Answer:
a) The x-position of the skateboarder is 0.324 meters.
b) The y-position of the skateboarder is -2.16 meters.
c) The x-velocity of the skateboard is 1.08 meters per second.
d) The y-velocity of the skateboard is -3.6 meters per second.
Explanation:
a) The x-position of the skateboarder is determined by the following expression:
[tex]x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}[/tex] (1)
Where:
[tex]x_{o}[/tex] - Initial x-position, in meters.
[tex]v_{o,x}[/tex] - Initial x-velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a_{x}[/tex] - x-acceleration, in meters per second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:
[tex]x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}[/tex]
[tex]x(t) = 0.324\,m[/tex]
The x-position of the skateboarder is 0.324 meters.
b) The y-position of the skateboarder is determined by the following expression:
[tex]y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}[/tex] (2)
Where:
[tex]y_{o}[/tex] - Initial y-position, in meters.
[tex]v_{o,y}[/tex] - Initial y-velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a_{y}[/tex] - y-acceleration, in meters per second.
If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o,y} = -3.6\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{y} = 0\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:
[tex]y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}[/tex]
[tex]y(t) = -2.16\,m[/tex]
The y-position of the skateboarder is -2.16 meters.
c) The x-velocity of the skateboarder ([tex]v_{x}[/tex]), in meters per second, is calculated by this kinematic formula:
[tex]v_{x}(t) = v_{o,x} + a_{x}\cdot t[/tex] (3)
If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-velocity of the skateboarder is:
[tex]v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)[/tex]
[tex]v_{x}(t) = 1.08\,\frac{m}{s}[/tex]
The x-velocity of the skateboard is 1.08 meters per second.
d) As the skateboarder has a constant y-velocity, then we have the following answer:
[tex]v_{y} = -3.6\,\frac{m}{s}[/tex]
The y-velocity of the skateboard is -3.6 meters per second.
The energy transfer diagram for a piece of equipment is shown. You are using a hand-cranked flashlight. One minute of cranking typically provides about 30 to 60 minutes of light.
Describe ALL of the energy transformations that are taking place in this process.
A) mechanical energy to heat energy Eliminate
B) electrical energy to light energy
C) mechanical energy to light energy
D) chemical energy to mechanical energy
E) mechanical energy to electrical energy
Answer:
n/a
Explanation:
No diagram found. I am willing to help
In a hand-cranked flashlight first, mechanical energy produced with the hand is converted into electrical energy then the electrical energy is converted into light energy, therefore the correct answers are the option B and option E.
What is mechanical energy?The sum of all the energy in motion (total kinetic energy) and all the energy that is stored in the system (total potential energy) is known as mechanical energy.
The expression for total mechanical energy can be given as follows
ME= PE + KE
As given in the problem, the energy transfer diagram for a piece of equipment is shown. You are using a hand-cranked flashlight. One minute of cranking typically provides about 30 to 60 minutes of light.
The transformation that is taking place is the mechanical energy to electrical energy and then electrical energy to light energy.
Thus, the correct options are option are B and E.
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Four balls with different masses are dropped from the heights shown. Air resistance may be ignored. Which ball has the greatest average speed?
Answer:
The one falling from the greatest height will have the greatest speed.
h = 1/2 g t^2 time for ball to fall distance h
h2 / h1 = t2^2 / t1^2 dividing equations
h2 / t2^2 = h1 / t1^2
Let v be the average speed (v2 = h2 / t2)
1 / t2 * v2 = 1 / t1 * v1
v2 / v1 = t2 / t1 the one taking the longest to fall has the greater av. speed
Check:
h4 / h1 = t4^2 / t1^2 or
t4 / t1 = (h4 / h1)^1/2
In this case t4 / t1 = (4 / 1)^1/2 = 2 or twice the average speed
t1 = (2 h / g)^1/2 = .2^1/2 = .447 using g = 10
t4 = (2 h / g)^1/2 = .8^1/2 = .894
v1 = 1 / .447 = 2.24 m/s average speed
v4 = 4 / .894 = 4.47 or twice the average speed
Important parts of stydi g physics
Answer:
put your question in proper way' i hope you understand.
Explanation:
What is the amplitude of this wave ?
Hope you could get an idea from here.
Doubt clarification - use comment section.
Which is true of gamma radiation? O A. It increases the number of protons. O B. It is the heaviest of the three types. O C. It does not cause transmutation. O D. It has a positive charge.
Answer: Your answer Is A)
Explanation:
Its direction of deflection shows it possitively charged
It brings one element into another by bombardment(transmutation)
Would you please help me with this? I can't figure it out, please! I need to know what the E means!
