19. A beach ball is rolling in a straight line toward you at a speed of 9 m/sec. Its momentum is 3 kg'm/sec.
What is the mass of the beach ball? moss - 0.33
aroto from 10mle to speed of 14 m/s The mass of the bicycle

Answers

Answer 1

The mass of the beach ball is 0.33 kg

From the question,

We are to determine the mass of the beach ball.

Using the formula

ρ = mv

Where ρ is the momentum

m is the mass

and v is the velocity

From the given information,

ρ = 3 kg m/sec

v = 9 m/sec

Putting the parameters into the formula, we get

3 = m × 9

∴ m = 3 ÷ 9

m = 0.33 kg

Hence, the mass of the beach ball is 0.33 kg

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Related Questions

What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?

Answers

Hi there!

Recall Newton's Second Law:

[tex]\large\boxed{\Sigma F = ma}[/tex]

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We must begin by solving for the acceleration using the following:

a = Δv/t

In this instance:

Δv = 3 m/s

t = 2.5 sec

a = 3/2.5 = 1.2 m/s²

Now, plug this value along with the mass into the equation for net force:

[tex]\Sigma F = 5(1.2) = \boxed{6 N}}[/tex]

Answer:

[tex]\boxed {\boxed {\sf 6 \ Newtons}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration. The mass of the object is 5 kilograms, but the acceleration is unknown.

We can find acceleration using the following formula:

[tex]a= \frac{ \Delta v}{t}[/tex]

The velocity of the object increases by 3 meters per second. It accelerates in 2.5 seconds.

Δv= 3 m/s t= 2.5 s

Substitute the values into the formula.

[tex]a= \frac{3 \ m/s}{2.5 \ s}[/tex]

[tex]a= 1.2 \ m/s/s = 1.2 \ m/s^2[/tex]

Now we know the mass and acceleration.

m= 5 kg a= 1.2 m/s²

Substitute the values into the force formula.

[tex]F=ma[/tex]

[tex]F= (5 \ kg)(1.2 \ m/s^2)[/tex]

[tex]F= 6 \ kg*m/s^2[/tex]

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 6 kilogram meters per second squared is equal to 6 Newtons.

[tex]F= 6 \ N[/tex]

Please help if you can

Answers

Answer:

The answer is B

Explanation:

Because when the both sides aren't balanced one side has to cause motion. (fall down)

What two things related to sedimentary rocks do you see at the Grand Canyon ?

Answers

Answer:

sandstone and mudstone

Explanation:

Does the coefficient of kinetic friction depend on the speed?

Answers

Answer:

yes because if the speed/K.E. of body increases the friction will also apply more instantly

What would be the difference in your welght if you were four times farther from the center of Earth than you are now?

Answers

you would die.

the core of the earth is as hot as the sun surface so you would be dead

Help me pls both answers i’ll give brainly

Answers

Do u know the formula

A weightlifter lifts a 350 N set of weights over his head a vertical distance of 2.5 meters above the floor. Calculate the work the athlete does on the weights, assuming he lifts them at a constant speed.

A. 525 J

B. 750 J

C. 1005 J

D. 700 N•m

E. 875 J

Answers

Answer: D) 700 N•m

Answer: Took that test already
The answer is E! I use a calculator and I did 350 divided by 2.5 and the answer was 875!!

who has the best answer for this
what percentage more water is used to provide us with electricity vs for irrigation

Answers

Answer:

electricity because it has more percentage nd energy

Explanation:

mark me brainliest pl

Hi, please help! •_•

One workman is measured as having a power of 528W. His weight is 800N. He can develop the same power climbing a ladder, whose rungs are 30cm apart. How many rungs can he climb in 5s?

Answers

Answer:

He can climb 11 rungs

Explanation:

Power = 528W, Force = 800N, Time= 5s Total Distance moved = 30x cm or 0.3x m, where x is the number of rungs he climbed

Power = Work done/time

Work done = Power × time

= 528 × 5

= 2640Joules

Also Work done = Force × total distance moved

Total Distance moved = Work done/force

0.3x = 2640/800

0.3x = 33

x = 3.3/0.3

x = 11

Give proof that mass is constant and weight keeps changing.​

Answers


The mass of an object stays the same wherever it is, but its weight can change. This happens if the object goes where the gravitational field strength is different from the gravitational field strength on Earth, such as into space or another planet.

