a) If the mass of the car is doubled (by adding passengers and cargo) by what factor must the
force change so the car can make the turn? (2 pts)

How do you find the uncertainty of a meter stick?

Thus, L =5 . 7 cm measured with a meter stick implies an uncertainty of 0.05 cm. A common rule of thumb is to take one-half the unit of the last decimal place in a measurement to obtain the uncertainty

Answer:

Ans: a=0.056ms^-2  S=2500m

Explanation:

vi=0ms^-1

vf=60km/hr= [tex]\frac{60*1000}{3600}[/tex]=16.6ms-1

a=[tex]\frac{vf-vi}{t}[/tex]

 =[tex]\frac{16.6-o}{5*60}[/tex]

S=vit+[tex]\frac{1}{2}[/tex][tex]at^{2}[/tex]

S=0+[tex]\frac{1}{2}[/tex]*0.056*[tex](300)^{2}[/tex]

Answer:

Explanation:

B

A primary source is one that is written by the person who lived through an advenure or is one doing the action and recording it.

In this case, the answer is B.

Most frequently, for a historian, a primary source is something like a diary or a bunch of letter written by someone experiencing something.

It cannot be about someone else's work. Not C or D

A textbook is not the answer either. It is a bulky summary of scientific, or other kind of thought.

Answer:

It take him 1 hour to travel 90 meters

Explanation:

Answer:

90 m

Explanation:

30 ÷ 20 = 1.5 m/min x 60 min = 90 m

Answer:

[tex] \boxed{d = 120\: meters} [/tex]

Explanation:

Given:

We can use this equation with the variables given to solve:

[tex] d \: = \frac{t \: (v + v_{0})}{2}[/tex] →

[tex] d \: = \frac{(10s) \: ((16.0 \: m/s) + (8.0 \: m/s))}{2} [/tex] →

[tex] d \: = \frac{(10s) \: (24 \: m/s)}{2} [/tex] →

[tex] d \: = \frac{(10 \: s \: • \: 24 \: m/s)}{2} [/tex] →

[tex] d \: = \frac{240 \: m}{2} [/tex] →

[tex] d \: = 120 \: m [/tex]

The distance traveled by the object in 10 seconds is 120 meters.

Given to us,

Final velocity of the object, [tex]v= 16\ meter/sec[/tex]

Initial velocity of the object, [tex]u= 8\ meter/sec[/tex]

Time for traveling, [tex]t= 10\ sec[/tex]

Using the first equation of motion,  we can find out acceleration of the object [tex]a[/tex],

[tex]v=u+at\\16=8+a\times 10\\a= 0.8\ meter/sec^2[/tex]

Now using the second equation of motion and putting the value of [tex]a[/tex],

The distance[tex](S)[/tex] traveled by the object ,

[tex]\begin{aligned}S&= ut+\frac{1}{2}at^2\\&= 8\times 10+ \frac{1}{2}\times 0.8\times 10^2\\&= 120\ meters \\\end{aligned}[/tex]

Hence,The distance traveled by the object in 10 seconds is 120 meters.

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Explanation:

Answer:

True

Explanation:

It is true because scientists must use this method to get a correct and accurate solution.

Answer:

0

Explanation:

acceleration = velocity÷time

20/5=4m/s square

substitute it v= u+at

u will get final answer as 0m/s

distance- 14 meters

displacement- 125 kilometers South

speed- 45 mph

velocity- 75 m/s upwards

acceleration due to gravity on the earth- 9.8 m/s^2

net force- 22 Newtons to the left

weight- 14 Newtons down

acceleration- 128 m/s^2 West

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Velocity  →  75 m/s upwards

Velocity is the ratio of displacement and time, therefore, it has magnitude as well as direction.

Net force  →  22 Newtons to the left

Force is measured in Newton and can be in any direction

Acceleration due to gravity on Earth  →  9.8 m/s^2 or 9.81 m/s^2

It is the general value of Acceleration due to gravity on Earth.

Acceleration  →  128 m/s^2 West

Acceleration is measured in meters per second square.

Distance  →  14 meters

Distance is measured in meters, feet, cm, km, etc.

Speed  →  45 mph

Speed is the ratio of distance and time.

Displacement  →  125 kilometers South

Displacement has magnitude and direction both.

Weight  →  14 newtons down

Weight is also a force but it is in the downward direction.