Answer:
Without the full content of your question, I will have to GUESS at the context and assume
E = Energy
released when glucose is broken down.
A 12-kg block on a horizontal frictionless surface is attached to a light
spring (force constant = 0.80 kN/m). The block is initially at rest at its
equilibrium position when a force (magnitude P = 80 N) acting parallel to
the surface is applied to the block, as shown. What is the speed of the
block when it is 13 cm from its equilibrium position?"
The speed of the block at the displacement from the equilibrium position is 1.062 m/s.
The given parameters:
Mass of the block, m = 12 kgSpring constant, k = 0.8 kN/mExtension of the spring, x = 13 cm = 0.13 mApplied parallel force, F = 80 NThe speed of the block is calculated by applying the principle of conservation of mechanical energy as shown below;
[tex]\frac{1}{2} mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{800 \times 0.13^2}{12} } \\\\v = 1.062 \ m/s[/tex]
Thus, the speed of the block at the displacement from the equilibrium position is 1.062 m/s.
Learn more about conservation of mechanical energy here: https://brainly.com/question/6852965
Answer:
The speed of the block at the displacement from the equilibrium position is 1.1266 m/s.
Step-by-step explanation:
Solution :
Using principle of conservation of mechanical energy formula to find the speed of the block :
[tex]\begin{gathered} \longrightarrow{\pmb{\sf{\frac{1}{2} mv^2 = \frac{1}{2}kx^2}}}\end{gathered}[/tex]
»» m = Mass of the block, »» k = Spring constant,»» x = Extension of the spring»» F = Applied parallel forceAs per given data information in the question we have :
✧ Mass of the block = 12 kg✧ Spring constant = 0.8 kN/m✧ Extension of the spring = 0.13 m✧ Applied parallel force = 80 NSubstituting all the given values in the formula to find the speed of the block
[tex]\longrightarrow{\sf{ \: \:\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: \cancel{\dfrac{1}{2}}mv^2 = \cancel{\dfrac{1}{2}}kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: mv^2 = kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v^2 = \dfrac{kx^2}{m}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: \sqrt{{v}^{2} } = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13}^{2}}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13} \times 0.13}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times 0.0169}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{13.52}{12}}}}[/tex]
[tex]{\star{\underline{\boxed{\rm{\red{ v \approx 1.1266 \: m/s}}}}}}[/tex]
Hence, the speed of block is 1.1266 m/s.
[tex] \rule{300}{1.5}[/tex]
The mass of fifteen washers is _____ kg, which exerts a force of _____ N
Answer:
It could be related with the lesson from which this question belongs as far we did not read the lesson
Sorry
2 difference between calorimetry and calorimeter
Calorimetry :
the process of measuring the amount of heat released or absorbed during a chemical reaction.
Calorimeter :
device for measuring the heat developed during a mechanical, electrical, or chemical reaction, and for calculating the heat capacity of materials.
Which is an electromagnetic wave A. The waves that heat a cup of water in a microwave oven B. A flag waving in the wind C. Turning on a flashlight D.The changes in the air that result from blowing a horn
Answer:
The answer would be A. The waves that heat a cup of water in a microwave oven
How much energy does a 150 N child have sitting on a 35M high cliff
Answer:
150+35=185 Plsss Brainliest plssss
For northern hemisphere observers, which celestial object would be above the horizon for the greatest
amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is
70° above the celestial equator, or one that is 40" below the celestial equator? Which one would be above
the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.
Answer:
Explanation:
For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.
For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
What is the equator?The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.
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Which feature of a balanced chemical equation demonstrates the law of
conservation of mass?
O A. It has the same types of atoms on both sides of the reaction
arrow.
O B. It shows the reactants of a chemical reaction to the left of the
reaction arrow.
O C. It has coefficients to show how much of each substance a
chemical reaction uses.
Thing
D. It shows the products of a chemical reaction to the right of the
reaction arrow.
Answer: A) It has the same types of atoms on both sides of the reaction
arrow.
Explanation: A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.
Which of these is Newton's 3rd law of motion?
Equal and opposite forces
F = m x a
Inertia
Gravity
The more matter in an object, the _____.(1 point)
lower its weight and mass
lower its weight and mass
more likely it is to have gravity
more likely it is to have gravity
greater its gravitational attraction to Earth
greater its gravitational attraction to Earth
less likely it is to have weight on the moon
Answer:
more likely it is to have gravity
You illuminate the surface of some metal with green laser and observe the photocurrent. If you decrease the intensity of laser, the stopping potential will
Remain the same.
