Answer:
2
Explanation:
please mark me as brainlist i need it so sorry
A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N. (a.) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0° above the horizontal? (b.) How much work does the cable do on the tow truck in both cases of part (a)? (c.) How much work does gravity do on the car in part (a)?
I think 1980is the answer because you add???
Identify:
In each case the forces are constant and the displacement is along a straight line, so
[tex]$$W=F s \cos \phi \text {. }$$[/tex]
Set-Up:
In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.
Execute:
(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].
When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].
(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],
So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].
(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.
Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork
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Example 4.16
An object of mass 3 kg rests on a plane. The coefficient of static friction and that of kinein
friction are given by Hs = 0.3 and pk = 0.2.
The plane is inclined at angle o to the horizontal.
(i) Find the maximum value of 0 for which the object remains at rest on the plane.
(ii) Find the acceleration of the object if it started sliding from rest down the plane at
angle Omax to the horizontal.
(ii) How long does it take the object to move, from rest, a distance of Imetre under the
conditions of (ii).
Answer:
Explanation:
(i) μs = F/N = mgsinθ/mgcosθ = tanθ
tanθ = 0.3
θ = 16.7°
(ii) a = F/m
a = (mgsinθ - (μk)mgcosθ) / m
a = g(sinθ - (μk)cosθ)
a = 9.8(sin16.7 - (0.2)cos16.7)
a = 0.94 m/s²
(iii) s = ½at²
t = √(2s/a)
t = √(2(1)/0.94)
t = 1.5 s
I tossed a ball straight up into the air and timed how long it took to return to the height I tossed it from. It took 4.2s. How fast did I throw the ball into the air?
Explanation:
[tex]v = u + at \\ 0 = u + ( - 10 { ms}^{ - 2}) \times 4.2s \\ u = 42 {ms}^{ - 1} [/tex]
Light from the Sun takes about 8.0 min to reach Earth. How far away is the Sun? Answer in scientific notation
a wave travels at a constant speed. how does the wavelength change if the frequency is reduced by a factor of 3? assume the speed of the wave remains unchanged.
A. the wavelength decreases by a factor of 3
B. the wavelength does not change
C. the wavelength increases by a factor of 3
D. the wavelength increases by a factor of 9
Answer:
b
Explanation:
when the wavelength increase it doesnt affect the frequency of a wave.
Answer: The wavelength increases by a factor of 3
Explanation:
A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipes,
it is observed that the number of beats per second first diminishes to zero and then increases again to 4.
By how much has the temperature of the air in the pipe been altered?
The temperature of the air in the open orang pipe has been altered by 18.73° C
The frequency of an open orang pipe is estimated by using the formula:
[tex]\mathbf{f = \dfrac{v}{2L}}[/tex]
Then, the combination of the frequency of the tuning fork and the open orang pipe is:
[tex]\mathbf{254 - \dfrac{v}{2L} }[/tex]
These combinations of frequency produce 4 beats per sound.
i.e.
[tex]\mathbf{254 - \dfrac{v}{2L} =4}[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 254-4 }[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 250 ----(1)}[/tex]
When it is altered, the beats first diminish and increase again by 4.
i.e.
[tex]\mathbf{ \dfrac{v'}{2L} = 254+4 }[/tex]
[tex]\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }[/tex]
If we equate both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}[/tex]
However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.
Hence;
when the temperature of the pipe = unknown ???the temperature of the open orang pipe = 15∴
[tex]\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}[/tex]
By squaring both sides, we have:
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}[/tex]
[tex]\implies \mathbf{273 +T =306.726912 }[/tex]
T = 306.726912 - 273
T ≅ 33.73 ° C
∴
The change in temperature ΔT = 33.73° C - 15° C
The change in temperature ΔT = 18.73° C
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Three point charges are placed on the y-axis: a chargeqaty=a,a charge-2qat the origin, and a chargeqaty= -a.Such an arrangement is called an electricquadrupole.(a) Find the magnitude and direction of the electric field at points on thepositive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x >> a.Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.
When an alpha particle is emitted from an unstable nucleus, what happens to the atomic number of the nucleus
Answer:
The answer to your question is the number decreases by 4.
Explanation:
I hope this helps and have a great day!
