A cannon shoot a cannon ball. The momentum compared to the cannon ball is

Answers

Answer 1

Answer:

After the firing occurs, both the cannon and cannonball have the same momentum (big mass, small velocity vs. small mass, big velocity). But since the momentum for each is moving in the opposite direction, the momentums cancel out, causing the cannon-cannonball system's momentum to equal zero.

Explanation:


Related Questions

which graph best demonstrates the general relationship between mass and temperature similar to the trend of thermal energy absorbed by your sand and water samples

Answers

Answer: the first one

Explanation: got it right on edg

Answer:

The first graph

Explanation:

Correct on Edge 2022! :]


A ball is thrown upward in the air with an initial velocity of 40 m/s. How long does it take to
reach back to the point it was thrown from?

Answers

Answer:

You need the definition of acceleration (a=Vf-Vi/t) and 1 equation of linear motion (deltaX = Vi×t + 1/2×a×t^2). Since you know a is constant (gravity) and you know your initial Vi to be 40 m/s and your final velocity Vf to be zero (maximum height), then you can use thhe definition of acceleration to find time.

-9.81m/s^2 = (0-40m/s)/t

t = (-40)/(-9.81) s

t = 4.077s

Now that you have time, you should know all but deltaX in the equation of linear motion.

dX = (40m/s)(4.077s) + (1/2)(-9.81m/s^2)(4.077s)^2

dX = (163.099m) — (81.549m)

dX = 81.55m

This is about acceleration, velocity and all that mumbo jumbo

Which best depicts the data?

1, From 0 to 2 seconds, Anna experienced negative acceleration.
2, From 3 to 6 seconds, Elan accelerated while Anna did not.
3, From 6 to 8 seconds, Elan accelerated faster than Anna.
4, From 8 to 10 seconds, Elan experienced negative acceleration.

Answers

Answer:

[tex]3. \: From \: 6 \: to \: 8 \: seconds, \: Elan \: \\ accelerated \: \: faster \: than \: Anna. [/tex]

Which of the following choices is not part of the scientific method

Answers

Answer:

What are the choices?

Explanation:

(DUE IN FIVE MINUTES, QUICK)

Explain why your weight would change if you went to the moon, but your mass wouldn’t.

Answers

The moon's gravitation force is determined by the mass and the size of the moon. Since the moon has significantly less mass than the Earth, it will not pull objects toward itself at the strength that Earth will.

When a 600 g mass is suspended from a spring, the spring stretches 1.2 cm. What is the spring constant of the spring? ​

Answers

Well first you need to know the formula for the spring force. It’s -1/2k*d

K being the constant

D being the displacement

We can actually find the force on the spring by calculating the weight of the block. It would be mass multiplied by gravity, convert the grams to kilograms first tho.

600/1000 = 0.6kg

0.6kg * 9.8 m/s^2 = 5.88N

Now plug everything into the equation, also convert the centimeters to meters:

1.2/100 = 0.012 meters

5.88 = -1/2k*(0.012)

5.88 = k*(0.006)

We could drop the negative sign because we really just want the magnitude of the spring constant.

K = 980N

which produces wave particles that move and displace one another

Answers

Answer:

a guitar?

Explanation:

A 10 [kg] object is dropped from rest. a. How far will it drop in 2 [s]?

Answers

Answer:

y=0.5 g t^2

=0.5*10*2^2

=20 m

what are Newton's 3 laws

Answers

Answer:

Newton 3 laws

Explanation:

1.object will not change its motion unless a force acts on it . 2.the force on an object is equal to it's time mass on acceleration. 3. When 2 o jecys interact they apply forces to each other of equal.

Answer- In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Hope this helps:)

When a heavy football player and a light one run into each other, who exerts more force?

Answers

The light football player because he going against someone bigger than him

Answer:

When a heavy football player and a light one run into each other, does the lighter player really exert as much force on the heavy player s the heavy player exerts on the light one. Yes. The interaction between the two players, the force each exerts on the other have equal strength.

Explanation:

2
A crane has an arm to which is attached a frictionless pulley. A cable passes over the pulley and
supports a load of 10 kN as shown.
frictionless pulley
104N
30
30°
crane arm
load
10 KN
The crane arm exerts a force F on the pulley.
What is the value of F?
C
8.7 KN
10 KN
D
A
17 kN
B
5.0 KN

Answers

Answer:

17kN

Explanation:

The force of 10 kN in the cable and the load of 10 kN both have a component along the crane arm.

The force F exerted by the crane arm on the pulley is the sum of these 2 components.

Component of force in cable along the crane arm = 10 cos 30°

Component of load along the crane arm = 10 cos 30°

Considering the forces on the pulley,

F = 10 cos 30° + 10 cos 30°

Force F = 17.32 kN = 17 kN

PLEASE PLEASE HELP PICTURE INCLUDED

Answers

Answer:

a is the answer

Explanation:

need help ASAP. please and thank you ​

Answers

I believe the correct answer is 9

A yo-yo is attached to the end of a string. A professional yo yo is spinning it by the string around in a circular
motion. All of a sudden he lets go of the string. What direction is the yo yo moving in initially?

