Answer:
S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m
Explanation:
A local FM radio station broadcasts at a frequency of 100.8 MHz. Calculate the energy of the frequency at which it is broadcasting. Energy
Answer:
[tex]E=6.68\times 10^{-26}\ J[/tex]
Explanation:
Given that,
The frequency of FM ratio station, f = 100.8 MHz = 100.8 × 10⁶ Hz
We need to find the energy of the wave. We know that,
Energy, E = hf
Put all the values,
[tex]E=6.63\times 10^{-34}\times 100.8\times 10^6\\\\=6.68\times 10^{-26}\ J[/tex]
So, the energy of the wave is equal to [tex]6.68\times 10^{-26}\ J[/tex].
Một xe hơi nặng 1000kg đang kéo một toa mo1oc 300kg. Cả hai cùng tiến về phía trước với gia tốc 2.15m/s2. Bỏ qua lực cản không khí xác định: Tổng lực tác dụng lên xe hơi
Answer:
Một ô tô có khối lượng m=1000kg đang chạy với vận tốc 18km/h thì hãm phanh.Biết lực hãm là 2000N. Tính quãng đường xe còn chạy thêm trước khi dừng ... Chiếu phương trình của định luật II Newtơn mà →F=m. ... chuyển động ta có F=ma, suy ra gia tốc chuyển động của xe ( với F=2000N) ...
Explanation:
In the mirror diagram shown, which is the normal?
А
В
С
D
Answer:
C
Explanation:
The normal is the line which divides the angle between the incident ray (which is the ray of an object which strikes the mirror) and the reflected ray(the ray which is thrown back as the object hits the mirror surface) into two equal parts. The normal is always perpendicular to the surface. In the description agram Given , the Noa which is the line C, divides the reflected ray (line D) and the incident ray (line A) into two equal parts. The plane surface is line B and the other incident ray (line C) is perpendicular to B
____ are the foundation of psychoanalytic theory.
Answer:
an unconscious needs rooted in childhood are the foundation of psychoanalytic theory.
Explanation:
An object starts to rotate about an axis from rest wih a, uniform angular acceleration of 2pi rads-2 what is the no.of rotations it can complete in 5s
Answer:
θ = 12.5 rotations
Explanation:
The number of rotations can be found by using the second equation of motion:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2\\\\[/tex]
where,
[tex]\theta[/tex] = angular displacement = ?
ωi = initial angular speed = 0 rad/s
t = time = 5 s
α = angular acceleration = 2π rad/s²
Therefore,
[tex]\theta = (0\ rad/s)(5\ s)+\frac{1}{2}(2\pi\ rad/s^2)(5\ s)^2\\\\\theta = 78.54\ rad[/tex]
converting it to no. or rotations:
[tex]\theta = (78.54\ rad)(\frac{1\ rotation}{2\pi\ rad})[/tex]
θ = 12.5 rotations
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
a particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.46 N making an angle 30. during its motion between A and B which is subjected to a frictional force f=1.5 N
calculate V of B by applying the kinetic energy theorem
Answer:
The final speed is 5.78 m/s.
Explanation:
mass, m = 375 g = 0.375 kg
initial velocity, u = 4 m/s
Distance, s = 2.5 m
Angle, A = 30 degree
Force, F = 1.5 N
let the final velocity is v.
Use the work energy theorem
Work done = change in kinetic energy
[tex]W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s[/tex]
A dipole placed in an electric field will... Try to align along the direction of the field. Try to align antiparallel to the direction of the field. try to align perpendicular to the field. none of these choices.
Answer:
The dipole always tries to align in the direction of electric field so that the net torque is zero.
Explanation:
When a dipole placed in an electric field at any arbitrary angle, it experiences a torque.
Let the dipole moment is p, electric field is E and the angle is A.
The torque is given by
Torque = p E sin A
where, A is the angle between the electric field and dipole moment.
So, the dipole always tries to align in the direction of electric field so that the net torque is zero.
The process of finding the age of fossil is called which answer carbondating,urinumdating ,both of them,none
Answer:
the process of finding the age of fossils is called carbon dating because they use samples of carbon and compare it's characteristics to the samples of carbon from each set time
Which of the following refers to a force acting toward the center of a circular
motion?
A. Centrifugal force
B. Circular force
C. Central force
D. Centripetal force
Answer:
D. Centripetal Force :)
A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in second. Find the acceleration of the particle at the instant when particle is at rest.
Answer:
The acceleration of the object is -69.78 m/s²
Explanation:
Given;
postion of the particle:
[tex]x = 32t - 38\frac{t^3}{3} \\\\[/tex]
The velocity of the particle is calculated as the change in the position of the particle with time;
[tex]v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s[/tex]
Acceleration is the change in velocity with time;
[tex]a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2[/tex]
Bonnie and Clyde are sliding a 289 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 387 N of force while Bonnie pulls forward on a rope with 341 N of force.
Required:
What is the safe's coefficient of kinetic friction on the bank floor?
