2.72 sec
Explanation
Step 1
to solve this we need to use the formula
[tex]h=v_0t+\frac{1}{2}gt^2[/tex]let
[tex]\begin{gathered} v_0=0 \\ h=36.4\text{ m} \\ g=9.8\text{ }\frac{\text{m}}{s^2} \\ t=\text{ unknown= t} \end{gathered}[/tex]replace
[tex]\begin{gathered} h=v_0t+\frac{1}{2}gt^2 \\ 36.4=0\cdot t+\frac{1}{2}(9.8\text{ }\frac{m}{s^2})t^2 \\ 36.4=+(4.9\frac{m}{s^2})t^2 \\ \text{divide both sides by 4.9} \\ \frac{36.4}{4.9}=t^2 \\ t=\sqrt[]{7.42} \\ t=2.72 \end{gathered}[/tex]therefore, the answer is
2.72 sec
I hope this helps you
hello can you help me with my AP physics assignment
1)
R is the required distance from the starting point. The right triangle representing this scenario is shown below
We would apply pythagorean theorem which is expressed as
hypotenuse^2 = one leg^2 + other leg^2
hypotenuse = R
one leg = 25
other leg = 18
Thus,
R^2 = 25^2 + 18^2 = 949
R = √949
R = 30.81
You're 30.81m from your starting point
2) We would find θ by applying the tangent trigonometric ratio which is expressed as
tanθ = opposite side/adjacent side
opposite side = 25
adjacent side = 18
tanθ = 25/18
θ = tan^-1(25/18)
θ = 54.25 degrees
Position = 54.25 degrees west of north
Why doesn't an object falling from an airplane continue to accelerate? (1 point)
O Gravity's force diminishes as the object nears the ground.
O Air resistance on the object will eventually equal the force of force of gravity.
O The object's weight varies as it nears the ground.
O Hitting the ground stops the object's acceleration.
A falling object accelerates as it descends. The quantity of air resistance rises in proportion to the speed. The pull of gravity eventually is balanced by the force of air resistance as it grows. The item will cease accelerating since there is no net force at this point in time (0 Newton).
Since the upward force of air resistance eventually equals the downward force of gravity, a falling item cannot continue to accelerate indefinitely before reaching its terminal velocity.In contrast to air resistance, which operates in the opposite direction and slows acceleration, gravity causes objects to accelerate downhill. Greater surface area falling objects encounter more air resistance. In the absence of air, or in a vacuum, all objects fall with the same precise rate of acceleration.To know more about gravity
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Answer: (B). Air resistance on the object will eventually equal the force of force of gravity.
Explanation:
A block has a velocity of 6 m/s to the East and 360 J of kinetic energy. The block is pushed West with a 30 N external force, while the block moves 3 m East. How much work is done by the force?
Work done will be 270 J
What is work energy theorem?
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
according to work force theorem
Work done = Force x direction = FD Cosθ
Even if the force is applied to the opposite direction, the box will move in the direction of East. Firstly it was already 6m towards East and after applying force, the box moves further 3m towards same direction i.e East.
= 30 N x (9) cos 0⁰
= cos 0⁰ is 1
= 270 J
270 J work is done by the force.
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What 2 factors affect the Kinetic Energy of an object? Which of the 2 factors has more of an influence? Explain. What are the 2 units?
Kinetic energy is a form of energy that an object or a particle has by reason of its motion.
[tex]K=\frac{1}{2}mv^2[/tex]m: mass
v: velocity
velocity has more influence
m: kg
v: m/s
Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 33.55 N when they are separated by 55.75 cm. What is the magnitude of the charges in microCoulombs ?
Given
Two equal, but oppositely charged particles are attracted to each other electrically.
Force of attraction is F=33.55 N
Distance between them, d=55.75 cm=0.5575 m
To find
What is the magnitude of the charges in microCoulombs ?
Explanation
Let the charge be q.