Answers

Answer:

counter-clockwise

Explanation:

There is not enough information here. The yo-yo will continue to move in a straight line at the point where it is released. Newton’s 1st Law.

i know this is a lot to do but i will give you the highest amount of points (100, but you get 50 from it.) and i will cash app you $20 just tell me your cash app in the comments. please don't post troll answers. please be legit! this is due at 11:59 today, so in about 5 hours. the questions are in the images below. thank you so much!

Answers

Answer:

mel mel 4 ever

Explanation:


Planets A and B have the same size, but planet A is half the mass of planet B.
Which statement correctly explains the weight you would experience on each
planet?
A. You would weigh the same on both planets because the planets
are the same size.
B. You would weigh less on planet A because it has less mass than
planet B.
C. You would weigh the same on both planets because your mass
would be the same on both.
D. You would weigh more on planet A because it has less mass than
planet B.

Answers

Answer:

B

Explanation:

The more mass an object has, the more gravity it has.

The statement which correctly explains the weight you would experience on each planet is: B. You would weigh less on planet A because it has less mass than  planet B.

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, the weight of an object is given by the formula;

[tex]Weight = mg[/tex]

Where;

m is the mass of the object.g is the acceleration due to gravity.

Hence, we can deduce that the weight and gravity acting on an object is highly dependent on the mass of an object.

Therefore, the higher the mass in a planet, the higher the gravity existing there.

Read more: https://brainly.com/question/18320053

Describe two similarities and two differences between electric and magnetic field lines. (Consider such things as where they originate and terminate, how they are related to the direction and strength of the field, whether they are closed curves or lines, and whether there's anything you can say about their flux through a closed surface.). Others

Answers

Answer:

Please see below as the answer is self- explanatory.

Explanation:

Similarities:The field vector is tangent to the field lines at any point of the trajectory.The density of field lines is proportional to the strength of the field.Differences:Electrostatic Field lines originate in charge distributions of positive charges and end in charge distributions of negative charges.Magnetic Field lines are closed lines, due to there are no isolated magnetic charges discovered yet.Electrostatic Field lines have the direction of the trajectory taken by a positive test charge.Magnetic Field lines have the direction defined by right-hand rule.Vector net flux of electrostatic field lines through a closed surface is equal to the net charge enclosed by the surface.Vector net flux of magnetic field lines through a closed surface is always zero.

Similarities

Field vectorDensity of field lines

Differences

Dispersions of positive charges .No separated attractive charges.

Electric and Magnetic field lines.

Similarities:

The field vector is tangent to the field lines at any point of the trajectory. The density of field lines is relative to the quality of the field.

Differences:

Electrostatic Field lines begin in charge dispersions of positive charges and conclusion in charge dispersions of negative charges.

Magnetic Field lines are closed lines, due to there are no separated attractive charges found yet.

Learn more about "Magnetic Lines":

https://brainly.com/question/17012638?referrer=searchResults

Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to u

Answers

Answer:  Option A is your answer

Explanation:

Why is it important to warm up your muscles?
Warming up the muscles increases lactic acid build up
Warming up the muscles increases temperature and flexibility of your muscles Warming up the muscles increases your risk of injury
All of the above

Answers

Answer: Warming up the muscles increases temperature and flexibility of your muscles
Hey :)

The answer would be

B. Warming up the muscles increase temperature and flexibility of your muscles.

Because, lactic acid buildup is from overexerting yourself when working out, and warming up muscles does not increase your risk in injury, rather, it decreases your risk of injury

Hope this helps!

why cats cannot run on smoth table​

Answers

Answer:

bc of their paw pads, they don't get the traction they need in order to be able to get a firm grip to push off of on the table for each bound

Explanation:

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10 m/s2 for the downward acceleration due to gravity.

Answers

Answer:

v_y = 10 - 10t

v_x = 10 m/s

Explanation:

This is a projectile launch where the x axis and y axis will be treated independently.

Now, towards the north on the x-axis, there will be no acceleration and so the speed is constant

So, vₓ = v₀ₓ

Whereas, on the vertical y - axis, the acceleration due to gravity with be negative since it's in a downward direction.

Thus, the equation is;

v_y  =  v_oy - gt

Now, the initial velocity component will be;

cos 45 = v₀ₓ/v₀

And sin 45 = v_₀y/v₀

Thus, we have;

v₀ₓ = v₀(cos 45)

Also, v_oy = v₀(sin 45)

Now, the initial velocity would be gotten from the equation of range which is;

R = (v₀² × sin 2θ)/g

Making v₀ the subject, we have;

v₀ = √(Rg/sin 2θ)

We are given;

R = 20 m

g = 10 m/s²

θ = 45°

Thus;

v₀ = √[20 × 10/(sin (2 × 45))]

v₀ = √200

v₀ = 14.14 m/s

Thus;

v₀ₓ = 14.14(cos 45) = 10 m/s

v_oy = 14.14(sin 45) = 10 m/s

Earlier we saw that;

v_y  =  v_oy - gt

Thus;

v_y = 10 - 10t

Also,we saw that;

vₓ = v₀ₓ

Thus;

v_x = 10 m/s

For the graph, we will use times of  t = 0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2

Thus, for each of those values of t, we will have the following values of v_oy

t (s)       v_oy (m / s)

0           10

0.2         8

0.4          6

0.6         4

0.8          2

1.0         0

1.2          -2

Graph is attached

A race car driver travelling at 15m*s^-1 accelerates at a constant value for 10.0s. She is now driving at a speed of 35 m*s^-1 What was her acceleration? Give your answer in m*s^-2 to one significant figure . Do not include units with your answer.

