Answer:
t = 1.32 s
Explanation:
We are given;. Frequency of C4 note; F_c = 262 Hz
In conversions, we know that 1 Hz = 1 cycle/s
Thus, F_c = 262 cycles/s
Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.
346 air pressure maxima denotes that the air pressure maxima is 346 cycles.
Thus, time will be;
t = 346 cycles/262 cycles/s
t = 1.32 s
The time taken for the musical note to pass the stationary listener is 1.32 s.
The given parameters:
frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346The frequency of a sound wave is defined as the number of cycles completed per second by the wave.
[tex]F = \frac{n}{t}[/tex]
where;
t is the time to compete the maximum cycleThe time taken for the musical note to pass the stationary listener is calculated as follows;
[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]
Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.
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What is the picture called that shows ALL the forces acting on an object?
Answer:
Free Body Diagram
Explanation:
Such picture is called the "free body diagram"
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons is this charge?
Answer:
[tex]q\approx 6.6\cdot 10^{13}~electrons[/tex]
Explanation:
Coulomb's Law
The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:
[tex]\displaystyle F=k\frac{q^2}{d^2}[/tex]
Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:
[tex]\displaystyle q=\sqrt{\frac{F}{k}}\cdot d[/tex]
Substituting values:
[tex]\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1[/tex]
[tex]q = 1.05\cdot 10^{-5}~c[/tex]
This charge corresponds to a number of electrons given by the elementary charge of the electron:
[tex]q_e=1.6 \cdot 10^{-19}~c[/tex]
Thus, the charge of any of the spheres is:
[tex]\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}[/tex]
[tex]\mathbf{q\approx 6.6\cdot 10^{13}~electrons}[/tex]
A track star runs a 100m race in 12s what is the velocity of the runner?
Answer:
8.33 m/s
Explanation:
v=d/s, velocity = displacement/ time
a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic system?
Answer:
The answer is 7 PaExplanation:
The pressure transmitted in the hydraulic system can be found by using the formula
[tex]p = \frac{f}{a} \\ [/tex]
f is the force
a is the area
From the question we have
[tex]p = \frac{35}{5} \\ [/tex]
We have the final answer as
7 PaHope this helps you
Which energy transformation occurs after a skydiver reaches terminal velocity? Gravitational potential energy transforms into thermal energy. Gravitational potential energy transforms into kinetic energy. Kinetic energy transforms into thermal energy. Kinetic energy transforms into gravitational potential energy. The answer is A. just took it
Answer:
a
Explanation:
The energy transformation occurs after a skydiver reaches terminal velocity is follows as;
A. Gravitational potential energy transforms into thermal energy.
B. Gravitational potential energy transforms into kinetic energy.
What is the gravitational potential energy?The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:
U = mgh
where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground.
The skydiver gravitational potential energy decreases as the altitude decreases and his kinetic energy store increases as his speed increases.
When a skydiver jumps out of a plane, the energy transfers take place as;
The skydiver's kinetic energy store increases as their speed increases and the thermal store of the air and the skydiver increases, as there is friction between the skydiver and the air particles.
In the given situation, both options A and B are correct.
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What was used as a basket at each end of the gym?
Group of answer choices
bucket
hula hoop
apple basket
peach basket
Answer: apple basket ✨
Explanation:
Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?
Answer:
The value is [tex]w = 0.1167 \ rev/second[/tex]
Explanation:
From the question we are told that
The rate at which the plate rotates is [tex]w =7.0 \ rev/min[/tex]
Generally the revolution per second is mathematically represented as
[tex]w = \frac{7.0}{60}[/tex]
=> [tex]w = 0.1167 \ rev/second[/tex]
A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 3.0 seconds. What is the speed of the cart at the end of this 3.0 second interval? A. 1.5 m/s B. 5.5 m/s G. 3.0 m/s D. 7.0 m/s
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)
The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
Given the following data:
Initial velocity = 4 m/sMass of cart = 6 KgAcceleration = 0.5 [tex]m/s^2[/tex]Time = 3 secondsTo find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;
[tex]V = U + at[/tex]
Where:
U is the initial velocity.V is the final velocity. a is the acceleration. t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]
Final velocity, V = 5.5 m/s.
Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
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If a person Travels 100 metre due east and then returns to the same place his total displacement is 200. (needed ASAP)
A. True
B. False
Distance is the total path covered by the object
Here, 200 m is the distance covered by the person and NOT the displacement
Displacement of an object is nothing more than the shortest path between the initial and the final point
If the person travelled 100m and came back, his initial and final point will remain the same which means that he will have a displacement of 0 m
For satellite travelling on circular orbit if radius of the orbit increased 4 times then the period of the satellite increased *
2
4
8
none of the above
Answer:
8
Explanation:
The components of vector Upper A Overscript right-arrow EndScripts are Ax and Ay (both positive), and the angle that it makes with respect to the positive xaxis is θ. Find the angle θ if the components of the displacement vector Upper A Overscript right-arrow EndScripts are:
(a) Ax = 12 m and Ay = 12 m,
(b) Ax= 19 m and Ay = 12 m, and
(c) Ax = 12 m and Ay = 19 m.
(a) θ = Number____________ Units____
(b) θ = Number____________ Units____
(c) θ = Number ____________Units____
Answer:
(a) θ = 45° = 0.78 rad
(b) θ = 32.27° = 0.56 rad
(c) θ = 57.27° = 1 rad
Explanation:
When a vector is resolved into its rectangular components, the formula for the direction angle of the vector with positive x-axis is given as:
tan θ = Ay/Ax
θ = tan⁻¹(Ay/Ax)
(a)
Ax = 12 m
Ay = 12 m
θ = tan⁻¹(12 m/ 12 m)
θ = tan⁻¹(1)
θ = 45° = 0.78 rad
(b)
Ax = 19 m
Ay = 12 m
θ = tan⁻¹(12 m/19 m)
θ = tan⁻¹(0.6315)
θ = 32.27° = 0.56 rad
(c)
Ax = 12 m
Ay = 19 m
θ = tan⁻¹(19 m/12 m)
θ = tan⁻¹(1.58333)
θ = 57.27° = 1 rad
Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?
A. 165 m/s
B. 1.65 m/s
C. 10.4 m/s
D. 1040 m/s
Answer:
Maximum speed ( v ) = 10.4 m/s (Approx)
Explanation:
Given:
Amplitude A = 15.0 cm = 0.15 m
Frequency f = 11.0 cycles/s (Hz)
Find:
Maximum speed ( v )
Computation:
Angular frequency = 2πf
Angular frequency = 2π(11)
Angular frequency = 69.14
Maximum speed ( v ) = WA
Maximum speed ( v ) = 69.14 x 0.15
Maximum speed ( v ) = 10.371
Maximum speed ( v ) = 10.4 m/s (Approx)
An object has a mass of 15 kg and is accelerating to the right at 16.3 m/s2. The free-body diagram shows the horizontal forces acting on the object. A free body diagram with 2 forces. The first vector is pointing right, labeled F Subscript a Baseline 250 N. The second vector is shorter pointing left, labeled F Subscript f Baseline. What is the frictional force, Ff, acting on the object?
Answer:
A. -5.5 N
Explanation:
I can confirm that the answer is A.
An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?
Answer:
2.22 kPaExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
From the question we have
[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]
We have the final answer as
2.22 kPaHope this helps you
A cyclist is riding along at a speed of 20.7 when she decides to apply the brakes which gave a deceleration applied was a rate of -3.4 m/s2 over the span of 7.8 s. What distance does she travel over that period of time.
Answer:
The distance is 58.03 m
Explanation:
Constant Acceleration Motion
It occurs when the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
[tex]v_f=v_o+at[/tex]
The distance traveled by the object is given by:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]
The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.
The final speed is:
[tex]v_f=20.7+(-3.4)\cdot 7.8[/tex]
[tex]v_f=20.7-26.52[/tex]
[tex]v_f=-5.82\ m/s[/tex]
Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:
[tex]\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}[/tex]
[tex]\displaystyle x=161.46-103.43[/tex]
x=58.03 m
It has been suggested that rotating cylinders about 12.5 mi long and 6.25 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?