ПОМОГИТЕ ПЖЖППЖЖППЖЖПЖЖПЖПЖПЖ ООООООЧЕНЬ СРОЧНО!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! С ДАНО И РЕШЕНИЕМ!!!!!!
4. У сталевій коробці
масою 250 г
розплавляють 100 г
свинцю. Яка кількість
теплоти витратилася
на теплові процеси,
якщо початкова
температура тіл
становила 27 °С?
Answer:
thanks for the points dia
What is the possible resultant of a 2-unit vector and 4-unit vector?
Answer:
Explanation:
resultant could have magnitude 4 - 2 ≤ V ≤ 4 + 2 or 2 ≤ V ≤ 6
If θ is the angle of the 4 unit vector, the resultant could have angle φ in the range of
φ = θ ± arcsin (2/4)
θ - 30° ≤ φ ≤ θ + 30°
The possible resultant of a 2-unit vector and 4-unit vector is 6 and 2.
What is a Unit vector?The quantity which has both the magnitude and direction is known as vector.The Examples for Vector quantity is Acceleration, velocity, force and displacement, etc.,.The quantity which has only the magnitude but has no direction, (examples include speed, time and distance) can be known as scalar.If the magnitude of vector is 1, then it can be the Unit vector.
In the Unit vector, x axis will have the vector of i, y axis will have the vector j, z axis will have the vector k.
When two vectors are in same direction then we add up to get their resultant and in this case resultant would be maximum
R = A + B
R = 4 + 2
R = 6
If vectors are in opposite direction then they provide minimum resultant and in this case we subtract them
R = A - B
R = 4 - 2
R = 2
Thus, the possible resultant are 6 and 2.
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PLEASE HELP BRAINLIEST TO THE RIGHT ANSWER!!!!!
A 7.8 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s^2.
A. Determine the tension in the string if it is accelerating upward at a rate of 1.5 m/s^2. Answer in units of N.
B. Determine the tension in the string if it is accelerating downward at a rate of
1.5 m/s^2
Gravitational pull downward on the object: 7.8 x 9.8 = 76.44N
A. 7.8 x 1.5 = 11.7N upward force
Tension = 76.44 + 11.7 = 88.14 N
Answer: 88.14 N
B. 76.44 - 11.7 = 64.74N
Answer: 64.74 N
Answer:
88.14 N
64.74 N
Explanation:
Question below...........................
[tex]\boxed{\sf PE=mgh}[/tex]
[tex]\boxed{\sf KE=\dfrac{1}{2}mv^2}[/tex]
[tex]\boxed{\sf ME=KE+PE}[/tex]
#1
[tex]\\ \sf\longmapsto PE=60(0)(10)=0J[/tex]
[tex]\\ \sf\longmapsto KE=\dfrac{1}{2}(60)(8)^2=30(64)=1920J[/tex]
[tex]\\ \sf\longmapsto ME=1920+0=1920J[/tex]
#2
[tex]\\ \sf\longmapsto PE=60(10)(1)=600J[/tex]
[tex]\\ \sf\longmapsto KE=600J[/tex]
[tex]\\ \sf\longmapsto ME=1200J[/tex]
Now
[tex]\\ \sf\longmapsto \dfrac{1}{2}mv^2=600\implies 30v^2=600\implies v^2=20\implies v=4.2m/s[/tex]
An object weighs 573.0 N on planet Xyleneer. If the object's mass is 92.1 kg, what is the acceleration due to gravity on planet Xyleneer?
Answer:
a = 6.22 m//s²
Explanation:
F = ma
a = F/m
a = 573.0 / 92.1
a = 6.221498...
Answer:
[tex]\boxed {\boxed {\sf 6.22 \ m/s^2}}[/tex]
Explanation:
We are asked to find the acceleration due to gravity on another planet.
Weight is the measure of the force of gravity. Therefore, we can use the following version of the force formula:
[tex]F_g=mg[/tex]
In this formula, [tex]F_g[/tex] is the weight, m is the mass, and g is the acceleration due to gravity.
The object weights 573.0 Newtons (or 573.0 kg*m/s²) on the planet. The object has a mass of 92.1 kilograms.
[tex]F_g[/tex]= 573.0 kg* m/s²m= 92.1 kgSubstitute these values into the formula.
