Answer:
A
Explanation:
weight of the tree =normal force
Horizontal force =coefficient of friction x Fnormal
0.5×430=215
A 7.5-kg rock and a 8.9 × 10-4-kg pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.
Answer:
F' = 73.7 N
F = 8.749×10⁻³ N
a' = a = 9.83 m/s²
Explanation:
(a)
For the rock
Applying
F' = Gm'm/r²................... Equation 1
Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.
From the question,
Given: m' = 7.5 kg
Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m
Substitute these values into equation 1
F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²
F' = 7.37×10¹ N
F' = 73.7 N
Also, For the pebble,
F = GMm/r².............. Equation 2
Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth
Given: M = 8.9×10⁻⁴ kg,
Substitute into equation 2
F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²
F = 8.749×10⁻³ N
(b)
For the rock,
a' = F'/m'
Where a' = magnitude of the acceleration of the rock
a' = 73.7/7.5
a' = 9.83 m/s²
For the pebble,
a = F/M
Where a = acceleration of the pebble
a = (8.749×10⁻³)/(8.9×10⁻⁴)
a = 9.83 m/s²
1. Lifting an elevator 18m takes 100kJ. If doing so takes 20s, what is the average power of the elevator during the process?
2. How much work can a 0.4 hp electric mixer do in 15 s?
Answer:
1. Power = 5000 Watts
2. Workdone = 11185.5 Joules
Explanation:
Given the following data;
1. Distance = 18 m
Energy = 100 KJ = 100,000 Joules
Time = 20 seconds
To find the average power of the elevator;
Power = energy/time
Power = 100000/20
Power = 5000 Watts
2. Power = 0.4 HP
Time = 15 seconds
Conversion:
1 horsepower = 745.7 Watts
0.4 horsepower = 0.4 * 745.7 = 298.28 Watts
To find the amount of work done by the electric mixer;
Work done = power * time
Workdone = 745.7 * 15
Workdone = 11185.5 Joules
The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00\times 10^32.00×10 ^3 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s^2 .
What is the moment of inertia of the boxer's forearm?
Answer:
Explanation:
From the given information:
The torque produced due to the force can be expressed as:
[tex]\tau = F \times r[/tex]
where;
[tex]\tau[/tex] = torque
F = force exerted
r = lever's arm radius
[tex]\tau[/tex] = [tex]2.00 \times 10^3 \times 0.03 m[/tex]
[tex]\tau[/tex] = 60 N.m
However, equating the torque with the moment of inertia & angular acceleration, we use the equation:
[tex]\tau[/tex] = I∝
60 Nm = I × 120 rad/s²
I = 60 Nm/120 rad/s²
I = 0.5 kg.m²
How can a wire become magnetic?
add a resistor
point it north
heat it up
run a current through it
Answer:
Moving electrons always create a magnetic field. Electrons moving along a wire make a magnetic field that goes in circles around the wire. When you bend the wire into a coil, the magnetic fields around each loop of the coil add up to make a long , thin magnet with north at one end and south at the other.
Explanation:
PLEASE HELP MEE THIS IS DUE IN 45 MINS
Answer:
The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
Explanation:
This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:
[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)
Where:
[tex]m[/tex] - Mass of the car, in kilogram.
[tex]v[/tex] - Initial velocity, in meters per second.
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]s[/tex] - Travelled distance, in meters.
Then we derive an expression for the distance travelled by the vehicle:
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]
[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]
As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.
Answer:
The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.
Explanation:
Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
Which of the following scenerios fits all of the criteria for the two-source interference equations to be valid?
a. An observer is standing far away from two red LED signal lights.
b. Light from an incandescent bulb shines onto a screen with a single slit; then the light shines onto a screen with two slits in it and the light from the two slits finally shines onto a far-away screen.
c. An observer stands on a road far away from two neighboring radio towers for different radio stations.
d. Light from an incandescent bulb shines onto a screen with a single slit; then the light shines onto a screen with two slits in it and the light from the two slits finally shines onto a nearby screen.
e. An observer stands on a road that runs five kilometers away from the two synchronized transmitting towers for a radio station.
