multiply the mass with the velocity before touching the ground and after. And subject zero. That's the momentum after touching the ground. So we multiply mass 55 kg. At 7.41 velocity we got 407.45.  kilograms meters per second of impulse. we know that the force will be larger for stiff stiff flex compared to bend flex. Because the stiff flex. Bring the object to rest in a short amount of time and distance. So in part C, that's calculate for stiff flex. The stopping distance, Delta Y. Is 0.1 m. This one centimeter. So we can calculate the acceleration of the body when it comes to rest.  From the moment it touches the ground.  equals v squared over two times Delta Y. So  we use the square is 54.88. The 7.41 square divided by two times 0.01 gives us an explosion of 2744 m/s square. So you multiply this acceleration by the mass to get the force so force equals mass times acceleration gives us a large force,  159 20 newtons. Then for bent legs, repeat the same process with a different delta Y, delta Y of 200.5 m. So that's 50 centimeters and calculate acceleration the same process. Just use a different value of delta Y. And we get as much smaller acceleration 54.88 m per second square. And the force experience would just be mass times acceleration. That gives us 3018.4 newton's a lot smaller than for stiff flex.

Which of the following types of stars is hottest?

A
B
C
D

Answers

Answer:

graph b

Explanation:

A roller coaster has a mass of 585 kg when fully loaded with passengers.
A)If the vehicle has a speed of 18.0 m/s at point a, what is the force (in N) of the track on the vehicle at this point?
B) What is the maximum speed (in m/s) the vehicle can have at point B in order for gravity to hold it on the track?
Point A says the radius is 10m. Point B says radius is 15m.

Answers

Answer:

Explanation:

I'm going to ASSUME that point A is at the bottom of a circular curve.

The track must support the weight of the car and also supply the needed centripetal force

F = mg + mv²/R

F = 585(9.81 + 18.0²/10) = 24,692.85

F = 24.7 kN

I'm going to ASSUME that point B is at the top of a circular curve.

The track must support the weight of the car and also supply the needed centripetal force.

At a minimum, the weight can be reduced to zero and all the gravity acceleration acts as centripetal acceleration.

0 = mg - mv²/R

        g = v²/R

         v = √Rg = √(15(9.81) = 12.1305399...

        v = 12.1 m/s

A crane lifts 1800kg mass through a vertical height of 6cm in 9 seconds, Taking (g) as 10N/kg. what is the cranes power output?​

Answers

Explanation:

power output=(1800×10×0.06)/9=120watts

You had an amazing race. You finished the 300kms race in 1hr 40 minutes. What is the average velocity?

Answers

Average Velocity

Level : JHS

v = s/t

v = 300 km/(1 hour + 40/60 hour)

v = 300 km/(1.66 hour)

v = 180 km/hour

#LearnWithEXO

What is a common noun?

Answers

I HOPE IT WILL HELP YOU.

Thank you.

TEDDYLOVER

If a person weighs 160 Newtons on Earth, what is their weight in pounds on Earth?
A. 27
B. 36
C. 712
D. 534

Answers

Answer:

B. 36 lb

Explanation:

m = 160/9.8 = 16.3 kg

16.3 kg ( 2.205 lb/kg) = 36 lb

A racecar begins at rest and accelerates to 25 m/s at a rate of 6.25 m/s2. What distance does the racecar cover?

a.
2 m

b.
156 m

c.
4 m

d.
50 m

Answers

Answer:

d.

50 m

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s =(25² - 0²) / (2(6.25))

s = 50 m

Which of the following is an example of an acceptable time to ignore the rules and etiquette of a sport?
A.
when no one else notices
B.
when nobody really gets hurt
C.
when the outcome of the game isn't affected
D.
it's never acceptable to ignore rules and etiquette

Answers

Answer:

D-it's never acceptable to ignore rules and etiquette

Explanation:

We should know that it's never acceptable to ignore rules and etiquette.

What are the rules of etiquette in sports?

The rules of etiquette in sports are the rules of the sports that have been document by the various sports federations.