We know the force of attraction is given by
[tex]\begin{gathered} F=k\frac{q^2}{d^2} \\ \Rightarrow33.55=9\times10^9\times\frac{q^2}{(0.5575)^2} \\ \Rightarrow q=3.39\times(10)^{-5}C \\ \Rightarrow q=\pm33.9\mu C \end{gathered}[/tex]Conclusion
The charges are:
[tex]+33.9\mu C,-33.9\mu C[/tex]The magnitude of equal charges are:
[tex]\lvert{q}\rvert=33.9\mu C[/tex]Jontell finds a giant spring that has a 350N/m spring constant. Jontell figures out how to compress is 1.5m. If Jontell and his spring propelled cart's mass is 85kg, how fast is he going after the push fromthe spring? PEspring1/2*k*x?^2
Given data
*The given mass of the propelled cart's is m = 85 kg
*The given spring constant is k = 350 N/m
*The spring compresses at a distance is x = 1.5 m
The formula for the speed is given by the conservation of energy as
[tex]\begin{gathered} U_{p.e}=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=\sqrt[]{\frac{kx^2}{m}} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{\frac{350\times(1.5)^2}{85}} \\ =3.04\text{ m/s} \end{gathered}[/tex]Point p, to me, doesn't look like a conjunction point where I can apply Kirchhoff's first law, which is the sum of the current in is equal to the sum of the current out. And how do I explain using charges that flow through?
ANSWER
a) The current at P is 1.0A and it is moving towards the node.
b) A total of 1C charges flow through P in 1.0 seconds
EXPLANATION
a) To find this, we have to apply Kirchoff's node law which states that:
Therefore, we have that:
[tex]4.0+3.0-8.0+P=0[/tex]Solve for P:
[tex]\begin{gathered} 4.0+3.0-8.0+P=0 \\ -1.0+P=0 \\ \Rightarrow P=1.0A \end{gathered}[/tex]This implies that the current at point P is 1A and it is moving towards the node (in the left direction).
b) To find the electric charge that flows through P in 1.0s, we have to use the formula for current:
[tex]I=\frac{q}{t}[/tex]where I = current, q = electric charge and t = time
Therefore, we have to find q:
[tex]\begin{gathered} q=I\cdot t \\ \Rightarrow q=1.0\cdot1.0 \\ q=1.0C \end{gathered}[/tex]Therefore, a total of 1C charges flow through P in 1.0 seconds.
7. A 50 kg woman standing on a polished floor (with no friction) has a cat jump into her armsand she catches the cat. If the cat has a mass of 5 kg and jumps with a speed of 12 m/sa) How much momentum does the cat have before the woman catches it?b) What happens to the woman/cat combination after the cat is caught?c) What is the total momentum of the woman/cat combination after the cat is caught?d) What is the speed of the woman/cat combination after the cat is caught?
a) The momentum of the cat before the woman catches it = 60 kgm/s
b) After the cat is caught, the woman and the cat move with an equal velocity
c) Total momentum of the woman/cat combination after the cat is caught =
60 kgm/s
d) The speed of the woman/cat combination after the cat is caught = 1.09 m/s
Explanation:Mass of the woman, m₁ = 50 kg
The woman was standing before she catches the cat. This means that she was at rest and the initial velocity is 0
u₁ = 0 m/s
The mass of the cat, m₂ = 5 kg
The cat's speed, u₂ = 12 m/s
Momentum = Mass x Velocity
a) The momentum of the cat before the woman catches it = m₂u₂
The momentum of the cat before the woman catches it = 5(12)
The momentum of the cat before the woman catches it = 60 kgm/s
b) After the cat is caught, the woman and the cat move with an equal velocity
c)
Total momentum of the woman/cat combination before the cat is caught= Total momentum of the woman/cat combination after the cat is caught
The initial momentum of the woman = m₁u₁
The initial momentum of the woman = 50(0) = 0 kgm/s
Total momentum of the woman/cat combination after the cat is caught = 0 + 60
Total momentum of the woman/cat combination after the cat is caught =
60 kgm/s
d) The speed of the woman/cat combination after the cat is caught
m₁u₁ + m₂u₂ = (m₁ + m₂)v
50(0) + 5(12) = (50+5)v
60 = 55v
v = 60/55
v = 1.09 m/s
The speed of the woman/cat combination after the cat is caught = 1.09 m/s
A student conducts an investigation into the motion of objects dropped from rest in the absence of air resistance. The student releases objects of various masses in a vacuum chamber and records the speed and distance fallen every 0.1 seconds for each object. What experimental evidence would lead the student to conclude that free-fall motion is constant acceleration? Select two answers. A. All objects released from rest at the same time fall together.B. The distance fallen is proportional to the square of the falling time.C. A graph of velocity vs. time is linear. D. Objects speed up as they free-fall. One of your answers can be explained using a basic equation of motion. Which answer is it and which equation is it? One of your answers can be explained using a rule about how acceleration appears on a certain graph. State the graph rule, then explain why “constant acceleration” looks the way that it does on this certain graph.
