Physics College

which of the following examples is not considered a pure substance?
Answer asap

Answers

Answer 1

Answer:

Fruit Punch

Explanation:

A pure substance is whereby there is only one type of element or a compound in the periodic table in the substance.

Carbon: Well its just C in the periodic table, so this is definitely pure.

Water Molecule: What makes water? H2O right? Contains Hydrogen and Oxygen, and as we all know H2O is a compound, therefore this is a pure substance.

Fruit Punch: What makes fruit punch, water and fruits. Fruits may contain citric acid(a compound itself), and is mixed with water with already has a compound, so having 2 different compounds will result in a mixture and therefore it will not be pure.

Related Questions

Question Completion Status: 1 points QUESTION 1 The following excerpt is taken from the Governors Highway Safety Association (GHSA) website (https://www.ghsa.org): Setting speed limits has traditionally been the responsibility of states, except for the period of 1973-1994. During that time, the federal government enacted mandatory speed limit ceilings on interstate highways and similar limited access roads through a National Maximum Speed Limit. ✓ Sav Congress repealed the National Maximum Speed Limit in 1995. Since then, 41 states have raised speed limits to 70 mph or higher on some portion of their roadway systems. In many states, maximum speeds vary depending on vehicle type (car or truck), roadway location (urban or rural), or time of day. GHSA tracks state maximum speed limits for both urban and rural interstates, as well as other limited access roads. In a few states, speed limits are not set by law. Your textbook includes the following physics problem: If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time. To complete the Civic Engagement component of this course, write an essay about the National Speed Limit. In this essay, do the following: 1. Solve the above physics problem. 2. Use your solution to the above physics problem and the information contained in the above excerpt to argue for the reinstatement of a National Maximum Speed Limit. Your orrny should be 3000 to 5000 characters long (not including spaces) You are expected to format all variables/equations using the math editor, which you access

Answers

There should be a reinstatement of the National Maximum Speed Limit for the safety of road users because the minimum braking distance is 2,25m.

Calculations and Parameters

Hence, we can see that the minimum braking distance for the car would be:

d'/d= (v'1/v1)^2 = (1.5/1)^2

= 2.25m

The speed limit ensures that drivers do not exceed a set speed limit in order to reduce road accidents and keep road users safe and also the pedestrians would be able to navigate easier.

Answers

The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

Average power produced by the rocket

The average power produced by the rocket is calculated as follows;

P = FV

where;

P is the average powerF is the force exerted by the rocketV is the velocity of the rocket

Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

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importance of states of matter

Answers

Answer:

STATES OF MATTER The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.

Explanation:

the sound of a lamb is ....... due to its pitch depending upon its ....... frequency?

Answers

The sound of a lamb is grave due to its high pitch depending on its low frequency. Option B is correct.

What is the frequency of the sound?

A sound pressure wave's frequency is the number of times it repeats itself every second.

The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.

Pitch and the frequency of sound are inverse to each other. A lamb's sound is grave because of its high pitch and low frequency.

The sound of a lamb is grave due to its high pitch depending on its low frequency.

Hence, option B is correct.

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Q1) Write True or False for each statement:

Equal volumes of the two different substances have equal masses.

Answers

Answer:

True

Explanation:

it is true because blabla bla

Which expression is equivalent to 6(a - 3)?

Answers

Answer:

it's 6a - 18

Explanation:

because 6 is multiplied witk a and -3

How do the Sun, Earth, and Moon systems interact

Answers

The sun is the mother star of the solar system, which only emits light to half of the planet, while the other part is always dark.

The sun emits light towards the earth, which dominates all life on earth. The movements of the Moon around the Earth and of the Earth around the Sun are complex. Movements of rotation around their own axes are superimposed on movements of orbital translation. The Earth and the Moon rotate around their own axes: This is rotation.

Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 65.3 N, then what is the tension in rope B?
46.17 N is incorrect.

Answers

The tension in the rope B is determined as 10.9 N.