Answer:
answer the correct is B
Explanation:
For the interference phenomenon to occur, some conditions must be met.
* You must have a light in phase and coherent, for this you can for a light from an incandescent source through a single slit, the light that comes out is coherent
* This light must strike two slits
* the light that passes through the slits must hit a distant screen and be able to see the phenomenon of constructive interference
when examining the different answer the correct one is B
For the criteria for the two-source interference, Option B is correct.
What is two-source interference?Wave interference happens when two waves collide while traveling through the same medium.
Wave interference causes the medium to take on a form determined by the total effect of the two distinct waves on the medium's particles.
The scenarios that fit all of the criteria for the two-source interference equations will be light from an incandescent bulb shining onto a screen with a single slit.
Then the light shines onto a screen with two slits in it and the light from the two slits finally shines onto a far-away screen.
Hence for the criteria for the two-source interference, Option B is correct.
To learn more about the two-source interference refer to the link;
https://brainly.com/question/14420120
45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
the acceleration due to gravity is 9.8 m/s2, what is the crate's potential energy
at this point?
Answer:
[tex]\boxed {\boxed {\sf 29,400 \ Joules}}[/tex]
Explanation:
Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.
[tex]E_P= m \times g \times h[/tex]
The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.
m= 150 kg g= 9.8 m/s²h= 20 mSubstitute the values into the formula.
[tex]E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m[/tex]
Multiply the three numbers and their units together.
[tex]E_p=1470 \ kg*m/s^2 \times 20 m[/tex]
[tex]E_p=29400 \ kg*m^2/s^2[/tex]
Convert the units.
1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.
[tex]E_p= 29,400 \ J[/tex]
The crate has 29,400 Joules of potential energy.
Answer:
29,400 J
Explanation:
did the quiz <3
As part of a safety investigation, two 1300 kg cars traveling at 17 m/s are crashed into different barriers. Find the average forces exerted on:
a. the car that hits a line of water barrels and takes 1.5 s to stop
b. the car that hits a concrete barrier and takes 0.10 s to stop.
Answer:
a. F = 14,733.33 N
b. F = 221,000 N
Explanation:
Given;
mass of the cars, m = 1300 kg
velocity of the cars, v = 17 m/s
time taken for the first car to stop after hitting a barrier, t = 1.5 s
time taken for the second car to stop after hitting a barrier, t = 0.1 s
The average forces exerted on each car is calculated as follows;
a. the car that hits a line of water barrels and takes 1.5 s to stop
[tex]F = ma = m\times \frac{v}{t} = 1300 \times \frac{17}{1.5} = 14,733.33 \ N\\\\F = 14,733.33 \ N[/tex]
b. the car that hits a concrete barrier and takes 0.10 s to stop
[tex]F = ma = m\times \frac{v}{t}= 1300 \times \frac{17}{0.1} = 221,000 \ N\\\\F = 221,000 \ N[/tex]
Study the position-time graph for a bicycle. Which statement is supported by the graph? Position vs Time O The bicycle has speed but not velocity. O The bicycle is moving at a constant velocity. O The bicycle has a displacement of 3 m. O The bicycle is not in motion. 3 Position (m) 0 1 2 3 4 5 Time (s) Next Submit Save and Exit Mark this and return tViewers/AssessmentViewer/Activit. 0 M M
Answer:
D) The bicycle is not in motion.
Explanation:
Study the position-time graph for a bicycle.
Which statement is supported by the graph?
A) The bicycle has speed but not velocity.
B) The bicycle is moving at a constant velocity.
C) The bicycle has a displacement of 3 m.
D) The bicycle is not in motion.
Solution:
Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.
Speed is the time rate of change of distance. It is the ratio of distance to time taken.