Which of the following choices correctly describes the orbital relationship between Earth and the sun?
a.
The sun orbits Earth in a perfect circle
c.
The sun orbits Earth in an ellipse, with Earth at one focus
b.
Earth orbits the sun in a perfect circle
d.
Earth orbits the sun in an ellipse, with the sun at one focus
The correct statement which describes the relationship between earth and the sun is that earth orbits the sun in an ellipse, with the sun at one focus.
What is an Orbit?An orbit in celestial mechanics is the curved path taken by an object, such as the path taken by a planet around with a star, a celestial body around a planet, or a manufactured satellite around an object or location in space, such as a planet, satellite, meteorite, or Lagrange point.
Ordinarily, the physics term "orbit" implies a trajectory that repeats itself with a regular basis, though it can also denote a non-repeating trajectory.
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The wavelength of a light wave as it passes through a material is 657.61 nm with a speed of 232,536,355.7 m/s. What is the frequency of this wave?
Given:
The wavelength of light is
[tex]\begin{gathered} \lambda\text{ = 657.61 nm} \\ =657.61\times10^{-9}\text{ m} \end{gathered}[/tex]The speed is v = 232536355.7 m/s
Required: The frequency of the wave.
Explanation:
The frequency can be calculated by the formula
[tex]f=\frac{v}{\lambda}[/tex]On substituting the values, the frequency will be
[tex]\begin{gathered} f=\text{ }\frac{232536355.7}{657.61\times10^{-9}} \\ =3.536\times10^{14}\text{ s} \end{gathered}[/tex]Final Answer: The frequency of the wave is 3.536e14 s
Please help me create a table and graph for the experiment below.Mass of bob: 25g, Time noted: 0.54 s, 0.54 s, 0.54 sMass of bob: 50g, Time noted: 0.54 s, 0.56 s, 0.49 sMass of bob: 100g: Time noted: 0.53 s, 0.53 s, 0.46 s
Given:
Mass of bob: 25g, Time noted: 0.54 s, 0.54 s, 0.54 s
Mass of bob: 50g, Time noted: 0.54 s, 0.56 s, 0.49 s
Mass of bob: 100g: Time noted: 0.53 s, 0.53 s, 0.46 s
To find:
The average time period for each mass of the pendulum bob.
Explanation:
As the sensor records the time it takes for the pendulum to complete one-half of the period, the time period of the pendulum can be calculated by multiplying the recorded value by 2.
For the bob of mass 25 g,
Period (s) Trial 1: 0.54 s × 2 = 1.08 s
Period (s) Trial 2: 0.54 s × 2 = 1.08 s
Period (s) Trial 3: 0.54 s × 2 = 1.08 s
Average Period (s): T = (1.08 s + 1.08 s + 1.08 s)/3 = 1.08 s
For the bob of mass 50 g,
Period (s) Trial 1: 0.54 s × 2 = 1.08 s
Period (s) Trial 2: 0.56 s × 2 = 1.12 s
Period (s) Trial 3: 0.49 s × 2 = 0.98 s
Average Period (s): T = (1.08 s + 1.12 s + 0.98 s)/3 = 1.06 s
For the bob of mass 100 g,
Period (s) Trial 1: 0.53 s × 2 = 1.06 s
Period (s) Trial 2: 0.53 s × 2 = 1.06 s
Period (s) Trial 3: 0.46 s × 2 = 0.92 s
Average Period (s): T = (1.06 s + 1.06 s + 0.92 s)/3 = 1.013 s
Final answer:
The average value of the period is
For bob of mass 25 g, Average Period = 1.08 s
For bob of mass 50 g, Average Period = 1.06 s
For bob of mass 100 g, Average Period = 1.013 s
6. A van of mass 1200 kg was moving at a velocity of 8 m/s when it was involved in a head-on collisionwith a lorry moving in the opposite direction. Assuming that the van came to a stop after the collision...(a) calculate the momentum of the van before the collision;(b) calculate the momentum of the van after the collision(c) find the change in momentum of the van (d) if the van took .30 s to stop, calculate the force that acted on each driver
Given data:
* The mass of the van is 1200 kg.