Answers

The driver speeds up with acceleration a so that

35 m/s = 15 m/s + a (10.0 s)

Solve for a :

20 m/s = a (10.0 s)

a = (20 m/s) / (10.0 s)

a = 2 m/s²

define density with a suitable formula​

Answers

Answer:

ratio between m/v

or d= m/v

m= mass

v= volume

10. The electron dot diagram for the element Ne would have ​

Answers

Answer:

2,8

Explanation:

The first electron shell would have 2 electrons, the second shell would have 8 electrons. This is because Neon has a relative charge of 10.

the graph below represents the relationship between velocity and time for an object moving along a straight line.

What is the magnitude of the object's acceleration?

1. 5.0 m/s^2

2. 8.0 m/s^2

3. 10. m/s^2

4. 20. m/s^2​

Answers

Answer:

10. m/s^2

Explanation:

40-(-10)= 50

50/5 =10

This question can simply be answered by using the slope of the velocity-time graph.

The correct answer for the magnitude of the acceleration is "3. 10 m/s²".

Since the graph between the velocity and the time is in the form of a straight line. Hence, the acceleration can be calculated as the slope of this line. We will use the first and last point of the line to calculate the slope (acceleration):

[tex]Acceleration = Slope = \frac{v_2 - v_1}{t_2-t_1}[/tex]

where,

v₁ = initial velocity from the graph = - 10 m/s

v₂ = final velocity from the graph = 40 m/s

t₁ = initial time from the graph = 0 s

t₂ = final time from the graph = 5 s

Therefore,

[tex]Acceleration = \frac{40\ m/s\ -\ (-\ 10\ m/s)}{5\ s\ -\ 0\ s}[/tex]

Acceleration = 10 m/s²

The attached picture shows the types of a velocity-time graph.

Learn more about the velocity-time graph here:

https://brainly.com/question/11290682?referrer=searchResults

Glucose is a carbohydrate.
True
0r False

Answers

Answer: True

Explanation: Glucose is a carbohydrate.

Can you help me please! Love you​

Answers

Answer:

I think its B. wait...U love ME?????

8. In explosives, potential energy is generally
stored in the form of
a thermal energy
b. electromagnetic energy
c. nuclear energy
d. chemical energy

Answers

Answer:

In explosives,potential energy is in the stored form of chemical energy.

A rubber cord catapult has cross-sectional area
25mmand initial length of rubber cord is 10c.
Its is stretched to 5cm and then released to
project a missile of mass 5gm. Taking
Yubber = 5x108N/m² velocity of projected
missile is :
(1) 20ms -
(2) 100ms
(3) 250ms
(4) 200ms)​

Answers

Potential energy stored in the rubber cord catault will be converted into kinetic energy of mass
2
1

mv
2
=
2
1


L
YAl
2



where l=5×10
−2
m, A=25×10
−6
m
2
, L=10×10
−2
m, m=5×10
−3
kg
⇒v=
mL
YAl
2





v=
5×10
−3
×10×10
−2

5×10
8
×25×10
−6
×(5×10
−2
)
2





v=250ms
−1

1) A rigid tank contains 2 kg of 134a refrigerant at 800 kPa and 120°C. Determine tank volume and total internal energy.

Answers

Answer:

V = 0.07525 [m³]

U  = 655.74 [kJ]

Explanation:

To solve this problem we must use the superheated refrigerant table for refrigerant 134a. The attached image shows the specific volume and specific internal energy values for a pressure of 800 [kPa] and a temperature of 120 [°C].

m = mass =2 [kg]

the specific volume = v = 0.037625 [m³/kg]

The volume of the tank can be found using the following expression:

V = v*m

V = 0.037625*2 = 0.07525 [m³]

The Specific internal energy = u = 327.87 [kJ/kg]

U = u*m

U = 327.87*2 = 655.74 [kJ]

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Compute the equivalent units of production for materials and conversion costs for the month of September.MaterialsConversion CostsThe equivalent units of production2. Compute the unit costs for materials and conversion costs for the month. (Round unit costs to 2 decimal places, e.g. 2.25)MaterialsConversion CostsUnit Costs3. Determine the costs to be assigned to the units transferred out and in process. (Round unit costs to 2 decimal places, e.g. 2.25 and final answers to 0 decimal places.)Transferred Out$Ending work in process$ A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.In figure a, the block is to the left of the spring, and an arrow above the block points to the right.In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.(a)Find the distance of compression d (in m). m(b)Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d). m/s(c)Find the distance D (in m) where the object comes to rest. m(d)What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left? m