Can someone help me please

Can Someone Help Me Please

Answers

Answer 1
the answer to your question is distance
Answer 2

Answer:

distance i think im not sure


Related Questions

If you use the same force to push a motorcycle as you would push a bike which one would have more acceleration and why explain using Newton's second law

Answers

Answer:

The bike would have more acceleration

Explanation:

Accourding to newtons first law a force is equal to its mass multiplied by its acceleration (f=ma) therefore an object with a higher mass compared to an object with a lower mass would experience less acceleration.

Eg.

F=50N

Motorbike M=200kg

F=ma

50=200 x a

50/200=a

0.25m/s/s =a

Bike M=35kg

F=ma

50=35 x a

50/35= a

1.43m/s/s=a

Answer:

the bike

Explanation:

it has less mass than the motorcycle so so it would have more acceleration

The air pressure inside a car tire is

3.00 atm, and it creates 41,500 N

of force pushing out on the tire.

What is the surface area of the

inside of the tire?

Answers

Answer:

0.137m²

Explanation:

Pressure = Force/Area

Given

Force = 41,500N

Pressure = 3.00atm

since 1atm = 101325.00 N/m²

3atm = 3(101325.00)

3atm = 303,975N/m²

Pressure = 303,975N/m²

Get the area

Area = Force/Pressure

Area = 41500/303,975

Area = 0.137m²'

Hence the surface area of the  inside of the tire is 0.137m²

What is the mathematical relationship between height, mass, gravity, and gravitational potential energy?

Answers

The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity. The gravitational potential energy of an object near Earth's surface is due to its position in the mass-Earth system.

A sound wave produce by a chime 515 m away is heard 1.50 s later.What is the speed of the sound in air?

Answers

Answer:

the seed of the sound in the air is 343.3 m/s

Explanation:

The computation of the speed of the sound in the air is shown below:

As we know that

Speed = Distance divided by Time

So, here speed be 515m

And, the time is 1.50 second

So, the speed of the sound is

= 515m divided by 1.50 seconds

= 343.3 m/s

hence, the speed of the sound in the air is 343.3 m/s

3. Why is static electricity not useful as a power

source?

A. Because electrons aren't transferred in bursts of

static electricity

B. Because all energy is released at once in static

electricity

C. Because static electricity is not a real form of

electricity

D. Because static electricity only occurs in lightning

Answers

The answer is B, because all energy is released at once in static electricity.

There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.

Static electricity is not useful as a power source primarily because it releases all of its energy at once rather than providing a continuous and controlled flow of electrical energy. The correct answer is B.

When static electricity is discharged, such as in a spark or a sudden discharge of stored charge, it happens in a rapid and uncontrolled manner, resulting in a brief burst of energy rather than a steady and sustained flow.

Power sources typically require a continuous and controllable flow of electrical energy to be useful for various applications. Static electricity, in its nature, does not provide this continuous flow but instead releases energy in an instantaneous manner. Therefore, it is not suitable for most practical power needs and applications.

Option A is not correct because electrons can be transferred in bursts of static electricity. Option C is not correct because static electricity is a real form of electricity, even though it has unique characteristics. Option D is not correct because static electricity can occur in various circumstances, not just in lightning.

Therefore, Static electricity is not useful as a power source primarily because it releases all of its energy at once rather than providing a continuous and controlled flow of electrical energy. The correct answer is B.

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Answer using mass m= F divided by a

And also use kilograms {kg}

A skyrocket is launched with a force of 10 N and accelerates at 20 m / s2. What is the mass of this skyrocket?

Answers

Answer:

mass=force/acceleration

m=10/20

m=0.5kg

A bear scratches his claws on a big pine tree. Does the bear do work on the tree? why or why not?
(btw it has to do with physics) ITS NOT A GENERAL QUESTION.

Answers

Answer:

The bear is actually sharpening his claws. That's the only reason he scratches a tree with his claws

Explanation:

Ultraviolet rays from the sun are dangerous because they can damage skin. Why do ultraviolet rays cause damage?

Answers

Answer:

Skin damage

Explanation:

exposure can lead to sunburn, and long term exposure can lead to aged skin and much more

Thanks+ BRAINLIST only for correct answers

AM radio waves have a
a. Shorter broadcast range than FM waves
b. Longer broadcast range than FM waves
C. Better quality sound than FM waves
d. More music options

FM radio travel a shorter distance than AM radio waves.
a.true
b.false


Radio stations, police radios, and amateur radios don't interfere because they
have different
a. volumes
b. music options
C. frequencies

Answers

1st - ‘B’

2nd - ‘A’ ( given by question 1’s answer)

3rd -because of variations in frequency ‘C’

Joe and tim have the same mass. Tim stands on a desk that is very high in the air, while Joe stays on the ground. Who had higher potential energy?