* The velocity of the van before the collision is 8 m/s.
* The velocity of the van after the collision is 0 m/s.
Solution:
(a). The momentum of the van before the collision is,
[tex]p_i=mv_i[/tex]where m is the mass of van, p_i is the momentum of van before the collision, and v_i is the velocity of van before the collision,
[tex]\begin{gathered} p_i=1200\times8 \\ p_i=9600kgms^{-1^{}} \end{gathered}[/tex]Thus, the momentum of the van before the collision is 9600 kgm/s.
(b). The momentum of the van after the collision is,
[tex]p_f=mv_f[/tex]weere p_f is the final momentum, and v_f is the final velocity of the van,
Substituting the known values,
[tex]\begin{gathered} p_f=1200\times0 \\ p_f=0^{} \end{gathered}[/tex]Thus, the momentum of the van after the collision is 0 kgm/s.
(c). The change in the momentum of the van is,
[tex]\begin{gathered} dp=p_f-p_i \\ dp=0-9600 \\ dp=-9600kgms^{-1} \end{gathered}[/tex]Here, the negative sign indicates that the momentum of van is decreasing with time.
Thus, the change in the momentum of the van is -9600 kgm/s.
(d). According to the Newton's second law, the force acting on the van in terms of the change in momentum is,
[tex]F=\frac{dp}{dt}[/tex]where dt is the time interval in which the momentum of the van changes,
Substituting the known values,
[tex]\begin{gathered} F=-\frac{9600}{0.30} \\ F=-32000\text{ N} \\ F=-32\times10^3\text{ N} \\ F=-32\text{ kN} \end{gathered}[/tex]Here, the negative sign is indicating the direction of force acting on the van is opposite to the direction of motion of van before the collision.
Thus, the force acting on the van is -32 kN.
Jenna manipulated the equation 4x+7=10 by adding -7 to both sides. Which of the following properties justifies this manipulation?1. The associative property of addition.2. The addition property of equality.3. The commutative property of addition.4. The multiplication property of equality.
2. The addition property of equality.
Explanation
Step 1
[tex]\begin{gathered} 4x+7=10 \\ \text{Jenna added -7 ( or subtract 7) to both sides, so} \\ 4x+7-7=10-7 \\ 4x=3 \end{gathered}[/tex]she applied the addition property for inequalities. it states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.
so, the answer is
2. The addition property of equality.
A substance has a mass of 15,000 kg and a volume of 30 m 3 . What is the density of thesubstance?
ANSWER
500 kg/m³
EXPLANATION
Given:
• The mass of the substance, m = 15,000 kg
,• The volume of the substance, V = 30 m³
Find:
• The density of the substance, ρ
The density of a substance of mass m and volume V is,
[tex]\rho=\frac{m}{V}[/tex]Substitute the known values and solve,
[tex]\rho=\frac{15,000kg}{30m^3}=500kg/m^3[/tex]Hence, the density of the substance is 500 kg/m³.
If you ride a bicycle 36 m in 12 seconds, what is your speed?O A. 0.3 m/sB. 2 m/sC. 3 m/sD. 0.09 m/s
We are given that the distance is 36 m and the time is 12 seconds.
We are asked to find the speed.
Recall that the time, speed, and distance formula is given by
[tex]v=\frac{x}{t}[/tex]Where v is the speed, x is the distance, and t is the time.
Let us substitute the given values into the above formula.
[tex]\begin{gathered} v=\frac{36}{12} \\ v=3\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the speed is 3 m/s
Option C is the correct answer.
ball is launched with an initial speed of 30 m/s making an angle of 45° above the horizontal. How long does it take the ball to reach a vertical displacement Δy= +5 m for the first time?*1.12 s0.54s0.9s0.25s0.7 s
Answer: t = 0.25 s
Explanation:
To find the time it will take the ball to travel a vertical height of 5m, we would apply one of Newton's equation of motion shown below
h = ut + 1/2gt^2
where
h is the height or vertical displacement
u is the initial velocity
g is the acceleration due to gravity
t is the time taken to reach height h
From the information given,
h = 5
g = - 9.8 m/s^2 because the ball is decelerating while moving upwards.