[tex]573.0 \ kg*m/s^2 = 92.1 \ kg * g[/tex]
We are solving for g, so we must isolate the variable. It is being multiplied by 92.1 kilograms. The inverse of multiplication is division, so divide both sides of the equation by 92.1 kg.
[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}= \frac{92.1 \ kg*a}{92.1 \ kg}[/tex]
[tex]\frac {573.0 \ kg*m/s^2}{92.1 \ kg}=a[/tex]
The units of kilograms cancel.
[tex]6.22149837 \ m/s^2=a[/tex]
The original measurements of weight and mass have 4 and 3 significant figures. Our answer must have the least number of sig figs, or 3. For the number we found, that is the hundredth place. The 1 in the thousandths place tells us to leave the 2 in the hundredth place.
[tex]6.22 \ m/s^2=a[/tex]
The acceleration due ot gravity on planet Xyleneer is approximately 6.22 meters per second squared.
A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is attached to the side of the tank, 0.580 m above the ground. If the valve is opened, at what speed (in m/s) will water come out of the pipe
h =(3.7 - .58)m = 3.12m
Now put PE into KE and we have to use the formula:
√2gh (g = gravity and h = height) therefor:
√2 x 9.8 x 3.12
= 7.82m/s
I hope this helps!
2.(01.01 LC)
Which of the following is true for gravitational force? (3 points)
Decreases with increase in mass
Increases with increase in mass
Increases with increase in distance
Decreases with decrease in distance
Answer:
Increases with increase in mass
Explanation:
gravity is proportional to mass and inversely proportional to the square of the distance between them
F = GMm/d²
Which of the following statements are true?
(a) An object can move even when no force acts on it.
(b) If an object isn't moving, no external forces act on it.
(c) If a single force acts on an object, the object accelerates.
(d) If an object accelerates, a force is acting on it.
(e) If an object isn't accelerating, no external force is acting on it.
(f) If the net force acting on an object is in the positive x-direction, the object moves only in the positive x-direction.
I can't understand how A is true.
Answer: a, c, d
Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it
Would appreciate brainly <3
Explanation:
inertia ...
you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.
please consider : we are only talking about moving. not about acceleration.
so, yes, (a) is true.
(c) is true.
(d) is true.
I will mark brainlist
A wave is disturbance that transfers energy and matter.
true
false
Answer:
False
Explanation:
A wave is a disturbance that transfers energy from one place to another without transferring matter.
Answer:
I'm pretty sure it true sorry if I'm wrong
What is the momentum of a 3 kg bowling ball moving at 3 m/s?
.
O 1 kg. m/s
O 3 kg. m/s
O 6 kg. m/s
O 9 kg • m/s
Explanation:
p = mvp denotes momentumm denotes massv denotes velocity→ p = 3 kg × 3 m/s
→ p = 9 kg.m/s
Option D is correct.
9. The acceleration (a)-time (t) graph of a particle moving in a straight line is as shown in figure. At time t = 0, the velocity of particle is 10 m/s. What is the velocity at t = 8 s?
(1) 2 m/s
(2) 4 m/s
(3) 10 m/s
(4) 12 m/s
Answer:Acceleration - time graph for a particle moving in a straight line is as shown in figure. Change in velocity of the particle from t = 0 to t = 6s is:-.
1 answer
·
Top answer:
Change in velocity = (sum of area of graph) = ( 12 × 4 × 4 ) + ( 12 × ( + 2) ( - 1) ) - 4 = 8 - 4 = 4 x
Explanation:
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained by the hammer head, how much will its temperature increase
The increase in temperature of the metal hammer is 0.028 ⁰C.
The given parameters:
mass of the metal hammer, m = 1.0 kgspeed of the hammer, v = 5.0 m/sspecific heat capacity of iron, 450 J/kg⁰CThe increase in temperature of the metal hammer is calculated as follows;
[tex]Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}[/tex]
where;
c is the specific heat capacity of the metal hammer
Assuming the metal hammer is iron, c = 450 J/kg⁰C
[tex]\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C[/tex]
Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.
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Calculate the amount of work done (in joules) to raise the 0.2kg mass 0.5 m
Answer:
1
Explanation:
A=mgh=0.2*10*0.5=1 J
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Nm-1
. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple
harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.