From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.
From the position-time data provided, it can concluded that the bicycle is not in motion.
MotionMotion of a body involves a change in the position of that body with time.
A body in motion is constantly changing position or orientation as time passes.
The body may move with constant velocity/speed or changes in its velocity.
A position-time graph provides information about the motion of a body.
From the data provided:
At time 0, the bicycle is at position 3At time 1, the bicycle is at position 3At time 2, the bicycle is at position 3At time 3, the bicycle is at position 3At time 4, the bicycle is at position 3At time 5, the bicycle is at position 3The position of the bicycle remains the same for all time intervals.
Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.
Learn more about motion and position-time graph at: https://brainly.com/question/2356782
The hottest ordinary star in our galaxy has a surface temperature of 53,000 K. Part A What is the peak wavelength of its thermal radiation
Answer:
[tex]\lambda=5.46\times 10^{-8}\ m[/tex]
Explanation:
The hottest ordinary star in our galaxy has a surface temperature of 53,000 K.
We need to find the peak wavelength of its thermal radiation.
Using Wein's law,
[tex]\lambda T=2.898\times 10^{-3}\\\\\lambda=\dfrac{2.898\times 10^{-3}}{53000}\\\\=5.46\times 10^{-8}\ m[/tex]
So, the peak wavelength of its thermal radiation is equal to [tex]5.46\times 10^{-8}\ m[/tex].
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration
Answer:
i) the minimum acceleration to take off is 22500 km/h²
ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) required force that the engine must exert to attain acceleration is 625 kN
Explanation:
Given the data in the question;
mass of plane m = 360,000 kg
take of speed v = 300 km/hr = 83.33 m/s
i)
What should be the minimum acceleration to take off if the length of the runway is 2.00 km
from Newton's equation of motion;
v² = u² + 2as
we know that a plane starts from rest, so; u = 0
given that distance S = 2 km
we substitute
(300)² = 0² + ( 2 × a × 2 )
90000 = 4 × a
a = 90000 / 4
a = 22500 km/h²
Therefore, the minimum acceleration to take off is 22500 km/h²
ii) At this acceleration, how much time would the plane need from starting to takeoff.
from Newton's equation of motion;
v = u + at
we substitute
300 = 0 + 22500 × t
t = 300 / 22500
t = 0.0133 hrs
Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) What force must the engines exert to attain this acceleration
we know that;
F = ma
acceleration a = 22500 km/hr² = 1.736 m/s²
so we substitute
F = 360,000 kg × 1.736 m/s²
F = 624960 N
F = 625 kN
Therefore, required force that the engine must exert to attain acceleration is 625 kN
What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
A wave has a frequency of 87.00 Hz and has a wavelength of 74.62 m. What is its
velocity?
Answer:
v = 6491.94 m/s
Explanation:
We are given;
Frequency; f = 87 Hz
Wavelength;λ = 74.62 m
Formula for velocity(v) of waves from the wave equation is;
v = fλ
Thus;
v = 87 × 74.62
v = 6491.94 m/s
A rod of 3.0-m length and a square (2.0 mm X 2.0 mm) cross section is made of a material with a resistivitYof 6.0 X 10-8 Ω m. If a potential difference of 0.60 V is placed across the ends of the rod, at what rate is heat generated in the rod in watt.
Select one:
a. 24
b. 12
c. 4
d. 8
What does it mean when work is positive?
O Velocity is greater than kinetic energy
O Kinetic energy is greater than velocity
The environment did work on an object.
O An object did work on the environment.
d. An object did work on the environment.
Explanation:Work is defined in many contexts. Some of these are;
i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.
ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.
In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).
Answer:
its c
Explanation:
what is newtons 2nd law
According to the Newton's second law :- The acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.
Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons
Answer:
No of proton is 13 and nucleus is 13
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?
Answer:
W = 16.4 kJ
Explanation:
Given that,
There are 135 steps from the ground floor to the sixth floor.