Given that the initial velocity is 30m/s and it was launched at an angle of 45 degrees, we would find the y or vertical component of the velocity. Thus,
u = 30sin45
By substituting these values into the equation, we have
5 = 30sin45t - 1/2 x 9.8 x t^2
5 = 21.21t - 4.9t^2
4.9t^2 - 21.21t + 5 = 0
This is a quadratic equation. The standard form of a quadratic equation is expressed as
ax^2 + bx + c = 0
By comparing both equations,
a = 4.9
b = - 21.21
c = 5
We would solve the equation by using the quadratic formula which is expressed as
[tex]\begin{gathered} x\text{ = }\frac{-\text{ b }\pm\sqrt{b^2-4ac}}{2a} \\ x\text{ = }\frac{-\text{ - 21.21 }\pm\sqrt{-\text{ 21.21}^2-4(4.9\text{ }\times5}}{2\times4.9} \\ x\text{ = }\frac{21.21\pm\sqrt{449.8641\text{ - 98}}}{9.8} \\ x\text{ = }\frac{21.21\text{ }\pm18.758}{9.8} \\ x\text{ = }\frac{21.21\text{ + 18.758}}{9.8}\text{ or x = }\frac{21.21\text{ - 18.758}}{9.8} \\ x\text{ = 4.08 or x = 0.25} \end{gathered}[/tex]Replacing x with t, we have
t = 0.25 s
A book of mass m = 10kg is balanced on a table Determine the weight of the book and the component of the reaction of the table in each of the following cases: 1)The book is simply placed on the table2) Someone puts their hand on the book by exerting a vertical force of 40N3)Someone is pulling this book vertically upwards with a force of 40N
Given,
The mass of the book, m=10 kg
The weight of an object is given by the product of its mass and the acceleration due to gravity.
Thus the weight of the book is given by,
[tex]W=mg[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} W=10\times9.8 \\ =98\text{ N} \end{gathered}[/tex]Therefore the weight of the book is 98 N
1)
When the book is placed on the table reaction of the table will be equal to the weight of the book. Thus, the reaction of the table, in this case, will be 98 N
2)
The vertically downward force applied on the book, F=40 N
The reaction of the table when there is a vertically downward force is equal to the sum of the weight and the applied force.
Thus the reaction of the table, in this case, is given by,
[tex]N=W+F[/tex]On substituting the known values,
[tex]\begin{gathered} N=98+40 \\ =138\text{ N} \end{gathered}[/tex]Thus, the reaction of the table when a vertically downward force is acting on it is 138 N
3)
The vertically upward force acting on the book, F=40 N
As the book is in equilibrium, the net force acting on the book must be equal to zero. That is total upward forces acting on the book must be equal to the total downward forces. Thus,
[tex]\begin{gathered} N+F=W \\ \Rightarrow N=W-F \end{gathered}[/tex]Where N is the normal force.
On substituting the known values,
[tex]\begin{gathered} N=98-40 \\ =58\text{ N} \end{gathered}[/tex]Thus the reaction of the table, in this case, is 58 N
1. Battery voltage = 12v. the current in the circuit is 0.5A. the resistance R must be 2. A 12 volt battery is connected across resistor, and a current 1.5A flows in the resistor. What is the value of the resistor?
To find the resistance, we have to use the following
[tex]R=\frac{V}{I}[/tex]Then, we replace the values
[tex]R=\frac{12}{1.5}=8[/tex]Hence, the answer is A.Alexander's hobby is dirt biking. On one occasion last weekend, he accelerated from rest to 12.8 m/s in 4.43 seconds. He then maintained this speed for 8.83 seconds. Seeing a coyote run cross the trail ahead of him, he abruptly stops in 6.36 seconds. Determine Alexander's average speed for this motion?