We should know that it's never acceptable to ignore rules and etiquette. These will make the the sports a lawless act and full of chaos.

Learn more about etiquette in sports:https://brainly.com/question/979191

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I am completely braindead on how to calculate the time first.

A flowerpot falls from a window sill 31.3m from a sidewalk. What is the velocity of the flowerpot when it hits the ground? (hint: Use d=1/2at^2 to find "t" first, then find velocity using v=at. Remember, ay= -9.8 m/s2).

Answers

24.8 m/s

Explanation:

You don't have to solve for time first. You can use the equation below to solve for the velocity before impact:

[tex]v_y^2 = v_{0y}^2 + 2a_yy[/tex]

We know that [tex]v_{0y} = 0[/tex] so the equation above becomes

[tex]v_y^2 = 2a_yy \Rightarrow v_y = \sqrt{-2a_yy}[/tex]

Plugging in the numbers, we get

[tex]v_y = \sqrt{-2(-9.8\:\text{m/s}^2)(31.3\:\text{m})}[/tex]

[tex]\:\:\:\:\:=24.8\:\text{m/s}[/tex]

METHOD 2:

If you insist on using that equation for d, we can do that too. So solving for t from the equation for d. we get

[tex]t = \sqrt{\dfrac{2d}{g}} = \sqrt{\dfrac{2(31.3\:\text{m})}{9.8\:\text{m/s}^2}}[/tex]

[tex]\:\:\:\:= 2.53\:\text{s}[/tex]

So using the equation [tex]v_y = gt,[/tex] we get

[tex]v_y = (9.8\:\text{m/s}^2)(2.53\:\text{s}) = 24.8\:\text{m/s}[/tex]

So you get the same result regardless of the method you use.

Question 5 (4 points)
Piedra Vista High School is 35 degrees north of east, 3.31 miles away from FHS. How many miles north and east is it? Round to the nearest
tenth of a mile.
North:
miles
East:
miles
Blank 1:
Blank 2:

Answers

Answer:

ayussssaasssassbfjdbskgsvdksbjabsdvsjvsjhjsjajbdnjsjnsnsnsnbfndksbsjjjsjdhjjsjjsjkjsjsjbdhsjabdhdhksbshsjsjisisisjdbbdjsjsjjebwnnwjbsnbhejshvjrebbdjndnebensjHELLONODNDIDBIDDBDUDBDIjsjs

Explanation:

jshshjabsjbrnejdjdhjsndjdhfisjshsidsnsndgjsbdjdvsjsvdudhwjgdusbsuwgshidsbhsgsuydisksvdudh

Độ hấp thụ của dung dịch có độ truyền qua 30% ở bước sóng 640nm là bao nhiêu?

Answers

henwbwhehhijwjwkishdbe

Chọn câu sai :

A. Đồ thị vận tốc theo thời gian chuyển động thẳng đều là một đường song song với trục hoành Ot

B. Trong chuyển động thẳng đều ,đồ thị theo thời gian của tọa độ và của vận tốc đều là những đường thẳng

C. Đồ thị toạ độ theo thời gian của chuyển động thẳng đều bao giờ cũng là một đường thẳng

D. Đồ thị toạ độ theo thời gian của chuyển động thẳng đều là một đường thẳng xiên góc

Answers

P.s this language is: Viatnamese :)

Answer:

This is just to help the other readers help you. Or so that you understand:

Đây là để giúp bạn đọc giúp bạn!

What they tried to say:

Choose the wrong sentence:

A. The velocity-time graph of uniform linear motion is a line parallel to the horizontal axis Ot

B. In uniform linear motion, the time graphs of the coordinates and of the velocity are both straight lines

C. The time-coordinate graph of uniform rectilinear motion is always a straight line

D. The coordinate-time graph of uniform linear motion is an oblique line

Is what they asked you.

Thankyou and your welcome

Vì tôi đã giúp đỡ, tôi có thể vui lòng có cám ơn nào không?

Have a great day!

~Happy helper~

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