Answer:
= 0.257
Explanation:
Applied force on the safe by Bonnie and Clyde is F= 387 N + 341 N
= 728 N
Given safe slides with constant speed .So, force of friction =applied force
= 728 N
μ*normal force = 728 N
from this the safes coefficient of kinetic friction on the bank floor is
μ = 728 N / normal force
= 728 N / Mg
= 728 N / ( 289 * 9.8 )
= 0.257
A hole is made in a square tile of uniform thickness. The diagram shows the tile hanging loosely on a nail. Where is the centre of gravity of the tile?
Answer:
The center of gravity of the tile is at the center of the tile:
Examining 1/2 of the tile, each mass point will have an equal mass point
on the other side of the tile (considering a line thru the center of the tile)
with coordinates opposite and equal of the point being considered.
Each mass point will have an identical mass point on the opposite side of the tile, with coordinates opposite and equal to the point being analyzed, while looking at only half of the tile.
What is center of gravity ?
A hypothetical location where the gravitational force is said to appear to act is the human body's centre of gravity (COG). That is the location where the entire body's bulk appears to be concentrated[2]. The COG need not be located inside the actual boundaries of an object or a person because it is a hypothetical point. Imagine the thing balanced on one finger to get an idea of its COG (there are objective measurements).
The COG is positioned anatomically so that it is about anterior to the second sacral vertebra. The specific placement of the COG, however, continually shifts with each new posture of the torso and limbs since humans do not remain stable in the anatomical position. The dimensions of the body
The center of the gravity is at half of the length of tile.
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voltage is 50v and capacitance is 30c what is the charge
Explanation:
voltage is 50v and capacitance is 30c what is the charge
q=1500c
A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m
and slope angle 30 degree, suppose the coefficient of friction of the slope surface is
0.25, find (neglect air resistance,take g=10 m/s^2)
The friction force experienced by the box.
00) The acceleration of the box along the slope.
(1) The time T required for the object to reach the bottom of the slope from the slope top.
Answer:
Explanation:
The first thing we are asked to find is the Force experienced by the box. That is found in the formula:
F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:
[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:
First we need the weight of the box. Having the mass, we find the weight:
w = mg so
w = 25.14(10) so
w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:
251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so
F = 125.7 And in order to answer what a is equal to, we find f:
f = μ[tex]F_n[/tex] where Fn is the weight of the object.
f = .25(251.4) so
f = 62.85. Filling everything in now altogether to solve for a, the only missing value:
125.7 - 62.85 = 25.14a and
62.85 = 25.14a so
a = 2.5 m/s/s
Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that
a = -10 m/s/s
v₀ = 0 (it starts from rest)
Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:
Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:
[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and
[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so
t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.
1. It is an object's tendency to resist a change in motion.
2. The study of celestial object such as moon, planets stars and galaxies.
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If a duster is left on the top of the chalkboard, it is said to possess kinetic energy.
Answer:
In the above diagram, we move by horizontally with a velocity to the right. But due to gravity the hand falls a distance as we move .
Therefore the time it takes to fall by a distance is given by =.
If = as →∞, then from the laws of motion
=12()22
Therefore for the whole journey from left to right our muscle has to give a beat up at an amount equal to the gravity force. So, the work supplied by the muscle for one trip would be
∫=∫012()22
A man is going to rub the chalk off a blackboard, he is going to choose a way to rub off the chalk in two ways
Starting from the upper left corner of the board and moving horizontally to the right and moving slightly down and then again moving to the right and then when it reaches the left corner and so on
The second way is given in the following: He starts at the top and rubs down and then up again and so on.
He is going to choose the way that is less tiring in his arms. i.e., less work done by his muscles!
Somehow from vague intuitive notion I am inclined to agree that it would be less tiring to rub off the chalk if he moves the duster beginning at the top and rubbing horizontally and gradually decreasing the height when one reach the corners till everything is off.
I tried to calculate the muscle work by considering the following ideas(please correct me if I'm wrong):
The nerves inside our muscle has to fire its signal continuously throughout the entire interval of the process of rubbing and its has to oppose the gravitational force keeping the hand up the air.
When the hand is at the highest point on the blackboard, the muscle has to work against gravity =, as the muscle continues to work against gravity throughout the interval from left to right (the horizontal path of the first case) I couldn't find the total work withstood by our muscle.
Three forces are pulling on the same object such that the system is in equilibrium. Their magnitudes are F1 = 2.83 N.F= 3.35 N. and F3 = 3.64 N, and they make angles of 0, = 45.0°, 02 = -63.43 and 03 =164.05° with respect to the x-axis, respectively.
Required:
a. What is the x-component of the force vector F1?
b. What is the y-component of the force vector F1?
(a) 2.001N
(b) 2.001N
Explanation:A sketch of the scenario has been attached to this response.
Since only the force vector F₁ is required, the only force shown in the sketch is F₁.