The average speed of Alexander, given that he accelerated from rest to 12.8 m/s in 4.43 seconds is 12.8 m/s
How do I determine the average speed of Alexander?First, we shall determine the distance travelled in 4.43 seconds. This is shown below:
Speed = 12.8 m/sTime = 4.43 secondDistance =?Distance = speed × time
Distance = 12.8 × 4.43
Distance = 56.704 m
Next, we shall determine the distance travelled in 8.83 seconds. This is shown below:
Speed = 12.8 m/sTime = 8.83 secondDistance =?Distance = speed × time
Distance = 12.8 × 8.83
Distance = 113.024 m
Next, we shall determine the distance travelled in 6.36 seconds. This is shown below:
Speed = 12.8 m/sTime = 6.36 secondDistance =?Distance = speed × time
Distance = 12.8 × 6.36
Distance = 81.408 m
Finally, we shall determine the average speed of Alexander. This is shown below:
Total distance = 56.704 + 113.024 + 81.408 = 251.136 mTotal time = 4.43 + 8.83 + 6.36 = 19.62Average speed =?Average speed = Total distance / total time
Average speed = 251.136 / 19.62
Average speed = 12.8 m/s
Thus, the average speed is 12.8 m/s
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the heat in Yravels at the of lines. This is also radiation known as and/or write one phrase below each of these terms e heat conduction
Using the concept of Heat radiation, we have described conduction, convection, and radiation,
Heat can travel from one place to another in several ways. The different modes of the heat transfer include: Conduction, Convection, Radiation
Meanwhile, if the temperature difference exists between the two systems, heat will find a way to transfer from higher to the lower system.
Conduction is defined as:
The process of transmission of energy from one particle of the medium to another with the particles being in direct contact with a each other.
Following are the examples of the conduction:
Ironing of clothes is a example of conduction where the heat is conducted from the iron to the clothes.
Heat is transferred from hands to ice cube resulting in a melting of an ice cube when held in hands.
Convection is defined as the:
The movement of fluid molecules from higher temperature regions to lower temperature regions.
Examples of the convection include:
Boiling of water, that is molecules that are denser move at bottom while the molecules which are less dense move upwards resulting in a circular motion of the molecules so that water gets heated.
Warm water around a equator moves towards the poles while cooler water at the poles moves towards the equator.
Radiation is defined as:
Radiant heat is present in some or other form in our daily lives. Thermal radiations are referred to as the radiant heat. Thermal radiation is generated by the emission of the electromagnetic waves.
Following are the examples of the radiation:
Microwave radiation emitted in the oven is an example of radiation.
UV rays coming from the sun is a example of radiation.
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convert 675 nm to meters
Given,
The number needed to be converted is 675 nm
A nanometer is equal to one billionth of a meter.
Mathematically,
[tex]1nm^{}=1\times10^{-9}\text{ m}[/tex]Thus to convert any number from nanometer to meter, we need to multiply the number by 10⁻⁹ m.
Thus,
[tex]675\text{ nm}=675\times10^{-9}\text{ m}[/tex]Therefore 675 nm is equal to 675×10⁻⁹ m
Please help me solve From a previous question, the index of refraction of the liquid is 1.37
ANSWER
EXPLANATION
From the previous part, we have that the index of refraction of the liquid is 1.37, so we have to replace this in the equation and solve,
[tex]\sin\theta_c=\frac{n_{air}}{n_{liquid}}=\frac{1.00}{1.37}\approx0.73[/tex]And then, take the inverse of the sine to find the critical angle,
[tex]\theta_c=\sin^{-1}0.73\approx46.9\degree[/tex]Hence, the critical angle is 46.9°, rounded to the nearest tenth.
Question 3 of 5Molly wants to measure how hot water is when it begins to boil. What toolshould she use?O A. A graduated cylinderO B. A thermometerO C. CalipersD. A test tube
To find
Molly wants to measure how hot water is when it begins to boil. What tool should she use?
Explanation
A graduated cylinder is used to measure the volume of liquid.
Thermometer is a device which is used to measure temperature.
Calipers are used to measure dimensions of objects
A test tube is used to hold liquid during laboratory experiments
Conclusion
To measure the hotness of water she needs
B. A thermometer
Numerical problem on Coulomb's law 1. Find the magnitude of product of two charges kept 1 metre aparat when a active force of 9.10^9 newton is acting between them.
In order to calculate the product of charges, we can use the formula below for the force acting on two charges (Coulomb law):
[tex]|F|=k_e\frac{|q_1q_2|_{}}{r^2}[/tex]Where ke is the Coulomb constant (ke = 9 * 10^9), q1 and q2 are the charges and r is the distance between the charges.