Vertical angle of cable B

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

Angle between B and C

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

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Melanie gets out of her car at the park. She walks 25 m to the trail entrance.
She jogs around the trail until she reaches a pond, where she stops briefly.
She then continues to follow the trail around the pond. Which reference point
should be used to describe her motion?
A. Her car
B. The trail entrance
C. The pond
D. The trail

Answers

The reference point that should be used to describe Melanie motion is the pond (option C).

What is reference point?

Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.

According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.

However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.

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A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.
a) What is the slowest speed that the rock can travel and still maintain a circular path?
b) What is the tension in the string at the bottom of the swing?

Answers

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

[tex]\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg[/tex]

We can simplify and rearrange the equation to solve for 'v'.

[tex]\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}[/tex]

Plugging in values:

[tex]v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}[/tex]

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

[tex]\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg[/tex]

Rearranging for 'T":
[tex]T = \frac{mv^2}{r} + mg\\\\[/tex]

Plugging in the appropriate values:
[tex]T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}[/tex]

Hello!!
The correct answer would be number B)!! I took the test hope it helps

The form of energy that can move from place to place across the universe is

Answers

Answer:kinetic energy

Explanation:

Answer: kinetic energy

A container of PS5s with a mass of 451kg is loaded onto a Walmart truck using a ramp. The ramp is 6.12m long and
the bed of the truck is 1.53m above the ground. A force of 2025N applied parallel to the ramp moves the precious
consoles at a constant speed up the ramp. Find the efficiency of the ramp..
a) 54.6%
b) 32%
c) 46.4%
d) 54.65%

Answers

A. The efficiency of the ramp is 54.6 %.

Velocity ratio of the ramp

V.R = distance moved by effort/distance moved by load = L/h

V.R = (6.12)/(1.53) = 4

Mechanical advantage of the ramp

M.A = Load/Effort

M.A = (451 x 9.8)/(2025)

M.A = 2.183

Efficiency of the ramp

E = (M.A / V.R) x 100%

E = (2.183 / 4) x 100%

E = 54.6 %

Thus, the efficiency of the ramp is 54.6 %.

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●
Dean is skateboarding. He starts at a speed of 0.5 m/s and speeds up to 4.0 m/s
to go off a ramp over a time of 5.0 seconds. What is Dean's acceleration?

Answers

Answer:

[tex]\huge\boxed{\sf a = 0.7 \ m/s^2}[/tex]

Explanation:

Given Data:

Initial velocity = [tex]V_i[/tex] = 0.5 m/s

Final velocity = [tex]V_f[/tex] = 4 m/s

Time = t = 5 s

Required:

Acceleration = a = ?

Formula:

[tex]\displaystyle a =\frac{V_f-V_i}{t}[/tex]

Solution:

Put the givens in the formula.

[tex]\displaystyle a=\frac{4-0.5}{5} \\\\a = \frac{3.5}{5} \\\\a = 0.7 \ m/s^2\\\\\rule[225]{225}{2}[/tex]

calculate the pressure of water in a well if the depth of the water is 10m

Answers

Answer:

P= phg

so:- the height is 10m

density of water 1000 kg/m3

gravity is 9.8m/s2

P=1000 kg/m3*10m*9.8m/s2

=98000Pa

=98KPa

You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?

8427 J is Incorrect.

If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?

80.25 W is Incorrect.

Answers

The total work done is 5980 Joules and the power expended is 57 Watts.

What is work done?

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

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in order for work to be done wht must happen?
Answer asap please due soon

Answers

An object must move in the same direction as the force is applied for work to be accomplished. Option C is correct.

What is work done?

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative. it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.

The work done is found as;

W= F × d

Where.

F is the force

d is the displacement

W is the work done

In order for work to be done, an object must move in the same direction, and the force is applied

Hence option C is correct.

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A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 80.9 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
HINT

Answers

Answer:

Approximately [tex]4.61\times 10^{3}\; {\rm N}[/tex] upwards (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

External forces on this astronaut:

Weight (gravitational attraction) from the earth (downwards,) andNormal force from the floor (upwards.)