Answers

Answer:

Tim has a higher potential energy than Joe.

Explanation:

The higher an object or an organism is, the more potential energy it has

A string of density 0.01 kg/m is stretched with a tension of 5N and fixed at both ends. The length of the string is 0.1m. What is the first four resonance frequencies in the string?

Answers

Answer:

The first four resonance frequency in the string are;

1) 50·√50 Hz

2) 100·√50 Hz

3)150·√50 Hz

4) 200·√50 Hz

Explanation:

The given parameters of the string are;

The density of the string, ρ = 0.01 kg/m

The tension force on the string, T = 5 N

The length of the string, l = 0.1 m

Therefore the mass of the string, m = Length of string × Density of the string

∴ m = 0.01 kg/m × 0.1 m = 0.001 kg

The formula for the fundamental frequency, f₁, is given as follows;

[tex]f_1 = \dfrac{\sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{\sqrt{\dfrac{T}{\rho} } }{2 \cdot L}[/tex]

Where;

f₁ = The fundamental frequency in the string

T = The tension in the string = 5 N

m = The mass of the string = 0.001 kg

L = The length of the string = 0.1 m

ρ = The density of the string = 0.01 kg/m

By plugging in the values of the variables, we have;

[tex]f_1 = \dfrac{\sqrt{\dfrac{5}{0.01} } }{2 \times 0.1} = 50 \cdot \sqrt{5}[/tex]

The first four harmonics are;

f₁, 2·f₁, 3·f₁, 4·f₁

Therefore, we have the first four resonance frequency of the string are as follows;

1 × 50·√50 Hz = 50·√50 Hz

2 × 50·√50 Hz = 100·√50 Hz

3 × 50·√50 Hz = 150·√50 Hz

4 × 50·√50 Hz  = 200·√50 Hz

A 1600 kg car with the brakes applied comes to a stop in 4.20 seconds. During these 4.20 seconds the force of friction slowing down the car is 3900 N. What is the change in momentum of the car?

Answers

Answer:

The change in momentum of the car is 16380.8 kg.m/s

Explanation:

Given;

mass of the car, m = 1600 kg

time of motion, t = 4.20s

force of friction on the car, F = 3900 N

final velocity of the car after the brakes were applied, v = 0

The initial velocity of the car during the motion is calculated as;

[tex]F = ma = \frac{mu}{t} \\\\mu = Ft\\\\u = \frac{Ft}{m} \\\\u = \frac{3900\times 4.2}{1600} \\\\u = 10.238 \ m/s[/tex]

The change in momentum of the car is calculated as;

ΔP = mu

ΔP = 1600 x 10.238

ΔP = 16380.8 kg.m/s

The lamp has a resistance of 10Ω
Each resistor has a resistance of 10Ω (resistors are connected in parallel)
What is the total resistance of the circuit?

Tick one box

Between 20 and 30Ω?
Exactly 20Ω?
Exactly 30Ω?
Less than 20Ω?

Answers

D) Less than 20.

Explanation:

Equivalent resistance in a parallel combination is less than their individual value.

Derivation
law of
conservation
of momentum​

Answers

Derivation of Conservation of Momentum

Applying Newton's third law, these two impulsive forces are equal and opposite i.e. If the time of contact is t, the impulse of the force F21 is equal to the change in momentum of the first object. The impulse of force F12 is equal to the change in momentum of the second object.

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If two negative charges of -2 C and -2 C are put 32,692 m apart from each other how much force will push them apart?

Answers

Answer:

33.65 N

Explanation:

Force that will push them away is given from the equation;

F = Kq1•q2/r²

Where;

K is coulumbs constant = 8.99 × 10^(9) Nm²/C²

We are given;

q1 = q2 = -2 C

r = 32692 m

Thus;

F = 8.99 × 10^(9) × (-2) × (-2)/32692²

F ≈ 33.65 N

Escribir usando prefijos, en unidades del Sistema Internacional: longitud del ecuador, radios del núcleo y átomo, segundos de un milenio, edad de la Tierra, volumen de una pulga, masa del Sol, distancia de la estrella más cercana a la Tierra (después del Sol).

Answers

la característica de los elementos

Two identical conducting spheres are charged with a net charge of +5.0 q on the first sphere and a net charge of −8.0 q on the second sphere. The spheres are brought together, allowed to touch, and then separated. What is the net charge on each sphere now?

Answers

Answer:

The net charge on each sphere is -1.5 q

Explanation:

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

[tex]Q = \dfrac{Q_1 + Q_2}{2}[/tex]

Therefore, by substituting the known values of the variables, we have;

[tex]Q = \dfrac{5 \ q+ (-8 \ q)}{2} = -\dfrac{3 \ q}{2} = -1.5 \ q[/tex]

The net charge on each sphere, Q = -1.5 q.