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Answer:

1.Energy is a quantity that is transferred from system to system. Energy is the ability of a system to do work. A system has done work if it has exerted a force on another system over some distance. When this happens, energy is transferred from one system to another.

2.physical quantity is a property of a material or system that can be quantified by measurement. A physical quantity can be expressed as a value, which is the algebraic multiplication of a numerical value and a unit. ... One is numerical magnitude and the other is the unit in which it is measured.

Answer:

Remember 1 litre is equivalent to 1 cm³

[tex]density = \frac{mass}{volume} \\ \\ density = \frac{(2.08 \times 1000) \: g}{2 \: cm {}^{3} } \\ \\ { \boxed{ \boxed{density = 1040 \: g {cm}^{ - 3} }}}[/tex]

or: density is 1.04 kg/l

Answer:

You have no graph on the post.

Las dos líneas más largas del terreno de juego se denominarán líneas de banda. Las dos más cortas se llamarán líneas de meta. El terreno de juego será rectangular y la longitud de las líneas de banda deberá ser mayor que las de las líneas de meta. La longitud de la línea de banda debe tener como mínimo 90 metros y como máximo 120 metros. La anchura de la línea de meta debe de tener como mínimo 45 metros y como máximo de 90 metros. Todas las líneas del terreno de juego deberán se líneas continuas y tener la misma anchura, como máximo 12 cm., menos las líneas de meta que deben tener la misma anchura que los postes y larguero de la portería. Todas las líneas que delimitan al campo forma parte del terreno de juego. El terreno de juego puede ser de tierra, césped natural ó césped artificial de color verde según las normas del reglamento de la competición. El terreno de juego está dividido en dos partes iguales por una línea que unirá los puntos medios de las dos líneas de banda. El centro del campo estará marcado con un punto situado en la mitad de la línea media que divide las dos partes iguales del campo, tomando este punto como centro, haremos circulo de un radio de 9,15 metros. También se puede poner un banderín en cada extremo de la línea de media, a una distancia mínima de 1 m al exterior de la línea de banda.

Answer:

The temperature is [tex]$80.6$[/tex] degrees Fahrenheit.

Explanation:

The weather is predicted to be 27 °C.

[tex]c=\frac{5}{9}(F-32)[/tex]

[tex]27=\frac{5}{9}(F-32)[/tex]

[tex]$5(F-32)=243$[/tex]

[tex]$F-32=\frac{243}{5}$[/tex]

[tex]$F-32+32=\frac{243}{5}+32$[/tex]

[tex]$F=\frac{403}{5}$[/tex]

[tex]F=80.6[/tex]

The temperature is [tex]$80.6$[/tex] degrees Fahrenheit.

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She travelled in one third of time=160km

Total distance=160×3=480km

Time=6h

[tex]\\ \bull\tt\dashrightarrow Velocity=\dfrac{Displacement}{Time}[/tex]

[tex]\\ \bull\tt\dashrightarrow Velocity=\dfrac{480}{6}[/tex]

[tex]\\ \bull\tt\dashrightarrow Velocity=80km/h[/tex]

Answer: 160km

Explanation:

The absolute zero temperature is the smallest thermodynamic thermal limit, which is considered as a zero kelvin when the enthalpy, as well as the entropy of a cooled ideal gas, are of minimal value.

It is expressed by using the formula:

[tex]\mathbf{T_{abs} = T_N + k}[/tex]

where;

[tex]\mathbf{T_{abs } = absoulte \ temperature}[/tex]

[tex]\mathbf{T_N = temperature \ at \Neptune \ Scale}[/tex]

k = factor to convert temperature from absolute scale to Neptunian scale

N:B: An absolute temperature = zero (0) kelvin

From the information given, Using Ideal gas since it is used in Neptune:

PV = nRT

where;

From the question, provided that the gas is the same and no new mass is added;

Then, we can equate the moles of gas in the given conditions as;

[tex]\mathbf{\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}....\dfrac{P_nV_n}{T_n}}[/tex]

Using:

[tex]\mathbf{\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}}[/tex]

where the value of;

P₁V₁ = 28 dm³ atm

T₁ = 0⁰ N

P₂V₂ = 40dm³ atm

T₂ =  100⁰ N

∴

[tex]\mathbf{\dfrac{28}{(0+K)}=\dfrac{40}{(K+100)}}[/tex]

By cross multiply;

28(K+100) = 40(0+K)

28K + 2800  = 0 + 40K

40K - 28K = 2800

12 K = 2800

K = 2800/12

K = 233.33° N

Recall that:

[tex]\mathbf{T_{abs} = T_N + k}[/tex]

[tex]\mathbf{0= T_N+ 233.33 ^0 N}[/tex]

[tex]\mathbf{ T_N= - 233.33 ^0 N}[/tex]

Therefore, we can conclude that the value of the absolute zero of temperature on their temperature scale is -233.33° N

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Answer:

The object will float better.

