Explanation:
Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.
This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.
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.
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
Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object. Suppose a metal bar, initially at temperature 50 degrees Celsius, is placed in a room which is held at the constant temperature of 40 degrees Celsius. One minute later the bar has cooled to 40.18316 degrees . Write the differential equation that models the temperature in the bar (in degrees Celsius) as a function of time (in minutes). Hint: You will need to find the constant of proportionality. Start by calling the constant k and solving the initial value problem to obtain the temperature as a function of k and t . Then use the observed temperature after one minute to solve for k .
Answer:
Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),
Explanation:
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:
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.
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
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
I NEED THE ANSWER QUICK PLEASEE
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:
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].
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]
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 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
The diagram shows the molecular structure of ethane. What is the chemical
formula for ethane?
Ethane
H H
H-C-C-H
| |
H H
galileo was a contemporary of
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.
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
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.
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]
Need an answer in hurry u can make the pic big
A toy car of mass 600g moves through 6m in 2 seconds. The average kinetic energy of the toy car is
Answer:
12
Explanation:
I'm a beginner so am not sureeeeee
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
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the condenser. Suppose you have a discharging condenser and the instantaneous rate of change of the voltage is -0.01 of the voltage (in volts per second). How many seconds does it take for the voltage to decrease by 90 %?
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = [tex]e^{kt}[/tex]
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = [tex]e^{-0.01t}[/tex]
V = V₀[tex]e^{-0.01t}[/tex]
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀[tex]e^{-0.01t}[/tex]
0.1V₀ = V₀[tex]e^{-0.01t}[/tex]
0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]
0.1 = [tex]e^{-0.01t}[/tex]
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
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.
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?
Answer:
The angle is 4.1 rad.
Explanation:
The centripetal acceleration (α) is given by:
[tex] \alpha = \omega^{2} r [/tex] (1)
Where:
ω: is the angular velocity
r: is the radius
And the tangential acceleration (a) is:
[tex] a = \alpha r [/tex] (2)
Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:
[tex] \omega^{2} r = 8.2\alpha r [/tex]
[tex] \omega^{2} = 8.2\alpha [/tex] (3)
Now, we can find the angle with the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)
[tex]\Delta \theta[/tex]: is the angle
[tex] \omega^{2} = 2\alpha \Delta \theta [/tex] (4)
By entering equation (3) into (4) we can calculate the angle:
[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]
[tex] \Delta \theta = 4.1 rad [/tex]
Therefore, the angle is 4.1 rad.
I hope it helps you!
In air an object weighs 15N, when immersed in water it weighs 12N, when immersed in another liquid, it weighs 13N, Calculate the density of the object and that of the other liquid?
M1 = 15/g = 15/9.8 = 1.53 kg = mass of object in air. M2 = 12/9.8 = 1.22 kg = mass of object immersed. M1-M2 = 1.53-1.22 = 0.31 kg lost by object = mass of water displaced. ... Do = 4.94 g/cm^3 = density of object.
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
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
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 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