The net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

What is charge?

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

[tex]Q=\dfrac{Q_1+Q_2}{2}[/tex]

Therefore, by substituting the known values of the variables, we have;

[tex]\dfrac{5q+(-8q)}{2}=-1.5q[/tex]

Hence the net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

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Afootball is kicked at avelocity of 15 m/s at angle of 25 degree to the horizontal. What is the total flight time

Answers

Answer:

the total flight time is 1.29 s

Explanation:

The computation of the total flight time is given below:

The initial velocity is 15 m/s = v

And, the angle from horizontal is [tex]\theta = 25\ degrees[/tex]

Now the total time is

[tex]t = \frac{2\times v\times \sin \theta }{g} \\\\= \frac{2\times 15\times sin 25^{\circ}}{9.8} \\\\= 1.29 s[/tex]

Hence, the total flight time is 1.29 s

After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample

Answers

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

The half life of the sample will be "4 days".

According to the question,

Let,

The number of half life = n

→  [tex]\frac{N}{N_o} = (\frac{1}{2} )^n[/tex]

By substituting the values, we get

→  [tex]\frac{2}{128} = (\frac{1}{2} )^n[/tex]

→    [tex]\frac{1}{64} = (\frac{1}{2} )^n[/tex]

→ [tex](\frac{1}{2} )^6= (\frac{1}{2} )^n[/tex]

→     [tex]n=6[/tex]

So,

It takes 6 half life just to reduce from 128 mg - 2mg which is also "24 days".

→ [tex]6 \ half \ life = 24 \ days[/tex]

Hence,

→ [tex]Half \ life = \frac{24}{6}[/tex]

                   [tex]= 4 \ days[/tex]

Thus the answer above is right.

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Answer using mass m= F divided by a

And also kilograms {kg}

A 7.5 N force is applied to a football generating an initial acceleration of 15 m / s2. Calculate the mass of the football.

Answers

Answer:

from

force =mass x acceleration

mass = force/acceleration

m = f/a

m = 7.5/15

m=0.5kg

[tex]\huge\bf{\pink{\underline{\underline{\mathcal{AnSwer࿐}}}}}[/tex]

Given:-

acceleration = 1000 m/s²

force = 5000 N

To find:-

Mass of the cannonball

Formula to be used:-

[tex]\longrightarrow[/tex] [tex]\underline{\boxed{\sf mass = \dfrac{force}{acceleration}}}[/tex]

Solution:-

:[tex]\implies[/tex] [tex]\sf mass = \dfrac{5000}{1000}[/tex]

:[tex]\implies[/tex] [tex]\sf mass = 5 kg [/tex]

hence, the required answer is 5kg.

______________________________

PLEASE HELP WILL MARK BRAINLIEST
In the formula Q = m x C x ΔT, which symbol represents specific heat?
A. Q
B. ΔΤ
C. m
D. с

Answers

Answer:  D. с

Explanation:

Q = energy applied

m = mass

C = specific heat

ΔT = delta T = change in temperature

If a man with
with long-sighted eye want to
a' text book, what shoðld be the distance between
book and the lens? Give reason​

Answers

Lots of factors play a role. Firstly if he is presbyopic he won't be able to read without specs at normal reading distances. Many years ago a chap by the name of Donders published a table which shows a linear relationship between age and ability to focus. The ready made reader market relies on his findings to suggest the power you need for reading at different ages. Secondly if he is far sighted (needing correction at distance to see clearly) it will also influence his ability to see close. For example +3.00D distance Rx at an age of 40 years will be very different than the same prescription at age 20 years due to his ability to accommodate. My suggestion is to have comprehensive eye exam to find out what you need at the specific working distance (computer or laptop or book reading all have different reading distances).

How far must a spring with a spring constant of 64N/m be stretched to
store 0.32J of potential energy?

Answers

Answer:

0.1m

Explanation:

k = 64N/m

Ep = 0.32J

x = ?

Ep = 1/2 * kx²

make x the subject of the formula

x² = 2Ep/k

x = √(2Ep/k)

x = √(0.64/64)

x = √0.01

x = 0.1m

A car weighing 16,170 newtons is placed on a hydraulic lift. Piston B has an area of 5,005 cm?

and piston A has an area of 65 cm². What force must be exerted on piston A to lift the car?

P = F/A where Pequals pressure, Fequals force, and A equals area

Answers

Answer:

210N

Explanation:

According to pascal principle

Fb/Ab = Fa/Aa

Given

Fb = 16,170N

Ab = 5,005cm²

Fa = ?