Explanation:

If you have basic knowledge on how it works, this question is relatively easy. Let's start by reviewing something. We already know that this object (let's assume it's a spaceship of sorts) is in orbit, meaning there is already a certain amount of gravity from Earth tugging on it. We also know that space has an effect on how this moves, so the two forces "balance" each other, allowing the object to stay at orbit rather than just drifting off. Now let's see the answers to see which makes more sense.

A. The object will stop moving.

The object can't stop moving. With gravity increasing, it would be impossible for that to happen.

B. The object will fly off into space.

This could only happen if the gravity were to decrease by a decent amount.

C. The object will smash into another object.

There is no other object mentioned in the question.

D. The object will float better.

This seems most likely as gravity would be pulling on the object a bit more, but probably not enough to bring it back to the ground. It might drop a bit lower into the exosphere.

I hope I was right and I hope this helps!

Answer:

[tex]x=\frac{-y+\sqrt{y^2+4rt} }{2r}[/tex]

[tex]x=\frac{-y-\sqrt{y^2+4rt} }{2r}[/tex]

Explanation:

[tex]rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0[/tex]

Solve using the quadratic formula.

[tex]x=\frac{-y+\sqrt{y^2+4rt} }{2r}[/tex]

[tex]x=\frac{-y-\sqrt{y^2+4rt} }{2r}[/tex]

Answer:

x=-y+[tex]\sqrt{[/tex]y^2+4rt/2r

Explanation:

solve using the quadratic equation

(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,

Ce = 0.093 cal/g. °c

Q = m C ΔT

Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )

Q= 4×0.093 × 160

Q = 59.52 cal

I hope I helped you^_^

Answer:

14 m/s

Explanation:

a = v/t

subbing in:

2 = v/7

v = 14 m/s

No. A dog chasing its tail is just an animal running around in circles.

The dog's MOTION, however, IS accelerated motion, since the DIRECTION of its motion is constantly changing.

A a dog chasing its tail cannot be considered as acceleration.

Acceleration is the rate at which velocity changes over time, both in terms of speed and direction.

A point or object moving in a straight line is accelerated if it accelerates or decelerates. Even if the speed is constant, motion on a circle is accelerated because the direction is constantly changing.

For example, if a car accelerates as it turns a corner at constant speed, it is doing so because its direction is changing.

The faster you turn, the faster you accelerate. When velocity changes in magnitude (increase or decrease in speed), direction, or both, there is an acceleration.

Thus, as there is no change in either speed or direction in scenario in which a dog chases its tail, it is not considered as acceleration.

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Answer:

Explanation:

Well mathematically it would look like this

E = k * I

k is the constant of proportionality.

As the voltage goes up, so does the current.

As the voltage goes down, so does the current.

In physics and engineering,, the formula is written somewhat differently.

E = I * R

The R is a device which limits the current. If R = 0 (like it is just a switch closing) then the first thing you will notice is that the lights go out and you better know where the circuit breakers are and have a flashlight to see which one you shorted out.

So R has a value > 0 in practice.

Answer:

Electric is produce by water it is very helpful in our life

Answer:Sales tax, value-added tax, excise tax, and customs duties are examples of indirect taxes.

Explanation:

Answer:

the shape of the glass change and the size of the glass also change

Answer:

490.5

Explanation:

Formula

U = mgh

= 1000 x 9.81 x 0.05m

= 490.5

The velocity of the object in free fall is 39.2 m/s.

Given the following data:

We know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.

To find the final velocity (V) of the object, we would use the first equation of motion;

[tex]V = U + at\\\\V = 0 + 9.8(4)[/tex]

Final velocity, V = 39.2 m/s

Therefore, the velocity of the object in free fall is 39.2 m/s.

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