Each step is 16.6 cm tall.
The mass of a person, m = 73.5 kg
We need to find the work done by the person. We know that,
Work done = Fd
Where
d is the displacement, d = 135 × 0.166 = 22.41
So,
W = 73.5 × 10 × 22.41
= 16471.35 J
or
W = 16.4 kJ
So, 16.4 kJ is the work done by the person.
A small object with mass 0.200 kg moves with constant speed in a vertical circle of radius 0.500 m. It takes the object 0.500 s to complete one revolution. (a) What is the translational speed of the object
Answer:
6.28 m/s.
Explanation:
Given that,
The mass of the object, m = 0.2 kg
The radius of the circle, r = 0.5 m
It takes the object 0.500 s to complete one revolution.
We need to find the translational speed of the object. Let it is v. We know that,
[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]
So, the transalational speed of the object is 6.28 m/s.
How long does take for a freely falling object to reach 4.0 m/s
Answer :
Considering initial velocity is 0,
It takes about 0.4 seconds.
Use the equation v-u =at
4-0 = 9.8×t
t = 4/9.8 = 0.4 seconds approximately.
Need an answer in hurry u can make the pic big
A photon with a frequency of 5.02 × 1014 hertz is absorbed by an excited hydrogen atom. This causes the electron to be ejected from the atom, forming an ion. Calculate the energy of this photon in joules. [Show all work, including the equation and substitution with units.] Determine the energy of this photon in electron-volts. What is the number of the lowest energy level (closest to the ground state) of a hydrogen atom that contains an electron that would be ejected by the absorption of this photon?
Answer:
Explanation:
An atom emits a photon (particle of light) when transitioning from a ground state to its excited state. To obey conservation of energy, the energy gained by the atom when an electron moves to a lower energy level is equal to the energy it loses in emitting the photon. (The energy of a photon is E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the photon.) Conversely, when an atom absorbs a photon (as is the case in absorption spectra), the electron absorbing the photon moves to a higher energy level.
I NEED THE ANSWER QUICK PLEASEE
A vertical wall (8.7 m x 3.2 m) in a house faces due east. A uniform electric field has a magnitude of 210 N/C. This field is parallel to the ground and points 42o north of east. What is the electric flux through the wall
Answer:
[tex]\phi=4344.72Nm^2/c[/tex]
Explanation:
From the question we are told that:
Dimension of Wall:
[tex](L*B)=(8.7 m * 3.2 m)[/tex]
Electric field [tex]B=210 N/C[/tex]
Angle [tex]\theta =42 \textdegree North[/tex]
Generally the equation for electric Flux is mathematically given by
[tex]\phi=EAcos\theta[/tex]
[tex]\phi=210*(8.7*3.2)*cos 42[/tex]
[tex]\phi=4344.72Nm^2/c[/tex]
The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?
Answer:
[tex]I_2=0.50 w/m^2[/tex]
Explanation:
From the question we are told that:
initial Intensity [tex]I_1=0.020 w/m^2[/tex]
Final Electric field [tex]E_2=5E[/tex]
Generally the equation for Relation ship between intensity and Electric field is mathematically given by
[tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]
Therefore
[tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]
[tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]
[tex]I_2=0.50 w/m^2[/tex]
You are using a constant force to speed up a toy car from an initial speed of 6.5 m/s
to a final speed of 22.9 m/s. If the toy car has a mass of 340 g, what is the work
needed to speed this car up?
By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²
W ≈ 82 J
When a mass of 3.0-kg is hung on a vertical spring, it stretches by 0.085 m. Determine
the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
the period of oscillation of the given object is 0.14 s
Explanation:
Given;
mass of the object, m = 3 kg
extension of the spring, x = 0.085 m
The spring constant is calculated as follows;
[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]
The angular speed of a 4 kg object is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]
Therefore, the period of oscillation of the given object is 0.14 s