Answer:
w = 0.0436 rad/s
Explanation:
Radius of the cylinder is
3.125 * 1609m/mile = 5028 m
Using the angular acceleration formula, we have
a = v²/r = w²r.
Since we're interested in the angular speed, we make w the subject of the formula, so that
w² = a/r
w = √a/r
w = √(9.8/5028)
W = √0.0019049
w = 0.0436 rad/s
Then we can conclude that the angular speed of the rotation is w = 0.0436 rad/s
Based on the situation above, choose the CORRECT type of error.
The reading from the timer was not accurate because some of
the timer's display was missing and broken."
Answer:
what's your question I can't understand
A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light bulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb?
A) 15 W/m2
B) 4.8 W/m2
C) 2.4 W/m2
D) 0.60 W/m2
E) 1.2 W/m2
Answer:
a
Explanation:
A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light bulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb
so i did 60(50%)=30÷2=15
The radiation intensity is 15 W / m².
To find the radiation intensity, the values are given as,
Power = 60 W
Distance = 2 m
50% of the power input is emitted as electromagnetic radiation.
What is radiation intensity?The amount of energy emitted per unit solid angle by per unit area of the radiating surface can be said as radiation intensity.
As, when there is a power output, the input power will emit some energy as a kinetic energy and electromagnetic radiation. By the way, the power input emits 50 % as a radiation, so the power input given as,
P = 60 / 2 ( As it was 50 % )
= 30 Watt.
The radiation intensity is,
In = 30 / 2
= 15 W / m².
Thus, the radiation intensity is calculated as, 15 W/ m².
So, Option A is the correct answer.
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Which of the following descriptions best describe a liquid
A
takes the shape and volume of its container
B
matter is made of atoms so tightly packed together that they cannot move around
C
has a definite volume, but takes the shape of its container
Answer:
C
Explanation:
A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.120. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.559 m/s, what is her speed at the bottom?
Answer:
a
[tex]H =212.6 \ J[/tex]
b
[tex]v = 7.647 \ m/s[/tex]
Explanation:
From the question we are told that
The child's weight is [tex]W_c = 287 \ N[/tex]
The length of the sliding surface of the playground is [tex]L = 7.20 \ m[/tex]
The coefficient of friction is [tex]\mu = 0.120[/tex]
The angle is [tex]\theta = 31.0 ^o[/tex]
The initial speed is [tex]u = 0.559 \ m/s[/tex]
Generally the normal force acting on the child is mathematically represented as
=> [tex]N = mg * cos \theta[/tex]
Note [tex]m * g = W_c[/tex]
Generally the frictional force between the slide and the child is
[tex]F_f = \mu * mg * cos \theta[/tex]
Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as
[tex]F =m* g sin(\theta) - F_f[/tex]
Here F is the resultant force and it is represented as [tex]F = ma[/tex]
=> [tex]ma = m* g sin(31.0) - \mu * mg * cos (31.0)[/tex]
=> [tex]a = g sin(31.0)- \mu * g * cos (31.0)[/tex]
=> [tex]a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)[/tex]
=>[tex]a = 4.039 \ m/s^2[/tex]
So
[tex]F_f = 0.120 * 287 * cos (31.0)[/tex]
=> [tex]F_f = 29.52 \ N[/tex]
Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as
[tex]H = F_f * L[/tex]
=> [tex]H = 29.52 * 7.2[/tex]
=> [tex]H =212.6 \ J[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
=> [tex]v^2 = 0.559^2 + 2 * 4.039 * 7.2[/tex]
=> [tex]v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}[/tex]
=> [tex]v = 7.647 \ m/s[/tex]
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2.
**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**
Answer:
box 1 has larger mass than box 2 (which agrees with the first answer option given.
Explanation:
We need to consider the linear momentum of the boxes immediately before and after they crash.
Recall that momentum is defined as mass times velocity.
So for before the collision, the linear momentum of the system of two boxes is:
m1 * 4km/h - m2 * 8km/h
with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.
Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)
For after the collision, we have or the linear momentum of the system:
- m1 * 2km/h - m2 * 1km/h
Then, since the linear momentum is conserved in the collision, we make the initial momentum equal the final and study the mass relationship between m1 and m2:
4 m1 - 8 m2 = - 2 m1 - m2
combining like terms for each mas on one side and another of the equal sign, we get;
4 m1 + 2 m1 = 8 m2 - m2
6 m1 = 7 m2
therefore m1 = (7/6) m2
which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6
Therefore, the first answer option is the correct answer.
An artificial satellite in a circular orbit around the Sun has a period of 8 years. Determine the ratio of the satellite's orbital radius to that of the earth's orbital radius. Assume that the earth's orbit around the Sun is circular.
Answer:
The ratio is [tex]R_c:R_e = 4 : 1[/tex]
Explanation:
From the question we are told that
The period of the satellite is [tex]T_c = 8 \ years[/tex]
Generally the period of earth around the sun is [tex]T_e = 1 \ year[/tex]
Generally from Kepler's third law , which is mathematically represented as
[tex]\frac{T_c ^2}{T_e^2} = \frac{R_c^3}{R_e^3}[/tex]
Here [tex]R_c[/tex] is the radius of the orbit which the satellite rotate around the sun
[tex]R_e[/tex] is the radius of the orbit which the earth rotate around the sun
=> [tex]\frac{R_c^3}{R_e^3} = [\frac{8}{1} ]^2[/tex]
=> [tex]\frac{R_c}{R_e} = \sqrt[3]{[\frac{8}{1} ]^2}[/tex]
=> [tex]\frac{R_c}{R_e} = \frac{4}{1 }[/tex]
=> [tex]R_c:R_e = 4 : 1[/tex]
You are riding a bicycle. If you apply a forward force of 125 N, and you and
the bicycle have a combined mass of 82 kg, what will be the forward
acceleration of the bicycle? (Assume there is no friction.)
I WILL GIVE YOU POINTs
Answer:
1.52g
Explanation:
Given parameters:
Force = 125N
Mass combined = 82kg
Unknown:
Acceleration of the bicycle = ?
Solution:
From Newton second law of motion suggests that:
Force = mass x acceleration
Acceleration = [tex]\frac{force}{mass}[/tex] = [tex]\frac{125}{82}[/tex] = 1.52g
Answer:
The answer that was correct for me was A. 55 N pulling left, and 16 N, 17N p
pulling Right
Explanation:
A golfer hits the ball off the tee at an angle of thirty-five degrees from the horizontal with a speed of 46 m/s. It lands on the green, which is elevated 5.50 m higher than the tee. How much time elapsed from when the ball was hit to when it landed on the green?
Answer:
t = 5.16 seconds.
Explanation:
The flight time can be found using the following equation:
[tex] y_{f} = y_{0} +v_{0y}*t - \frac{1}{2}gt^{2} [/tex]
Where:
t: is the time
g: is the gravity = 9.81 m/s²
y₀: is the initial height = 0
[tex]y_{f}[/tex]: is the final height = 5.50 m
[tex]v_{0y}[/tex]: is the initial vertical velocity = v*sin(35)
v: is the speed = 46 m/s
[tex] 5.50 = 46*sin(35)*t - \frac{1}{2}9.81*t^{2} [/tex]
By solving the above cuadratic equation we have:
t₁= 0.22 s and t₂= 5.16 s
We will take the solution equal to 5.16 seconds, since in 0.22 seconds (very short time) the ball is going up and in 5.16 seconds it landed on the green.
Therefore, 5.16 seconds have passed since the ball was hit until it landed.
I hope it helps you!
A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0° below horizontal. How far from the base of the building will the ball land?
Answer:
Explanation:
horizontal component of velocity of throw = 20 cos20 = 18.8 m /s
vertical downwards component = 20 sin20 = 6.84 m /s
time to displace by height 30 m = t , initial velocity u = 6.84 m /s
h = ut + 1/2 gt²
30 = 6.84 t + .5 x 9.8 t²
4.9 t² + 6.84 t - 30 = 0
t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9
= - 6.84 ±√( 46.78 + 588 ) / 9.8
= - 6.84 ±√(634.78 ) / 9.8
= - 6.84 ±25.2 / 9.8
= 1.87 s
horizontal displacement in 1.87 s
= 18.8 x 1.87
= 35.15 m .