So, for r = 1 and F = 9 * 10^9, we have:
[tex]\begin{gathered} 9\cdot10^9=9\cdot10^9\frac{|q_1q_2|}{1^2} \\ |q_1q_2|=1 \end{gathered}[/tex]Therefore the product of the charges is 1 Coulomb.
If I have a USB 2.0 port and it provides a maximum current of 0.5A at a voltage of 5 volts, if I connect a device with a resistance of 3 Ohms, how much current would flow through it and what will the dissipated power be?
1. The amount of current that will flow through it is 1.67 A
2. The power dissipated is 8.35 watts
1. How do I determine the current that will flow through it?
Ohm's law states as follow
Voltage (V) = Current (I) × resistance (R)
V = IR
With the above formula, we can obtain the current. This is shown below
Voltage (V) = 5 VoltsResistance (R) = 3 ohmsCurrent (I) =?Voltage (V) = Current (I) × resistance (R)
5 = Current × 3
Divide both sides by 3
Current = 5 / 3
Current = 1.67 A
2. How do I determine the power dissipated?
The power dissipated can be obtained as follow:
Voltage (V) = 5 VoltsCurrent (I) = 1.67 APower dissipated (P) = ?Power (P) = voltage (V) × current (I)
Power dissipated = 5 × 1.67
Power dissipated = 8.35 watts
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Please help me I know it’s not the second choice
A 5000. kg elevator accelerates at 3.0m/s? as it begins to move upward. Calculate the tension in the cable
Given data;
* The mass of the elevator is 5000 kg.
* The acceleration of the elevator is,
[tex]a=3ms^{-2}[/tex]Solution:
The free body diagram of the elevator is,
The weight of the elevator is,
[tex]W=mg[/tex]where m is the mass of the elevator and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} W=5000\times9.8 \\ W=49000\text{ N} \end{gathered}[/tex]The net force acting on the elevator is,
[tex]F_{\text{net}}=ma[/tex]where a is the acceleration of the elevator moving upwards,
Substituting the known values,
[tex]\begin{gathered} F_{\text{net}}=5000\times3 \\ F_{\text{net}}=15000\text{ N} \end{gathered}[/tex]From the free body diagram, the tension acting on the cable is,
[tex]\begin{gathered} T-W=F_{\text{net}} \\ T=F_{\text{net}}+W \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} T=15000+49000 \\ T=64000\text{ N} \\ T=64\text{ kN} \end{gathered}[/tex]Thus, the tension acting in the cable is 64 kN.
A girl throws a ball vertically downward at 10m/s from the roof of a building 20m high. Whatwill its speed be when it strikes the ground?
Answer:
22.18 m/s
Explanation:
We will use the following equation:
[tex]v^2_f_{}=v^2_i+2ay[/tex]Where vf is the final velocity
vi is the initial velocity, so it is -10 m/s
a is gravity, so it is -9.8 m/s²
y is the change in the height so it is -20 m
Therefore, replacing the values, we get:
[tex]\begin{gathered} v^2_f=(-10)^2+2(-9.8)(-20) \\ v^2_f=100+392 \\ v^2_f=492 \\ v_f=\sqrt[]{492}=22.18\text{ m/s} \end{gathered}[/tex]So, the ball strikes the ground at 22.18 m/s
A student pushes a heavy box that is originally stationary across the floor with a force 112 N. What is the coefficient of static friction between the classroom floor and the 140 N box?
Given data:
* The force applied on the heavy box is,
[tex]F_a=112\text{ N}[/tex]* The weight of the heavy box is,
[tex]w=140\text{ N}[/tex]Solution:
The frictional force acting on the box is,
[tex]\begin{gathered} F_r=\mu\times w \\ \mu\text{ is the coefficient of friction,} \end{gathered}[/tex]The frictional force acting on the heavy box is equal to the applied force as the heavy box is not moving under the action of applied force.
Thus,
[tex]\begin{gathered} F_r=F_a \\ \mu\times w=112 \\ \mu\times140=112 \\ \mu=\frac{112}{140} \end{gathered}[/tex]By simplifying,
[tex]\mu=0.8\text{ }[/tex]Thus, the coefficient of friction between the floor and the box is 0.8.