a tiger leaps with an initial velocity of 55 km/hr at an angle of 13° with respect to the horizontal. what are the components of the tigers velocity?
Answer:
vₓ = 53.6 km/h
vy = 12.4 km/h
Explanation:
if we define two axis perpendicular each other with origin in the point represented by the tiger leaping (assuming we can treat it as a point mass) coincidently with the horizontal (x-axis) and vertical (y-axis) directions, we can obtain the components of the velocity in both independent directions.We can do it simply getting the projections of the velocity vector on both axes, using simple trigonometry, as follows:[tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]
[tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]
A Labrador retriever runs 50 m in 7.2 s to retrieve a toy bird. The dog then runs half way
back in 3.85 s. Determine the average speed and velocity of the dog
Answer:
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explanation:
From Physics we must remember the definitions of average speed and average velocity, both measured in meters per second. Velocity is a vectorial quantity, that is, it has both magnitude and direction, whereas speed is an scalar quantity, which is a quantity that is represented solely by its magnitude. We assume that dog moves at constant speed.
For the case of the dog, we get that average speed and average velocity of the animal are, respectively:
Average velocity:
[tex]\vec v_{avg} = \frac{1}{\Delta t}\cdot (\vec r_{B}-\vec r_{A})[/tex] (Eq. 1)
Where:
[tex]\Delta t[/tex] - Travelling time of the dog, measured in seconds.
[tex]\vec r_{A}[/tex] - Initial vector position of the dog, measured in meters.
[tex]\vec r_{B}[/tex] - Final vector position of the dog, measured in meters.
Average speed:
[tex]v_{avg} = \frac{1}{\Delta t} \cdot (s_{A}+s_{B})[/tex] (Eq. 2)
Where [tex]s_{A}[/tex] and [tex]s_{B}[/tex] are the travelled distances of each stage, measured in meters.
If we know that [tex]\Delta t = 11.05\,s[/tex], [tex]\vec r_{A} = 0\,\hat{i}\,\,\,[m][/tex] and [tex]\vec r_{B} = 25\,\hat{i}\,\,\,[m][/tex], [tex]s_{A} = 50\,m[/tex] and [tex]s_{B} = 25\,m[/tex], average velocity and average speed are, respectively:
[tex]\vec v_{avg} = \frac{1}{11.05\,s}\cdot (25\,\hat{i})\,\,\,[m][/tex]
[tex]\vec v_{avg} = 2.262\,\hat{i}\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]v_{avg} = \frac{75\,m}{11.05\,s}[/tex]
[tex]v_{avg} = 6.787\,\frac{m}{s}[/tex]
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are not quantized. When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. The energies of atoms are quantized.
Answer:
Explanation:
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. FALSE. The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.
The energies of atoms are not quantized. FALSE. The energies of the atoms are in specific levels.
When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. FALSE. During absorption, a specific wavelength of light is absorbed, not emmited.
Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. TRUE. Again, you can observe just the transition due the change of energy of an electron in the quantized energy level
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. TRUE. The electron decreases its energy releasing a specific wavelength of light.
The energies of atoms are quantized. TRUE. In fact, the energy of all subatomic, atomic, and molecular particles is quantized.
The reason why atoms emit only specific wavelengths is because the energy levels in atoms are quantized.
Max Plank introduced the idea of quantization of energy in the early 1900s. He introduced the idea that energy can only take on certain specific values. This idea was later extended to atoms by Neils Bohr.
The following statements explain why atoms only emit certain wavelengths of light when they are excited;
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are quantized.Learn more: https://brainly.com/question/24381583
What is the direction of the magnetic field if an electron moving in the positive x direction experiences a magnetic force in the positive z direction
Given :
An electron moving in the positive x direction experiences a magnetic force in the positive z direction.
To Find :
The direction of the magnetic field.
Solution :
We know, force is given by :
[tex]\vec{F}=q(\vec{v}\times \vec{B)}[/tex]
Here, q = -e.
[tex]\vec{F}=(-e)(\vec{v}\times \vec{B)}\\\\\hat{k}=(-e)(\hat{i}\times \vec{B})[/tex]
Now, for above condition to satisfy :
[tex]\hat{i}\times \vec{B}=-\hat{k}[/tex]
So, [tex]\vec{B}=-\hat{j}[/tex]
Therefore, direction of magnetic field is negative y direction.