At what speed does a 2000 kg compact car have the same kinetic energy as a 18000 kg truck going 21 km/hr?
Answer:
v = 17.5 m/s = 63 km/h
Explanation:
The general expression for the kinetic energy of one moving object is as follows:[tex]K = \frac{1}{2}*m *v^{2} (1)[/tex]
where m = mass of the object, v= speed of the object.
In order to get the value of the kinetic energy of the truck in Joules, we need to convert km/hr to m/s first, as follows:[tex]21 km/hr * \frac{1 hr}{3600s}*\frac{1000m}{1 km} = 5.83 m/seg (2)[/tex]
Now, replacing (2) and m = 18000 kg in (1), we get:[tex]K = \frac{1}{2}*18000 kg *(5.83m/s)^{2} = 306250 J (3)[/tex]
This value must be the same for the 2000 kg compact car, so we can write:[tex]K = 306250 J = \frac{1}{2}*2000 kg *v^{2} (4)[/tex]
Solving for v, we get:[tex]v = \sqrt{\frac{306250}{1000} (m/s)2} = 17. 5 m/s = 63 km/h (5)[/tex]
A ball is launched from ground level at 20 m/s at an angle of 40° above the
horizontal. A) How long the ball is in the air? B)What is the maximum
height the ball can reach?
(a) The ball's height y at time t is given by
y = (20 m/s) sin(40º) t - 1/2 g t ²
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :
0 = (20 m/s) sin(40º) t - 1/2 g t ²
0 = t ((20 m/s) sin(40º) - 1/2 g t )
t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 g t
t = (40 m/s) sin(40º) / g
t ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So
0² - ((20 m/s) sin(40º))² = 2 (-g) y
where y in this equation refers to the maximum height of the ball. Solve for y :
y = ((20 m/s) sin(40º))² / (2g)
y ≈ 8.4 m
A plane flying at a speed of 59.1 m/s is dropping a package 521 m above the intended target. How long does it take for the package to hit the ground?
A.) 10.3 seconds
B.) 0.1 seconds
C.) 106.3 seconds
D.) 8.8 seconds
Answer:
t = 10.31 seconds which agrees with answer option A)
Explanation:
Notice that the horizontal velocity of the plane imparted to the package, doesn't affect the vertical motion for which the original (vertical velocity) is zero.
Then the equation for the distance travelled by the package is:
d = (1/2) a t^2
in our case, d = 521 m, and a = g (acceleration of gravity 9.8 m/s^2)
then we can solve for t in the equation:
521 = 9.8/2 t^2
t^2 = 521/4.9
t^2 = 106.32 s^2
t = 10.31 seconds
7) A moving object is rolling on a surface that is 5 m off the ground. The object is moving at a constant speed of 4 m/s. If the object is 3.2 kg, what is the final energy of the ball after rolling for 10 m, assuming friction is negligible?
156.8 J
131.2 J
182.4 J
25.6 J
8) A spreadsheet application is used to create a computational model of the energy experienced by a pendulum. How do the energy values of the pendulum relate?
The sum of the potential energy and the kinetic energy is always constant.
The sum of the potential energy and the kinetic energy is always 0.
The potential energy is always greater than the kinetic energy.
The kinetic energy is always equal to the potential energy.
131.2 J and The last one on number 8
I gave the same answer and it passed.
7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.
8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.
What is energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.
Given:
A moving object is rolling on a surface that is 5 m off the ground,
The speed of the object, v = 4 m/s,
The mass of the object, m = 3.2 kg,
Calculate the kinetic energy after 10 meters as shown below,
KE = 1/2 × 4² × 3.2
KE = 25.6 J
Calculate the potential energy as shown below,
PE = 3.2 × 9.8 × 5
PE = 156.8 J
Thus, total energy = KE + PE
The total energy = 25.6 + 156.8
The total energy = 182.4 J
8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.
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A rock rolls down a hill. Which form of energy is this an example of? (2 points)
a
Chemical
b
Electrical
c
Mechanical
d
Thermal
Answer:
c.Mechanical
Explanation: