Answer:
c
Explanation:
A 1.47-newton baseball is dropped from a height of 10.0 meters and falls through the air to the ground. The kinetic energy of the ball is 12.0 joules the instant before the ball strikes the ground. What is the maximum amount of mechanical energy converted to internal energy during the fall?
Answer:
See the explanation below.
Explanation:
First, we must determine the mass of the baseball, we know that the weight of a body is defined as the product of mass by gravity.
[tex]w=m*g[/tex]
where:
w = weight = 1.47[N]
m = mass [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
[tex]m=w/g\\m = 1.47/9.81\\m = 0.149[kg][/tex]
We now know that kinetic energy is converted to potential energy as the ball descends. By means of the following equation, we can determine the potential energy when the baseball is 10 meters high.
[tex]E_{pot}=m*g*h\\[/tex]
where:
Epot = potential energy [J]
m = mass = 0.149[kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 10 [m]
Now replacing:
[tex]E_{pot}=0.149*9.81*10\\E_{pot}=14.7[J][/tex]
In theory, this same energy must be converted to kinetic energy just before the ball hits the floor. But we see that we only have 12 [J] of kinetic energy.
That is to say, that of the 14 [J] that were had as potential energy (mechanical energy) 2 [J] was converted to internal energy, and the rest was converted to kinetic energy (mechanical energy)
[tex]E_{int}=14-12\\E_{int}=2[J][/tex]
Note: Potential, kinetic and elastic energies are forms of mechanical energy.
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PLEASE HELP!!!!! which statements correctly conpare the masses of protons,neutrons,and electrons
An AC voltage source, with a peak output of 200 V, is connected to a 50-Ohm resistor. What is the effective (or rms) current in the circuit
Answer:
2.8 A
Explanation:
We know that;
Vrms = Vo/√2
Where;
Io = peak value of voltage = 200V
Hence;
Irms = 0.707 × 200
Irms = 141.4 V
Irms = Vrms/R = 141.4/50 = 2.8 A
The sun is located in the center of the system model because the sun: __________. *
1. is the least massive object in the system, and more massive objects orbit less massive objects.
2.is the most massive object in the system, and less massive objects orbit more massive objects
3. has the lowest temperature, and hotter objects orbit cooler objects.
4. has the highest temperature, and cooler objects orbit hotter objects.
Answer:
2.is the most massive object in the system, and less massive objects orbit more massive objects
is considered as a single focus by the objects describing elliptical paths about the sun.
An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free-fall acceleration on the moon?
Answer:
the free-fall acceleration on the moon is 1.68 m/s^2
Explanation:
recall the formula for the gravitational potential energy (under acceleration of gravity "g"):
PE = m * g * h
replacing with our values for the problem:
46 J = 91 * g * 0.3
solve for the "g" on the Moon:
g = 46 / (91 * 0.3)
g = 1.68 m/s^2
(6.MS-ETS2-1(MA).) The electrons in __________ move about more freely than the electrons in insulators which is why this type of material can be used to create electric circuits.
A) conductors
B) insulators
C) magnets
Answer:
A) conductors
Explanation:
A conductor can be defined as any material or object that allows the free flow of current or electrons (charge) in one or more directions in an electrical circuit. Some examples of a conductor are metals, tungsten, copper, aluminum, iron, graphite, etc.
Basically, the main purpose of a conductor in physics is to provide a low-resistance path between electrical circuits or components. This low-resistance path is to ensure that the electrical components allows the free flow of electrons and thus, enabling charge transfer.
Hence, the electrons in conductors move about more freely than the electrons in insulators which is why this type of material can be used to create electric circuits because it would significantly provide a low-resistance path between the electric circuits.
A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exert on the table if the brick is resting on its side?
Answer:
The pressure exerted by the brick on the table is 18,933.3 N/m².
Explanation:
Given;
height of the brick, h = 0.1 m
density of the brick, ρ = 19,300 kg/m³
acceleration due to gravity, g = 9.81 m/s²
The pressure exerted by the brick on the table is calculated as;
P = ρgh
P = (19,300)(9.81)(0.1)
P = 18,933.3 N/m²
Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².
1. A race car moves along a circular track of radius 100m at a velocity of 25m/s. (a) What is
the time taken to complete one lap of the circular track. (b) What is the time taken for 10
laps.
Given :
A race car moves along a circular track of radius 100m at a velocity of 25m/s.
To Find :
(a) What is the time taken to complete one lap of the circular track.
(b) What is the time taken for 10 laps.
Solution :
Circumference of circular track,
[tex]C = 2\pi r\\\\C = 2\times \pi \times 100\ m\\\\C = 628.32 \ m[/tex]
a) Time taken to complete one lap is :
[tex]t= \dfrac{628.32}{25}\ s\\\\t =25.13 \ s[/tex]
b) Time taken to complete 10 laps is :
[tex]t_{10} = 10\times t\\\\t_{10} = 10\times 25.13\ s\\\\t_{10} = 251.3\ s[/tex]
Hence, this is the required solution.
According to the workshop, habits, attitudes, feelings and thoughts
contribute to 60% of your college success.
contribute to 30% of your college success.
contribute to 70% of your college success.
contribute to 100% of your college success.
Answer:
60%
Explanation:
what are the two conditions for work done
2 conditioner are
Explanation:
force applied and displacement produced
Two bolls with a masses m and 2m are connected by a rod of negligible mass. The rod can be rotated in several positions along the rod. To obtain the greatest angular accretion for a fix torque applied Determine the position of the torque applied to obtaim the greatest angular acceleration 2m.
Answer:
\alpha = \frac{2F}{3m} \ \frac{1}{r}
maximmun x =r
Explanation:
In this exercise we are asked for the maximum angular acceleration, let's start by writing the second law of / newton for rotational motion. Let's fix our reference system at the midpoint of the bar that has a length 2r
Σ τ = (I₁ + I₂) α
where I₁ and I₂ moment of inertia of the capsule with masses m and 2m, respectively. Let's treat these capsules as point particles
I₁ = m r²
I₂ = 2m r²
the troque of a pair of force is the force times the distance perpendicular to the point of application of the force which is the same for both forces, we will assume that the counterclockwise rotation is positive
Στ = F x + F x
the angular acceleration is the same because they are joined by the bar of negligible mass, let us substitute
2 F x = (m r² + 2m r²) α
α = [tex]\frac{2F x}{3m r^{2} }[/tex]
α = [tex]\frac{2F }{3m } \ \frac{x}{r^{2} }[/tex]
let's analyze this expression
* for the application point in the center (x = 0) at acceleration is zero
* for the point of application of the torque at the ends the acceleration is
[tex]\alpha = \frac{2F}{3m} \ \frac{1}{r}[/tex]
this being its maximum value
what is formulae method
Answer:
Explanation:
The formula method is used to calculate termination payments on a prematurely ended swap, where the terminating party compensates the losses borne by the non-terminating party due to the early termination.
The world’s largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds. What is the velocity for one of the blades? (THIS IS physics, CIRCULAR MOTION)?
Answer:
The free end of the blade has a tangential velocity of about 88.19 m/s
Explanation:
The angular velocity of the blades is [tex]2 \pi /5.7\,\,rad/sec[/tex]
since the blades are 80 m long, then the tangential velocity of the free end of the blade is:
[tex]v_{tan} \approx 88.19\,\,m/s[/tex]
camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive
The question is incomplete. Here is the complete question.
The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.
Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?
Answer: v = 6.5 m/s
Explanation: The question asks for velocity of the car. Velocity is given by:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:
Δx = 7(5.3)
Δx = 37.1 m
The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
Then, the car was driving:
[tex]v=\frac{37.1}{5.4}[/tex]
v = 6.87 m/s
The car drove at, approximately, a velocity of 6.87 m/s
The velocity of the car will be 6.5 m/s.The rate of change of displacement is defined as speed.
What is velocity?The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
t is the time for camera deliver= 1.3 frames per second
l is the length = 5.3 meters
The instantaneous velocity is given as;
[tex]\rm v = \frac{\triangle x }{\triangle t } \\\\ \rm \triangle x = 7 \times 5.3 \\\\ \rm \triangle x = 37.1 m[/tex]
The time engaged is find as;
1.3 frames = 1 s
[tex]\rm \triangle t= 7 \ frames \\\\ \rm \triangle t=5.4 sec[/tex]
Hence the velocity of the car driving;
[tex]\rm v= \frac{37.1}{5.4} \\\\ \rm v= 6.87 m/sec[/tex]
Hence the velocity of the car will be 6.5 m/s.
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URGENT
A force of 35 N is used to stretch a spring 15 cm beyond its normal length. What is the
increase in the spring's energy?
Answer:
5.25 J
Explanation:
W = PE = (f)(x)
PE = 35N*0.15m
PE = 5.25 N*m
1 N*m = 1 J
PE = 5.25 J
A lizard accelerates from 12.0 m/s to 40.0 m/s in 6 seconds. What is the lizard’s average acceleration?
Answer:
[tex]\boxed {\boxed {\sf \frac{14}{3} \ or \ 4.6667 \ m/s^2}}[/tex]
Explanation:
Acceleration can be found using the following formula.
[tex]a=\frac{v_f-v_1}{t}[/tex]
where [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity and t is the time.
The lizard started at 12.0 m/s and accelerated up to its final velocity of 40.0 m/s in 6 seconds.
Therefore:
[tex]v_f= 40.0 \ m/s \\v_i= 12.0 \ m/s \\t= 6 \ s[/tex]
Substitute the variables into the formula.
[tex]a=\frac{40.0 \ m/s - 12.0 \ m/s}{6 \ s}[/tex]
Solve the numerator first and subtract.
40.0 m/s - 12.0 m/s= 28 m/s[tex]a=\frac{ 28 \ m/s}{6 \ s}[/tex]
Divide.
[tex]a= \frac{14}{3} \ m/s/s= \frac{14}{3} \ m/s^2[/tex]
[tex]a=4.66667 \ m/s^2[/tex]
The lizard's average acceleration is 14/3 or 4.66667 m/s²
Answer:
4 2/3 m/s
Explanation:
first thing to find the average acceleration is to figure out what the increase in speed was, we can do that by subtracting the original speed from the speed after accelerating, that looks like:
40 - 12 = 28
so the lizard accelerated 28 m/s in 6 seconds, to find the average increase in m/s every second, we divide the m/s by the seconds, which gives us:
28 / 6 = 4.66 = 4 2/3 m/s
8. What are the two types of mechanical waves?
Otransverse, electromagnetic
O longitudinal, infrared
Otransverse, longitudinal
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Answer:
the blue one
Explanation:
A boy on a swing set has a speed of 4.5 m/s and a centripetal acceleration of 8.1 m/s2 at the bottom of his swing. How long are the ropes of the swing?
Pls help!!!!
Answer:
R = 1.8 m
Explanation:
This is a simple harmonic movement exercise, at the bottom of the swing the acceleration is vertical upwards and the speed is tangential to the trajectory, that is horizontal; the expression for the centralized acceleration is
[tex]a_{c}[/tex] = v² / R
R = v² /a_{c}
where the radius is equal to the length of the swing
let's calculate
R = 8.1 / 4.5
R = 1.8 m
Grizzly bears are known to run really fast. What is the momentum of a grizzly bear (about 420 pounds) running at 17 m/s ?
How fast should a 250 lb human run to match the momentum of the grizzly bear ?
Answer:
A. Momentum of the grizzly bear is 3241.56 Kgm/s
B. The human will run at 28.56 m/s
Explanation:
A. Determination of the momentum of the grizzly bear.
Mass of grizzly bear = 420 lb
Velocity of grizzly bear = 17 m/s
momentum of the grizzly bear =?
Next, we shall convert 420 lb to Kg. This can be obtained as follow:
1 lb = 0.454 kg
Therefore,
420 lb = 420 lb × 0.454 Kg / 1 lb
420 lb = 190.68 Kg
Thus, 420 lb is equivalent to 190.68 Kg
Finally, we shall determine the momentum of the grizzly bear as follow:
Mass of grizzly bear = 190.68 Kg
Velocity of grizzly bear = 17 m/s
momentum of the grizzly bear =?
Momentum = mass × velocity
Momentum = 190.68 × 17
Momentum = 3241.56 Kgm/s
Thus, the momentum of the grizzly bear is 3241.56 Kgm/s
B. Determination of how fast the human will run.
Mass of human = 250 lb
Momentum of human = momentum of the grizzly bear
Momentum of the grizzly bear is 3241.56 Kgm/s
Momentum of human = 3241.56 Kgm/s
Velocity of human =?
Next, we shall convert 250 lb to Kg. This can be obtained as follow:
1 lb = 0.454 kg
Therefore,
250 lb = 250 lb × 0.454 Kg / 1 lb
250 lb = 113.5 Kg
Thus, 250 lb is equivalent to 113.5 Kg
Finally, we shall determine how fast the human will run as follow:
Mass of human = 113.5 Kg
Momentum of human = 3241.56 Kgm/s
Velocity of human =?
Momentum = mass × velocity
3241.56 = 113.5 × velocity
Divide both side by 113.5
Velocity = 3241.56 / 113.5
Velocity = 28.56 m/s
Therefore, the human will run at 28.56 m/s to match the momentum of the grizzly bear
A pulley system consists of two fixed pulleys and two movable pulleys that lift a load that has a weight of 300 N. If the effort force used to lift the load is 100 N, what is the mechanical advantage of the system
Answer:
Mechanical advantage = 3
Explanation:
Given:
Weight = 300 N
Force load = 100 N
Find:
Mechanical advantage:
Computation:
Mechanical advantage = Weight/Force Load
Mechanical advantage = 300/100
Mechanical advantage = 3
Mechanical advantage is the ratio of output force to the input force.It is also used to find the efficiency of the system.The value of mechanical advantage is 3.
What is mechanical advantage?Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of forces in levers and pulley.
It is an effective way of amplify the force in simple machines like lever The theoretical mechanical advantage is defined as ratio of the force responsible for he useful work in system to the applied force.
The theoretical value may be grater than the actual value of mechanical advantage because in the theoretical mechanical advantage friction is assumed to be absent.
The given data in the problem is;
weight that lifted (output)=300
Effort force(input) =100
Mathematically it is given as;
[tex]\rm M A= \frac{F_O}{F_I} \\\\\rm M A= \frac{300}{100}\\\\\rm M A=3[/tex]
Hence the value of mechanical advantage is 3.
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while snowboarding mr. brooks (100 kg) takes a chair lift up to 800 m to the top of the mountain. if he snowboards down the mountain without stopping of falling, how fast will he be going when he get back to the bottom of the mountain
Given that,
Mass of Mr. Brooks = 100 kg
A chair lift up to 800 m to the top of the mountain.
To find,
How fast will he be going when he get back to the bottom of the mountain.
Solution,
Let v is the speed when he get back to the bottom of the mountain. Using the conservation of energy to find it.
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 800} \\\\v=125.21\ m/s[/tex]
So, the required speed is 125.21 m/s.
A massive truck of 1200N moving with a velocity of 2m/s hits a stationary mass of 30N. if the both bodies move together after the collision, determine their common velocity.