Hence, this is the required solution.
Which statements about potential energy are true?
▫ Gaining potential energy is always associated with a force field.
▫ A change in position always means that an object gains potential energy.
▫ There's only one kind of potential energy.
▫ Some kinds of potential energy are related to electric forces exerted by atoms and molecules.
Answer:
the answer is 1 and 4
Explanation
Plato users
For the potential energy, statement 1 and statement 4 are correct.
The potential energy of the object the energy of the object in its steady position. When the object is at rest, the energy of the object in that condition is called potential energy.
Let us consider an electron having charge [tex]e[/tex] is moving the distance [tex]d[/tex] in uniform electric field E.
Its potential energy can be written as,
[tex]P = eEd[/tex]
Where P is the potential energy and E is the electric field.
Hence, the potential energy of the electron is associated with the electric field.
The electric force can be written as,
[tex]F =eE[/tex]
Where [tex]F[/tex] is the electric force, [tex]E[/tex] is the electric field and [tex]e[/tex] is the charge on the electron.
So, the potential energy can be written as,
[tex]P=Fd[/tex]
Hence, the potential energy is related to electric force.
For more information, follow the link given below.
https://brainly.com/question/1413008.
A mass (m = 30 g) falls onto a spring (k = 7.3 N/m) from a height (h = 25 cm). The spring compresses an additional amount x before temporarily coming to a stop. What is the value of x?
Answer:
x₁ = 0.1878 m
Explanation:
For this exercise we will use conservation of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point with fully compressed spring
Em_f = K_e + U
Em_f = ½ K x² + m g x
energy is conserved
Em₀ = Em_f
m g h = ½ K x² + m g x
½ K x² + mg (x- h) = 0
let's substitute
½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0
3.65 x² + 0.294 (x- 0.25) = 0
x² + 0.080548 (x- 0.25) = 0
x² - 0.020137 + 0.080548 x = 0
x² + 0.080548 x - 0.020137 = 0
let's solve the quadratic equation
x = [0.080548 ±√ (0.080548² + 4 0.020137)] / 2
x = [0.080548 ± 0.29502] / 2
x₁ = 0.1878 m
x₂ = -0.1072 m
These are the compression and extension displacement of the spring
Complete each statement by dragging the forms of energy into their appropriate boxes.
wind turbine
roller coaster going downhill
toaster
car
A
converts electrical energy into thermal energy.
A
converts rotational energy into electrical energy.
A
converts gravitational energy into mechanical energy.
A
converts rotational energy into mechanical energy.
Statements 1,2,3 and 4 match statements B, C, A, and D respectively.A wind turbine converts rotational energy into electrical energy.
What is the law of conservation of energy?According to the Law of conservation of energy. Energy can not be created nor be destroyed, it can transfer from one to another form.
1.A wind turbine converts rotational energy into electrical energy.
2.A roller coaster going downhill converts gravitational energy into mechanical energy
3. Toaster converts electrical energy into thermal energy
4.A car converts rotational energy into mechanical energy.
Hence,statements 1,2,3 and 4 match statements B, C, A, and D respectively.
To learn more about the law of conservation of energy refer:
https://brainly.com/question/20971995
#SPJ1
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same. Describe how they are different.
Answer:
he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Explanation:
In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.
In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.
From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
What are two ways that an object can have kinetic energy?
Answer:
The object has to have mass and speed
Explanation:
You can increase both speed and mass to increase the kinetic energy, hope this answers your question.
Happy Halloween!
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.8 s. The readout on the display indicates that the average power you are producing is 90 W. What is the magnitude of the force that you exert on the handle?
Answer:
147.27N
Explanation:
Power = workdone/time
Power = Force*distance/time
Given
Power = 90Watts
Distance = 1.1m
Time = 1.8secs
Force = ?
Substitute the given parameters into the formula:
[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]
Hence the magnitude of the force that you exert on the handle is 147.27N
If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.