Explanation:
The kinetic energy of the ejected electrons, and thus the resulting photocurrent, does not depend on the intensity of the incident radiation. Instead, it depends on the frequency. So since the stopping potential is used to reduce the photocurrent to zero and the photocurrent does not depend on the intensity, the stopping potential remains the same.
Why is acceleration of an object moving at a constant velocity always zero?
Answer:
If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).
Explanation:
In the middle, when the object was changing position at a constant velocity, the acceleration was 0. This is because the object is no longer changing its velocity and is moving at a constant rate.
Can you solve this question?
Hi there!
In this instance, the object's centripetal force is provided by the horizontal component of the tension, so:
Tsinθ = mv²/r
**We use sine because in this situation, the angle is with the vertical**
We can plug in the known values for tension and theta:
60sin(60) = mv²/r
51.96 = mv²/r
The radius is equivalent to the sine of the string in respect to theta:
sin(60) = O/H = r/L
2sin(60) = 1.732 m
Now, solve for the velocity:
51.96 = mv²/r
51.96r / m = v²
51.96(1.732)/.400 = v²
v² = 225
v = 15 m/s
25 gram saturated solution of potassium nitrate at 95 C is cooled down to 55 C then how much gram of crystals of potassium nitrate will be separated if the solubility of potassium nitrate at 95 c is 100 and 55 C is 25 correspondingly
The mass of potassium nitrate (KNO₃) crystals that will be separated is calculated as 6.25 g.
The given parameters:
Mass of KNO₃ = 25 gInitial temperature = 95 ⁰CFinal temperature = 55 ⁰CSolubility at 95 ⁰C = 100 MSolubility at 55 ⁰C = 25 MThe mass of KNO₃ at 95 ⁰C is calculated as follows;
[tex]m = \frac{25\ g \times 100\ g}{100\ g} \\\\m = 25 \ g[/tex]
mass of water = 100 g - 25 g = 75 g
The mass of KNO₃ at 55 ⁰C is calculated as follows;
[tex]m = \frac{75 \ g \times 25 \ g}{100 \ g} \\\\m = 18.75 \ g[/tex]
The mass of potassium nitrate (KNO₃) crystals that will be separated is calculated as;
[tex]m= 25\ g \ - \ 18.75 \ g\\\\m = 6.25 \ g[/tex]
Thus, the mass of potassium nitrate (KNO₃) crystals that will be separated is calculated as 6.25 g.
Learn more about saturated solution and temperature here: https://brainly.com/question/4529762
If you apply a net force of 100 N to the hoverboard, and it accelerates
2m/s/s, how much mass does it have?
Answer:
50 kgExplanation:
The mass of an object given only the force acting on it and it's acceleration can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force in N
a is the acceleration in m/s²
From the question
f = 100 N
a = 2 m/s²
We have
[tex]m = \frac{100}{2} = 50 \\ [/tex]
We have the final answer as
50 kgHope this helps you
A child on ice skates is given a small push from behind by a parent. How are forces used to describe the resulting motion?(1 point)
Because of equal and opposite reactions, there will be a force opposing the push from the parent, and they will not move.
Because of equal and opposite reactions, there will be a force opposing the push from the parent, and they will not move.
There is an unbalanced force on the child, and the child’s motion will change direction or increase speed.
There is an unbalanced force on the child, and the child’s motion will change direction or increase speed.
There is a balanced force on the child, and the child’s motion will change direction or increase speed.
There is a balanced force on the child, and the child’s motion will change direction or increase speed.
Because of equal and opposite reactions, the child will move in the opposite direction to the force.
A child on ice skates is given a small push from behind by a parent. There is an unbalanced force on the child, and the child’s motion will change direction or increase speed.
A child on ice skates is given a small push from behind by a parent. How are forces used to describe the resulting motion?
Because of equal and opposite reactions, there will be a force opposing the push from the parent, and they will not move. FALSE. According to Newton's third law of motion, if the parent applies force on the child, there will be a reaction applied by the child on the parent. These forces are applied to different objects so they will not cancel and the child will move. There is an unbalanced force on the child, and the child’s motion will change direction or increase speed. TRUE. The child will have an acceleration as a consequence of the unbalanced force. The acceleration will be responsible for the change in the direction or speed of the child. There is a balanced force on the child, and the child’s motion will change direction or increase speed. FALSE. If the forces were balanced, that is, there was no net force, the movement of the child would not change, as stated by Newton's first law of motion. Because of equal and opposite reactions, the child will move in the opposite direction to the force. FALSE. The child will move in the direction of the net force.A child on ice skates is given a small push from behind by a parent. There is an unbalanced force on the child, and the child’s motion will change direction or increase speed.
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"A soccer ball is kicked horizontally off a 22 m high hill and lands a distance of 35 m from the edge of the hill." Which variable is 35 m? *
Answer:Let’s assume that, after the soccer ball is kicked and moves through its trajectory, it first makes contact with level ground a horizontal distance of 35 meters from where it was kicked. Let’s also assume that we can neglect air resistance. The time, t, that the soccer ball is in the air until it first contacts the ground can be found from the equation h = (1/2)gt^2 which can be rewritten as t = sqrt(2h/g) where h is the vertical distance the ball falls which is the height of the hill since the ball was kicked horizontally, and g is the acceleration of gravity which is 9.8 m/s^2. So t = sqrt(2(22)/9.8) = 2.12 seconds. In that time, the ball travelled 35 meters so its horizontal velocity was 35 meters/2.12 seconds = 16.5 meters/second.
Explanation:
7. What is the velocity of an object with a distance of 90m south and a time of
5s?
Answer:
Explanation:
v= s/t
V =90m/5s
V = 8m/s
the distance between two charges q a and q b is r and the force between them is F. What is the force between them if the distance between them is doubled?
The force will be reduced to 1/4 of the original
Explanation:
According to Coulomb's law, the force between two charges [tex]q_a\:\text{and}\:q_b,[/tex] separated by a distance r is given by
[tex]F = k\dfrac{q_aq_b}{r^2}[/tex]
where k is the Coulomb constant.
Now let F' be the force between the two charges when their separation distance is doubled. We can write this force as
[tex]F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}[/tex]
[tex]\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F[/tex]
Therefore, the force will be reduced to a quarter of its original value.
Hope you could understand.
If you have any query, feel free to ask.
You are in Paris, 50. m up on an Eiffel Tower support leg observation deck. If you throw a euro downward at a velocity of -1.0 m/s, how long would it take the coin to hit the ground?
Here’s my work to your question. I used kinematic equations to solve. :)
half-life questionnnnn:
a heat engine is a device that uses to produce useful work
A heat engine is a device that uses to produce useful work.
Definition - a device for producing motive power from heat, such as a gasoline engine or steam engine.
So..
If this is a true or false question.. Your answer is:
TRUE
Answer:
HEAT
Explanation:
22 Newton force is working on a 1,901 gram object. What is the acceleration in
meter/s^2 unit
Answer:
11.573
Explanation:
f = m*a
where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration
so, we solve for a
a = f/m
a = 22/1.901
a = 11.573
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time
Answer:
[tex]\huge\boxed{\sf a = 1200\ m/s\²}[/tex]
Explanation:
Given Data:
Initial Velocity = Vi = 40 m/s
Final Velocity = Vf = 80 m/s
Distance = S = 200 m
Required:
Acceleration = a = ?
Formula:
2aS = Vf² - Vi² (THIRD EQUATION OF MOTION)
Solution:
2a (200) = (80)² - (40)²
400a = 6400 - 1600
400a = 4800
Divide 400 to both sides
a = 4800 / 400
a = 1200 m/s²
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807The force shown in the figure(Figure 1) moves an object from x = 0 to x = 0.75 m.
1/How much work is done by the force?
2/How much work is done by the force if the object moves from x = 0.20 m to x = 0.55 m ?
Answer:
(a) The force changes its magnitude with respect to displacement, hence the total work will be sum of increment of work in three steps:-
step 1 . from 0 to 0.25m .
force = 0.6 N
displacement= 0.25m
work done =( force × displacement) = (0.25 × 0.6 ) = 0.15 joule.
step 2:- .
work done in moving from 0.25 to 0.50 m.
work done = ( force × displacement) = ( 0.4 × 0.25) = 0.10 Joule. .
step 3 :-
work done in moving from 0.50 to 0.75 m
work done = 0.8 × 0.25 = 0.200 joule.
hence total work done = ( 0.20+0.10+0.15) = 0.45 joule. ans
(b) similar concept you have to use here also.
step 1:
from 0.20 to 0.50, force of magnitude 0.4 N acts on the object.
Work done = ( 0.50-0.20)× 0.4 = 0.30 × 0.4 = 0.12 joule.
step :- 2
from 0.50 to 0.55 , force of magnitude 0.8 N acts on the block.
work done = 0.8× ( 0.55- 0.50) = 0.04 joule
total work done = 0.04 + 0.12 = 0.16 joule. ans