Convert 8 m/s to miles per hour

Answers

Answer:

17.8954903 or 17.9 mph

Explanation:

m/s is meters per second. 1609.344 meters is one mile. There are 3,600 seconds in an hour.  8 meters *3600 seconds= 28,800.  28,800/1609.344 = 17.8954903. Simplify to 17.9

free tingsssssssssss

Answers

Explanation:

yayyy thx u soo muchihuhuhu

Answer:

thx

Explanation:

PLEASE HELP!! 10 points!!!! brainliest!!!
Why is a sound wave considered to be a mechanical wave? What properties of sound
waves are responsible for pitch and volume

Answers

Answer:

Mechanical Sound Waves - A sound wave moves through air by displacing air particles in a chain reaction. ... Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave. For pitch the higher the frequency, the higher the pitch. Frequency is measured in hertz, or cycles per second. Frequency also affects loudness, with higher-pitched sounds being perceived as louder. Amplitude and frequency of sound waves interact to produce the experiences of loudness and pitch.

Mechanical Sound Waves - A sound wave moves through air by displacing air particles in a chain reaction. ... Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave. For pitch the higher the frequency, the higher the pitch. Frequency is measured in hertz, or cycles per second. Frequency also affects loudness, with higher-pitched sounds being perceived as louder. Amplitude and frequency of sound waves interact to produce the experiences of loudness and pitch.

Consider a car of mass m moving with an initial speed Vo on a straight, flat road. At time t=0, the driver fully applies the breaks to avoid colliding with debris in the road in front of the car. The car’s wheels lock, causing the car to slide on the roadway until the car stops, before running over the debris. The distance that the car slides is D. The coefficient of kinetic friction between the car’s tires and the roadway is a constant value of Uk.

A) Sketch a free body diagram for the car sliding to rest and write Fnet equations

A) does the value D increase or decrease with increasing initial speed?

B) does the value D increase or decrease with increasing the coefficient of friction?

C) derive an equation for the acceleration of the car in terms of Uk and physical consonants

Answers

the answer is B :)))

Someone answer? Please

Answers

I don’t know ask your parents

What did Ernest Rutherford expect to happen when he aimed a beam of particles at a thin gold foil

Answers

Answer:

he expected a fire to happen I think

Answer:

When Ernest Rutherford aimed a beam of particles at a thin gold foil he expected that the particles would be deflected slightly after passing through the foil.

Explanation:

hope it helped

Do two indentical objects become statically charged when you rub them together? Explain why they do or do not.

Answers

Answer:

Yes, they do. When two different materials are rubbed together, there is a transfer of electrons from one material to the other material. This causes one object to become positively charged and the other object to become negatively charged.

Explanation:

hope this helps

How to find final velocity and initial velocity and acceleration ? :(

Answers

Explanation:

Final velocity=Initial velocity+(acceleration×time)

4 ways to find initial velocity:

1) Initial velocity=Final velocity-(acceleration×time)

2) Initial velocity=(Distance/Time)-((acceleration×time)/2)

3) Initial velocity=√Final velocity-(2×(acceleration×distance))

4) Initial velocity=2(distance/time)-Final velocity

Total force = Mass×Acceleration

(F=ma)

A 2-kg object is pulled to the left with a force of 30 N and to the right with a force of 9 N. What is the acceleration of the object?

Answers

An object that weighs  75  N is pulled horizontally to the right with a force of  50  N. The force of friction on this object is  30  N to the left. What is the acceleration of the object?

The force on the object is  50  N to the right and  30  N to the left.

⇒  The net force on the object is,  =50−30=20  N to the right.

The weight of the object is,  =75  N.

⇒  The mass of the object is  ==759.8  kg.

⇒  The acceleration of the object is  ==20759.8=2.613  m/s 2.

The acceleration of the object is 10 [tex]m/s^2[/tex].

What is Acceleration?

Acceleration is defined as the rate of change of velocity of an object with respect to time which are vector quantities i.e. both magnitude and direction. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

It can be expressed as:

a= v/t

where, a is the acceleration

v is the velocity

t is the time taken

It can also be expressed as F=ma

where, F= force applied

m= mass of the object

So, for above given information

The force on the object is 9N to the right and  30N to the left. The net force on the object is 21N to the left

The mass of the object is 2kg

So, a= 21N/ 2 kg = 10.5 [tex]m/s^2[/tex]

Thus, the acceleration of the object is 10 [tex]m/s^2[/tex].

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Find the central force acting on the electron as it revolves in a circular orbit of radius 4.63 x 10-11 m.
Answer in units of N.

Answers

The central force acting on the electron as it revolves in a circular orbit is [tex]9.52 \times 10^{-8} \ N[/tex].

The given parameters;

speed of electron, v = 2.2 x 10⁶ m/sradius of the circle, r = 4.63 x 10⁻¹¹ m

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

[tex]F = \frac{M_e v^2}{r} \\\\[/tex]

where;

[tex]M_e[/tex] is mass of electron = 9.11 x 10⁻³¹ kg

[tex]F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N[/tex]

Thus, the central force acting on the electron as it revolves in a circular orbit is [tex]9.52 \times 10^{-8} \ N[/tex].

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Physicists call any change in energy an impulse true or false?

Answers

False
.!.7:&//‘is sis sis did is

write of the following does not represent electrical power in a circuit ?

(a) I²R


(b) IR²

(c)VI

(d) V²/R​

Answers

(b) IR^2 does not represent electrical power in a circuit.


The equations for electrical power are as follows:

P=I^2R

P=VI

P=V^2/R
(You can get this equation by rearranging V=IR, making I the subject to get I=V/R and substituting that into P=VI)

P=E/T


(a) One medium is air, in which the speed of light is 3.00 x 108 m/s. The other
medium is ice, in which light has a speed of 2.29 x 108 m/s. Use this information to
identify medium A and medium B. Explain

Answers

Answer:

WARNING WRONG SPELLING.that is a ternadow and very dangeruse air .

I am unique even though there are billions of me no two are the same what am I


AND IT'S NOT A FINGER-PRINT

Answers

I think the answer is Snow flakes

Since every ice crystal takes a different route to the ground. They will float through various clouds with various temperatures and moisture contents, which will cause the ice crystal to develop in a distinctive manner.

What make snowflakes unique?

A snowflake's shape changes as it flies through the atmosphere, thus no two are ever the same. Even two flakes floating side by side will each be blown through various vapour and humidity levels to produce a genuinely unique shape.

One of the most recognizable representations of winter weather are snowflakes. There are an unlimited amount of conceivable shapes for snowflakes, and each one is distinct.

Therefore, snowflakes unique even though there are billions.Even two floating flakes will each be exposed to different vapour and humidity conditions, creating a really distinctive form.

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will mark brainliest if answer​

Answers

I won't touch the object because there is a warning sign and that means you stay out if it because it is a danger sign that High Resistance Material that means it is dangerous.

A new moon is discovered orbiting Neptune with an orbital speed of 9.3 x103 m/s. Neptune's mass is 1.0 x1026 kg. A) What is the radius of the new moon's orbit? B) What is the orbital period? Assume that the orbit is circular. (G = 6.673 x10-11 N-m²/kg?​

Answers

The radius of the new moons orbit is R= 7.715 x 10^7 m
The orbital period of the moon is T= 14.48 hr

A father places his daughter in a swing that is 0.60\,\text{m}0.60m0, point, 60, start text, m, end text above ground. Then he raises the swing to a height of 1.3\,\text{m}1.3m1, point, 3, start text, m, end text and lets go. The girl and the swing have a combined mass of 14\,\text{kg}14kg14, start text, k, g, end text. Assume friction is negligible and use g = 9.8\,\text{m/s}^2g=9.8m/s
2
g, equals, 9, point, 8, start text, m, slash, s, end text, squared.
Calculate the girl’s fastest speed.

Answers

This question involves the concepts of the law of conservation of energy and kinetic energy.

The girl's fastest speed is "3.7 m/s".

According to the law of conservation of energy, the girl will have the fastest speed at mean position, which will be calculated as follows:

Loss in Potential Energy = Gain in Kinetic Energy

[tex]mg\Delta h=\frac{1}{2}mv^2\\\\v=\sqrt{2g\Delta h}[/tex]

where,

v = maximum speed = ?

g = acceleration due to gravity = 9.81 m/s²

Δh = change in height = 1.3 m - 0.6 m = 0.7 m

Therefore,

[tex]v=\sqrt{2(9.81\ m/s^2)(0.7\ m)}[/tex]

v = 3.7 m/s

Learn more about the Law of Conservation of Energy here:

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I need help asap! Will give brainliest

Answers

[tex]\\ \tt\hookrightarrow \dfrac{(3\times 10^4kg)(4\times 10^4m)}{6\times 10^4s)}[/tex]

[tex]\\ \tt\hookrightarrow \dfrac{12\times 10^8kgm}{6\times 10^4s}[/tex]

[tex]\\ \tt\hookrightarrow 2kgm/s[/tex]

We have calculated momentum

A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.
a) What is the range of values of the initial velocity so that the projectile falls between points M and N?

Answers

Answer: A

Explanation:

I want my points so yea

If an input force of 202N is applied to the handles of the wheelbarrow in the sample problem how large is the output force that just lifts the load?

Answers

For this case, the relationship between the two forces is given by:

F1 = nF2

Where,

F1: output strengthF2: input forcen: mechanical advantage

Then, replacing the values we have:

F1 = (2.2)(202)

Having the calculations we have:

F1 = 444.4N

Answer: The output force that only lifts the load is F1 = 444.4N.

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