The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;
x = 0.026 m
Given parameters
The mass of the body m = 220 g = 0.220 kg The force constant k = 7.0 N / m The initial displacement or amplitude xo = 5.2 cm = 0.052 mTo find
The point where scientific and potential energy are equal.
The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. With maximum compression.
Em₀ = U = ½ k x²
Final point. Where the kinetic and potential energy are equal.
[tex]Em_f = K +U[/tex]
Since the mechanical energy is constant at this point K = U, therefore we can write the energy.
[tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]
Energy is conserved.
[tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf
½ k x² = 2 (½ k xf²)
[tex]x_f = \frac{x_o}{2}[/tex]
let's calculate.
[tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]
In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;
x = 0.026 m
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HELP ME PLEASE I DONT GET THIS
How did the study of the atom contribute to our understanding of the periodic table of the elements? (1 point) Atoms are representative of elements, so scientists scaled up O atomic char- derstand elemental characteristics, allowing sc Highlight e elements in a periodic table. The determination of electron charge led to an understanding of O how atoms interact with one another, which facilitated the organization of the periodic table. Elements are made of atoms, so understanding atoms provided O information about elements, which led to the organization of the periodic table. Experiments that identified characteristics of atoms provided O scientists with atomic weights and atomic numbers, which were used to organize the periodic table.
The study of the atom contribute d to our understanding of the periodic table of the elements by virtue of the following;
Elements are made of atoms, so understanding atoms provided information about elements, which led to the organization of the periodic tableExperiments that identified characteristics of atoms provided scientists with atomic weights and atomic numbers, which were used to organize the periodic table.As we know; the periodic table is an array of elements in order of their atomic number.
In which case, the periodic table is made up of 7 rows otherwise called Periods and 8 columns otherwise called Groups.
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70 POINT!!!
IF YOU DON'T KNOW THE ANSWER DO NOT PUT A COMMENT BELOW
THERE ARE TWO QUESTIONS YOU HAVE TO ANSWER
QUESTION 1: Identify any variables that are present as dependent variables, independent variables, and constants in your experimental group and your control group.
QUESTION 2: How does knowing the properties of matter help you separate the substances in mixtures?
Answer:
1) Dependent variable: Type of separation method used
Independent variables: Substance that separated out
Constants: Sand, pepper, and salt
2) you can use the properties of matter to decide on a method to separate out a particular substance based on its unique properties. For example, knowing pepper has a very small mass allows you to use the charge of the static electricity to separate out those particles. Additionally, knowing the solubility of salt allows you to add water to the mixture to be able to remove it from the sand, since the salt dissolves in the water which is poured away.
A car of mass 200 kg, moving with a forward acceleration of 3 m/s-ıs acted upon by
constant resistive force of 500 N. Calculate the force exerted from the engine to
maintain this forward acceleration.
Answer:
600 N
Explanation:
The force exerted from the engine to maintain this forward acceleration is 600 N.
What is force?Force is defined in physics as: the push or pull on an object with mass that causes it to change velocity. Force is an external agent that can change the state of rest or motion of a body.
The term "force" has a specific meaning in science. At this level, it is perfectly acceptable to refer to a force as a push or a pull.
A force is not something that an object possesses or possesses. Another object applies a force to another.
We know that,
F = ma
Here, it is given that
Acceleration = 3 m/s
Mass = 200Kg.
So, by keeping the value
F = 200 x 3
F = 600 N
Thus, the force is 600 N.
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Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of research question o add another research question o use the very first source you find for your project O change the scope of your project
Answer:
"Narrow the focus of research question"
Explanation:
O Narrow the focus of research question
This is good! You can still use your question, but focus in on something so you have a proper research project.
O Add another research question
Would adding another question to an already broad question help? No.
O Use the very first source you find for your project
If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question
O Change the scope of your project
You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
PLEASE HELP ME GET THIS RIGHT
Explanation:
I'm not sure, but I would go for the more than A since its orbital speed is at its fastest and the sweep occurs in about the same period of days.
An electromagnet does not attract a piece of iron.Is it true ? Give reason
Answer:
False..
Explanation:
An electoMagnets attract iron due to the influence of their magnetic field upon the iron. ...
Is the acceleration change or constnt?