A 6.13-g bullet is moving horizontally with a velocity of 361 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1233 g, and its velocity is 0.741 m/s after the bullet passes through it. The mass of the second block is 1646 g. (a) What is the velocity of the second block after the bullet imbeds itself
Answer:
v₃ = 0.786 m/s
Explanation:
Here, we will use the law of conservation of momentum, which states the following:
Total Momentum of System Before Collision =
Total Momentum of System After Collision
m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃
where,
m₁ = mass of bullet = 6.13 g = 0.00613 kg
m₂ = mass of 1st block = 1233 g = 1.233 kg
m₃ = mass of 2nd block = 1646 g = 1.646 kg
u₁ = speed of first bullet before collision = 361 m/s
u₂ = speed of first block before collision = 0 m/s
u₃ = speed of 2nd block before collision = 0 m/s
v₁ = speed of bullet after collision
v₂ = speed of 1st block after collision = 0.741 m/s
v₃ = speed of 2nd block after collision = ?
Therefore,
(0.00613 kg)(361 m/s) + (1.233 kg)(0 m/s) + (1.646 kg)(0 m/s) = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)
2.2129 kg m/s + 0 kg m/s + 0 kg m/s - 0.9136 kg m/s = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)
1.2992 kg m/s = (0.00613 kg)(v₁) + (1.646 kg)(v₃)
since, the bullet is embedded in 2nd block after collision. Thus, there velocities will become same. (v₁ = v₃)
Therefore,
1.2992 kg m/s = (0.00613 kg)(v₃) + (1.646 kg)(v₃)
v₃ = (1.2992 kg m/s)/(1.6521 kg)
v₃ = 0.786 m/s
If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer
Answer:
The charge pass through your hair dryer is 3000 C.
Explanation:
Given that,
Power = 1200 W
Voltage = 120 V
Flow time = 5 min
We need to calculate the current
Using formula of power
[tex]P=VI[/tex]
[tex]I=\dfrac{P}{V}[/tex]
Put the value into the formula
[tex]I=\dfrac{1200}{120}[/tex]
[tex]I=10\ A[/tex]
We need to calculate the charge pass through your hair dryer
Using formula of current
[tex]I=\dfrac{Q}{t}[/tex]
[tex]Q=It[/tex]
Put the value into the formula
[tex]Q=10\times5\times60[/tex]
[tex]Q=3000\ C[/tex]
Hence, The charge pass through your hair dryer is 3000 C.
A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavelength numerically in nanometers.
Answer:
The wavelength is [tex]\lambda = 1.2 * 10^8 nm[/tex]
Explanation:
From the question we are told that
The frequency of operation of the microwave is [tex]f = 2.50 GHz = 2.50 *10^{9} \ Hz[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{3.0 *10^{8}}{ 2.50 *10^{9}}[/tex]
=> [tex]\lambda = 0.12 \ m [/tex]
converting to nanometer
[tex]\lambda = 1.2 * 10^8 nm[/tex]
An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?
Answer:
(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]
Explanation:
Given that,
The frequency of FM radio station, f = 93.4 MHz
(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]
(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]
Hence, this is the required solution.
Using component notation, enter the vector B⃗ B→B_vec in the answer box. Enter your answer as a pair of vector components, separated by a comma. You should not enter any parentheses.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]\vec B = 2, -3[/tex]
Explanation:
Looking at the graph in the diagram we see each unit is equal to 1 both in the x axis and in the y- axis
Now the value of B along the x axis is
[tex]B_x = 2[/tex]
and along the y axis the value is
[tex]B_y = -3[/tex]
Hence the vector B is
[tex]\vec B =(B_x , B_y)= ( 2, -3)[/tex]
If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by
Answer:
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
Explanation:
Let [tex]\vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})[/tex] and [tex]\vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex], both measured in meters. The resultant vector [tex]\vec R[/tex] is calculated by sum of components. That is:
[tex]\vec R = \vec A+\vec B[/tex] (Eq. 1)
[tex]\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex]
[tex]\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}[/tex]
[tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex]
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
If a projectile hits a stationary target, and the projectile continues to travel in the same direction, the mass of the projectile is less than the mass of the target. the mass of the projectile is equal to the mass of the target. the mass of the projectile is greater than the mass of the target. nothing can be said about the masses of the projectile and target without further information. this is an unphysical situation and will not actually happen.
The correct arrangement of the question is;
If a projectile hits a stationary target, and the projectile continues to travel in the same direction,
A) the mass of the projectile is less than the mass of the target.
B) the mass of the projectile is equal to the mass of the target.
C) the mass of the projectile is greater than the mass of the target.
D) nothing can be said about the masses of the projectile and target without further information.
E) this is an unphysical situation and will not actually happen.
Answer:
Option C: The mass of the projectile is greater than the mass of the target.