Let [tex](\text{normal force})[/tex] denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

[tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex].

Let [tex]m[/tex] denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be [tex](\text{weight}) = m\, g[/tex].

Let [tex]a[/tex] denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be [tex](\text{net force}) = m\, a[/tex].

Rearrange [tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex] to obtain an expression for the magnitude of the normal force on this astronaut:

[tex]\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}[/tex].

Theory of uniform accelerated motion lab reports

Answers

The theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.

What is the theory of uniform accelerated motion?

The theory of uniform accelerated motion is a scientific statement regarding the acceleration of an object in steady conditions.

This theory (uniform accelerated motion) is fundamentally applied in physic and astrophysics.

In conclusion, the theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.

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You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?
If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?

Answers

The work done is 8427 J while the energy expended is 80.25 W.

What is work done?

Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;

W = mgh

Total mass of the water bucket and chain = 13.9 kg + 19.3 kg = 33.2Kg

Distance covered = 25.9 m

W = 33.2Kg * 9.8 m/s^2 * 25.9 m

W = 8427 J

Recall that the work done = Energy expended

Power = Energy expended/ Time

Power = 8427 J/ 1.75 * 60 seconds

Power = 80.25 W

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physical properties of matter help describe a substance which of the following is not a physical property of a soil sample?
A. the container in which the soul is held
B. the color of the soil
C. the volume of the soil
D. the texture of the soil
Answer asap this is an exam il give brainlist

Answers

A the container the soil is held in is not a feature of the soil itself

What states the purpose of an experiment

Answers

Answer:

The purpose of an experiment is to test out your hypothesis. If your hypothesis is correct, then it is a theory that could work every single time the experiment has been performed by scientists.

A person of mass 100kg runs up a staircase with a vertical height of 10m. If the trips takes 5 seconds to complete, calculate the person's power ( acceleration due to gravity =10m/s²)

Answers

Answer:

Power = 2000J/s

Explanation:

Power= Total Work done/Time

Given

mass = 100kg

distance(height) = 10m

time = 5s

acceleration due to gravity = 10m/s²

Work done = force × distance

Finding Force

Force = mass × acceleration due to gravity

Force = 100kg × 10m/s²

Force = 1,000N

Finding work done

Work done = Force × distance

Work done = 1,000N × 10m

Work done = 10,000J

Finding Power

Power = work done/time

Power = 10,000J/5s

Power = 2,000J/s

Therefore Power is 2,000J/s

A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0°. What is the value of work being done on the object?

Answers

Answer:

The value of work being done on the object is 958J.

Explanation:

Work done is equal to force multiply by distance, but when the angle is between the force and distance work done=Force (cos theta) × distance

Work done = Force(cos theta) × distance

Work done = 25N(cos 40.0°) × 50m

Work done = 25N(0.7660) × 50m

Work done = 19.15N × 50m

Work done = 957.5J = 958J

Therefore the value of work being done on the object is 958J.

What is the size of the image on the retina of a size object 1.5 cm, placed at a distance (120 cm) away? Take the .lens-to-retina distance to be 2 cm

Answers

Answer: 0.025 cm

Explanation:

The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm

The magnification of the lens is given by

m = v/u = h'/h

Where h' is the height of the image

v/u = h'/ h

h' = ( v/u ) × h

h' = ( 2cm /-120cm ) × 1.5cm

h' = - 0.025cm

Therefore, the height of the image is 0.025 cm

The minimum takeoff speed for a certain airplane is 50 m/s. What minimum acceleration is required if the plane must leave a runway of length 2000 m? assume the plane starts from rest at one end of the runway.

Answers

Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that

[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]

Solve for [tex]a[/tex] :

[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]

A yoyo with a mass of m = 150 g is released from rest as shown in the figure.

The inner radius of the yoyo is r = 2.14 cm, and the outer radius is R = 4.00 cm, and the moment of inertia about the axis perpendicular to the plane of the yoyo and passing through the center of mass is I[tex]_{cm}[/tex] = 1.01×10-4 kgm2.