Answer:
The common speed is 1.95 m/s
Explanation:
Law Of Conservation Of Linear Momentum
It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of all of them:
[tex]P=m_1v_1+m_2v_2+...+m_nv_n[/tex]
If a collision occurs, the velocities change to v' and the final momentum is:
[tex]P'=m_1v'_1+m_2v'_2+...+m_nv'_n[/tex]
In a system of two masses, the law of conservation of linear momentum
is written as:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
If both masses stick together after the collision at a common speed v', then:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v'[/tex]
The common velocity after this situation is:
[tex]\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
The truck of m1=1200 N (weight) travels at v1=2 m/s and hits a stationary mass (v2=0) of m2=30 N (weight). After the bodies collide, they keep moving together. Before we can calculate the common speed, we need to calculate the masses of the bodies, since they are given as weights.
[tex]m_1=\frac{P_1}{g}=\frac{1200}{9.8}=122.45 Kg[/tex]
[tex]m_2=\frac{P_2}{g}=\frac{30}{9.8}=3.06 Kg[/tex]
Now calculate the common speed:
[tex]\displaystyle v'=\frac{122.45 * 2+3.06 * 0}{122.45+3.06}[/tex]
[tex]\displaystyle v'=\frac{244.9}{125.51}=1.95\ m/s[/tex]
The common speed is 1.95 m/s
A 100 kg cart is moving at 3 m/s. Calculate the cart’s kinetic energy.
Answer:
450
Explanation:
Given,
Mass= 100kg
Velocity= 3 m/s
Kinetic Energy= ?
Kinetic Energy= 1/2 mv^2
= 1/2× 100× 3^2
= 1/2× 900
= 450.
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The shape of Sugarloaf mountain, in Rio de Janeiro, Brazil, is such that, if you
were to kick a soccer ball hard enough, it could land near the base of the mountain
without hitting the mountain's side. Suppose the ball is kicked horizontally with an
initial speed of 9:37 m/s. If the ball travels a horizontal distance of 85.0 m, how tall
is the mountain?
Answer:
The mountain is 403 m tall
Explanation:
Horizontal Launch
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it hits the ground.
The range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
It's given that if a ball is kicked from the top of the Sugarloaf Mountain, it will land near its base without hitting the side. The initial speed is v=9.37 m/s and it reaches a horizontal distance of d=85 m. Solving the equation for h:
[tex]\displaystyle h=\frac{d^2g}{2v^2}[/tex]
Substituting:
[tex]\displaystyle h=\frac{85^2*9.8}{2*9.37^2}[/tex]
[tex]\displaystyle h=403 \ m[/tex]
The mountain is 403 m tall
a body is under going non uniform circular motion work done by tangential force on body is
Answer:
The tangential force will act as a torque on the body, increasing its angular velocity and thus also increasing its kinetic energy. By the work-kinetic-energy theorem, work has been done on the body. Yes, in non-uniform circular motion the work done on the object is non-zero, for the reason you stated.
Explanation:
Create an energy path in the Gizmo, starting at the Sun. For each step of the path, describe the energy conversion that takes place. The first one is done for you. Discuss your answers with your classmates and teacher.
Explanation:
Sun— Nuclear energy is converted to heat and light energy
Corn absorbs sunlight in order to produce sugars and oxygen for plant growth.
Corn is fermented to produce ethanol, which is burned to power generators to give an electrical current The speaker takes the electrical current and produces sound.
Answer:
sun, corn, corn, speaker
Explanation:
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Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7 r, respectively What is the ratio of their orbital speeds
Answer:
The ratio of their orbital speeds are 5:4.
Explanation:
Given that,
Mass of A = 5 m
Mass of B = 7 m
Radius of A = 4 r
Radius of B = 7 r
The orbital speed of satellite A,
[tex]v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}[/tex]......(I)
The orbital speed of satellite B,
[tex]v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}[/tex]......(I)
We need to calculate the ratio of their orbital speeds
Using equation (I) and (II)
[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}[/tex]
Put the value into the formula
[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}[/tex]
[tex]\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}[/tex]
Hence, The ratio of their orbital speeds are 5:4.
which of the following is an application of of the technology developed by Charles kuen kao
A. Producing shock-resistant cameras
B. Predicting earthquakes
C. Transmitting digital information
D. Developing safer spaceships