Explanation:
We want to find what will happen when a projectile continues in motion after it hits a target.
Now, for the projectile to keep moving in that direction after it hits the target, it means it had a force bigger than the force of the target to overpower it and force it to move with it.
Now, from law of inertia, Force = ma.
But in this case acceleration is 0 because the speed of the projectile is constant.
Thus, the force depends on the mass. So for a higher force, the mass of the projectile has to be more than that of the stationary object.
Thus, option C is correct
A pole-vaulter just clears the bar at 5.53 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3200 J. What is his weight?
Answer:
578.66 N
Explanation:
The first step is to calculate the mass
mgh= 3200J
3200/9.8×5.53
3200/54.194
m = 59.047 kg
Therefore the weight can be calculated as follows
Weight = m × g
= 59.047 × 9.8
= 578.66 N
What is the probability that a junior non-Physics major and then a freshman non-Physics major are chosen at random?
Answer:
Probability = 0.0244
Explanation:
Probability that Junior Non Physics Major & then a Freshman Non Physics Major are chosen:
Prob (Jr No-Ph Mjr) = Jr No-Ph Mjr / Total
= 18 / 82 = 0.2195
Prob (Fr No-Ph Mjr) = Fr No-Ph Mjr / Total (remaining)
= 9 / 81 = 0.1111
Prob [ Jr No-Ph Mjr & Fr No-Ph Mjr ] = 0.2195 x 0.1111 = 0.02439
≈ 0.0244
You are driving your car at 45 m/s, when a raccoon runs into the street in front of you. You slam on the brakes and come to a stop in 5 seconds. What is the acceleration of your car?
Answer:
-9m/s²
Explanation:
Given parameters:
Initial velocity = 45m/s
Final velocity = 0
duration = 5s
Unknown:
acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time;
Acceleration = [tex]\frac{v- u}{t}[/tex]
v is the final velocity
u is the initial velocity
t is the time taken
Input the parameters and solve;
Acceleration = [tex]\frac{0 - 45}{5}[/tex] = -9m/s²
The car accelerates at a rate of -9m/s² which is a deceleration
A plane travelling at 100 m/s accelerates at 5 m/s² for a distance of 125 m. What is the final velocity of the plane?
Analyzing the question:
We are given:
initial velocity (u) = 100 m/s
final velocity (v) = v m/s
distance (s) = 125 m
acceleration (a) = 5 m/s²
Solving for Final Velocity (v):
from the third equation of motion:
v² - u² = 2as
v² - (100)² = 2(5)(125)
v² - 10000 = 1250
v² = 1250 + 10000
v² = 11250
v = 106.06 m/s
I need help with this answer
decomposition
A decomposition reaction is just the opposite of combination reaction
If you are driving to see your cousins at a speed of 84.6 km/h and it took you 6.5 h to get there, how far did you travel?
Answer: 549.9 km
Explanation: 84.6km every hour so 84.6*6.5= 549.9
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
A force of 41 N acts on an object which has a mass of 2.4 kg. What acceleration (in m/s2) is produced by the force
Answer:
The acceleration is [tex] a = 17.083 \ m/s^2 [/tex]
Explanation:
From the question we are told that
The force is [tex]F = 41 \ N[/tex]
The mass of the object is [tex]m = 2.4 \ kg[/tex]
Generally the force is mathematically represented as
[tex]F = m* a[/tex]
=> [tex] 41 = 2.4* a[/tex]
=> [tex] a = 17.083 \ m/s^2 [/tex]
Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days. If it were to undergo gravitational collapse to a neutron star of radius 13 km , losing three-quarters of its mass in the process, what would its rotation speed be
Answer:
Its rotation will be 3.89x10⁴ rad/s.
Explanation:
We can find the rotation speed by conservation of the angular momentum:
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}\omega_{i} = I_{f}\omega_{f} [/tex] (1)
The initial angular speed is:
[tex] \omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d} [/tex]
The moment of inertia (I) of a sphere is:
[tex] I = \frac{2}{5}mr^{2} [/tex] (2)
Where m is 9 times the sun's mass and r is the sun's radius
By entering equation (2) into (1) we have:
[tex] \frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f} [/tex]
[tex]9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}[/tex]
[tex]\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s[/tex]
Hence, its rotation will be 3.89x10⁴ rad/s.
I hope it helps you!