1. Determine the linear acceleration of the yoyo.

2. Determine the angular acceleration of the yoyo.

3. What is the weight of the yoyo? (Hint: It's not 150 g)

4. What is the tension in the rope?

5. If a 1.27 m long section of the rope unwinds from the yoyo, then what will be the angular speed of the yoyo?

Answers

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

Linear acceleration of the yoyo

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

I is moment of inertiaα is angular accelerationT is tension in the roper is inner radiusR is outer radiusf is frictional force

rT - Rf = Iα ----- (1)

T - f = Ma -------- (2)

a = Rα

where;

a is the linear acceleration of the yoyo

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

Angular acceleration of the yoyo

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

Weight of the yoyo

W = mg

W = 0.15 x 9.8 = 1.47 N

Tension in the rope

T = mg = 1.47 N

Angular speed of the yoyo

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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An object moves in uniform circular motion at 50 m/s and takes 1.0 second to go a quarter circle. Calculate the
centripetal acceleration.

Answers

Answer:

The centripetal acceleration (ac)=314m/s²

Explanation:

look at the attachment ☝️

(15 points) A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
1. What is the horizontal component of the force exerted by the hi.nge on the beam? (Use the `to the right' as + for the horizontal direction.)
2. What is the magnitude of the force that the beam exerts on the hi.nge?

Answers

The horizontal component of the force is 336 N

The magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons

What is the horizontal component of the force exerted by the hi_nge on the beam?

The system of the beam, hi_nge and cable are in equilibrium.

The horizontal component of the force exerted by the beam is given below as:

Horizontal component of the force = mg × cos 31°

Horizontal component of the force = 40 × 9.8 × cos 31°

Horizontal component of the force = 336 Newtons

The magnitude of the force that the beam exerts on the hi_nge is given as:

F = mg × cos 31° + mg × sin 31°

F = 40 × 9.8 × cos 31° + 40 × 9.8 × sin 31°

F = 537.9 Newtons

Hence, the magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons.

In conclusion, the system of forces acting on the hi_nge and the beam are in equilibrium.

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Kim has her coffee cup on her car's dash when she takes a corner with radius 4 m and 20 km/h.
What is the minimum coefficient of static friction which would allow the coffee cup to stay there
without slipping?

Answers

The minimum coefficient of static friction is 0.807

What is static friction?

Static friction is friction between two or more solid objects that are not moving relative to each other.

According to the question,

When an object turns on a curve path at a specific speed, two forces will be acting on the object which balance each other:

Centripetal force and friction force.

The friction coefficient between the object and the path is independent of the object's mass.

Given,

The radius of a corner is, r=4.16m

The turning speed of the car is, v

[tex]20km/h * (\frac{\frac{5}{18}m/s }{1km/h} )[/tex]

= 5.72 m/s

To avoid slipping a coffee cup on the car's dashboard, the centripetal force would balance the friction force.

So, Fc=Ff

[tex]\frac{mv^2}{r}[/tex]=μmg

μ=[tex]\frac{v^2}{rg}[/tex]

Here,

Fc represents the centripetal force.

Ff represents the friction force.

μ represents the coefficient of static friction.

g represents the gravitational acceleration whose value is 9.8m/s²

By Substituting the values in the above formula, we get:

μ =[tex]\frac{(5.72m/s)^2}{(4m)(9.8m/s^2)} \\[/tex] ≈ 0.807

Therefore,

The minimum coefficient of static friction is 0.807

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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.

Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)

331.35 N is incorrect.

Answers

The magnitude of the tension in the string marked A and B is mathematically given as

A = 52.5 N

What is the magnitude of the tension in the string marked A?

Generally, the equation for is mathematically given as

The angle at A is...

[tex]tan\theta = 3/8[/tex]

When below the negative x

B= tan\phi

[tex]tan\phi = 5/4[/tex]

When below the negative x

[tex]tan \rho = 1/6[/tex]

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Hence, considering the Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

which of the following examples is not considered a pure substance? Answer asap

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which of the following examples is not considered a pure substance?
Answer asap