Aa = 65 cm².

Fa is the force that must be exerted on piston A to lift the car

Substitute

16170/5005 = Fa/65

Cross multiply

5005Fa = 65 * 16170

5005Fa = 1,051,050

Fa = 1,051,050/5005

Fa = 210N

Hence the required force is 210N

The force which must be exerted on piston A to lift the car is equal to 210 Newton.

Given the following data:

Force on piston B = 16,170 newtonsArea of piston B = 5,005 [tex]cm^2[/tex]Area of piston A = 65 [tex]cm^2[/tex]

To determine the force which must be exerted on piston A to lift the car:

Mathematically, the pressure acting on an area is given by the formula:

[tex]Pressure = \frac{Force}{Area}[/tex]

Since, the weight of the car remains constant, the pressure on piston A must be equal to the pressure on piston B.

Pressure A = Pressure B

Mathematically, this given by:

[tex]\frac{F_A}{A_A} = \frac{F_B}{A_B}[/tex]

Making [tex]F_A[/tex] the subject of formula, we have:

[tex]F_A =\frac{A_AF_B}{A_B}[/tex]

Substituting the given parameters into the formula, we have;

[tex]F_A = \frac{65 \times 16170}{5005} \\\\F_A = \frac{1,051,050}{5005}\\\\F_A = 210\;Newton[/tex]

Therefore, the force which must be exerted on piston A to lift the car is equal to 210 Newton.

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An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is given by v2 = 2gy. In an experiment, v and y are measured and the measured values are used to calculate g. If the percent uncertainty in the measured value of v is 3.69% and the percent uncertainty in the measured value of y is 5.00%, what is the percent uncertainty in the calculated value of g?

Answers

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

On earth a bag of sugar weight 10N. on mars a bag of sugar weight 4N. suggest the weight of the sugar is different on earth and mars

Answers

Answer:

gravity

Explanation:

there eould be a different gravitational strenght on both of the planets causing it to weigh more, or less

In mars, the effect of gravity on the object is weaker compared to while on earth. This weakness is what causes the reduction in the weight of the bag of sugar while on mars.

We must understand that the mass of an object may differ depending on its location on the planet. This is due to the effect of gravity on the object.

Gravity is simply defined as any force which attracts an object towards the centre of the earth.

From the question given, we are told that the weight of sugar is 10N on the earth and 4N on Mars. This difference in their mass is attributed to the effect of gravity on the object.

In mars, the effect of gravity on the object is weaker compared to while on earth. This weakness is what causes the reduction in the weight of the bag of sugar while on mars.

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The diagram below represents the movement of Earth on its axis. Which of 1 point
these four cities will receive sunlight first each.day? *


San Francisco
Phoenix
Chicago
Baltimore

Answers

Answer: Baltimore

Explanation:

There is no diagram included but the answer can be inferred from the rotation of the Earth.

The Earth rotates from east to west which is why places further to the east receive sunlight first each day. It is for this reason that Australia and New Zealand receive sunlight more than 15 hours before the United States does.

In the U.S. the furthest city to the east mentioned in the options is Baltimore. They will therefore receive sunlight before everyone else.

If a car is moving backward and has negative acceleration, what can be said about the speed of the car?

a. its speed is degreasing
b. its speed is increasing
c. its speed can either increase or decrease
d. its speed does not change ​

Answers

I think its A let me know its wrong or not

The first charge is pulling on the second. Is the second pulling on the first? Explain your reasoning

Answers

Answer:

law of action and reaction.

Explanation:

In Newton's three laws it is established that forces act in pairs, if one body interacts with another the second interacts with the first, this is the so-called law of action and reaction.

In this case, when the first load pulls on the second, the second pulls on the first, the two forces are not canceled because each one is applied to a different body.

Therefore the magnitude of the forces is the same, but the direction is opposite and each one is applied in a body

An electric force is an attractive or repulsive interaction. Newton's laws of motion define how it affects things, just like any other force. Yes, the second pulling on the first.

How does Newton's third law explain in terms of electricity?

An electric force is an attractive or repulsive interaction between any two charged things. Newton's laws of motion define how it affects things, just like any other force.

The motion of objects under the influence of such a force or combination of forces is studied using Newton's laws.

The analysis generally starts with the creation of a free-body diagram in which the kind and direction of the various forces are represented by vector arrows.

Newton's laws of motion define how it affects things, just like any other force. Hence Yes, the second pulling on the first.

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ms
2. The half-life of radium-226 is 1600 years. If a sample of radium-226 has an original activity of
200 Bq, what will it's activity be after:
i) 3200 years?
4800 years
iii) 6400 years

Answers

the activity after will be 4800 years
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