As shown in the sketch;
The x-component of the force vector F₁ = [tex]F_{x}[/tex]
The y-component of the force vector F₁ = [tex]F_{y}[/tex]
The magnitude of F₁ as given in the question = 2.83N
The angle that the force makes with respect to the x-axis = 45.0°
Using the trigonometric ratio, we see that;
(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]
=> [tex]F_{x}[/tex] = F₁ cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 x 0.7071
=> [tex]F_{x}[/tex] = 2.001N
(b) Also;
sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]
=> [tex]F_{y}[/tex] = F₁ sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 x 0.7071
=> [tex]F_{y}[/tex] = 2.001N
Therefore, the x-component and y-component of the force vector F₁ is 2.001N
The x and y component of vector F1 is mathematically given as
F_x = 2.001N
F_y= 2.001N
What is the x and y component of vector F1?Question Parameters:
Generally, the equation for the x-component is mathematically given as
x=Fsin\theta
Therefore
F_x = F₁ cos 45.0°
F_x = 2.83 x 0.7071
F_x = 2.001N
For y component
x=Fcos\theta
F_y = F₁ sin 45.0
F_y = 2.83 x 0.7071
F_y= 2.001N
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A car is travelling at 30m/s and decelerates [with normal car brakes, no trick] at 5m/s/s for 10 s. What is the car's final speed *
A) zero
B) -50m/s
C) -20m/s
D) Not possible to tell
show your full work
Answer:
B. -20 m/s
Explanation:
Given the following data:
Initial velocity = 30 m/s
Acceleration = 5 m/s²
Deceleration = -5 m/s² (deceleration is the negative of acceleration)
Time = 10 seconds
To find the final velocity, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the formula, we have;
[tex] V = 30 + (-5)*10[/tex]
[tex] V = 30 - 50[/tex]
Final velocity, V = -20 m/s
The figure shows three displacement vectors, which are
labeled a, 5, and c. What is the magnitude and direction
of the resultant vector found by adding 5 and ć?
N
4
S
15 m west
āt
2.0 m east
7.0 m east
Ĉ
A. 5.0 m east
B. 5.0 m west
ОО
C. 9.0 m west
0
D. 9.0 m east
Answer:
the correct answer is D
Explanation:
In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.
R = δ + ε
where
δ = 2.0 m
ε = 7.0 m
the positive sign indicates that it is heading east
R = 2.0 + 7.0
R = 9.0 m
the direction is east
the correct answer is D
A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.64 of her regular weight. Calculate the magnitude of the acceleration of the elevator.
Answer:
The downwards acceleration is 3.53 m/s2.
Explanation:
Let the true weight is m g.
The reading of the balance, R = 0.64 mg
Let the acceleration is a.
As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.
Use Newton's second law
m g - R = m a
m g - 0.64 m g = m a
0.36 g = a
a = 3.53 m/s2
A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff the aircraft travels at 63.0 m/s, so that the air speed relative to the bottom of the wing is 63.0 m/s. Given the sea level density of air to be 1.29 kg/m3, how fast (in m/s) must it move over the upper surface to create the ideal lift
Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² = [tex]\frac{\Delta F}{A \ \rho \ g} + v_1^2[/tex]
suppose that the area of the wing is A = 1 m²
we substitute
v₂² = [tex]\frac{1000}{1 \ 1.29 \ 9.8} + 63^2[/tex]
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s
explain how a lever can act as a force multiplier
Answer:
Example:Opening of a bottle cap by tool
when we hold a tool and open the bottle cap this is because , force x tool force .
The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.
What is second order levers?Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.
A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.
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On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs. Ramesh didn’t like the loud sound of DJs and asked his father to play it in a low volume so that their neighbours do not get much disturbed and people at the party can also enjoy the music. Ramesh’s father felt good for his wisdom and did as he said.
→Do you think when loud music is played at a party is acceptable to all the people living in, neighbourhood? Give a reason for your answer
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When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3m3 of water flows through the dam each second. The water is released 220 mm below the top of the reservoir. If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?
Answer:
The output electric power is 1338876 W.
Explanation:
Volume, V = 690 cubic meter
height, h = 220 mm = 0.22 m
efficiency = 90 %
time , t = 1 s
Let the mass is m.
m = volume x density
m = 690 x 1000 = 690000 kg
The input power is
P = m g h = 690000 x 9.8 x 0.22 = 1497640 W
The electric power out put is
[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]
A hot air balloon is moving at a speed of 10 meters/second in the +x direction. The balloonist throws a brass ball with a velocity of -2
meters/second with respect to himself. If the ball lands after 20 seconds, about how far does it land horizontally from the point at which it was
released?
OA 98 meters
OB. 160 meters
OC. 196 meters
OD. 420 meters
Answer:
160 meters
Relative of speed Vr = 10 -2 = 8 m/s (horizontal speed)
20 sec * 8 m/s = 160 m since ball travels 20 sec
Answer:
160 meters
Explanation:
____________is obtained from the fleece of animals.
Answer:
wool and fibers
Explanation:
What is the acceleration of a 0.30 kilogram ball that is hit
with a force of 27 N?
Answer: 90
Explanation: a=Fnet/M
=27/0.30
=90
the acceleration is 90
Answer:
The acceleration of the ball is 83.333ms2 [forward].
Explanation:
i hope it helps :)
Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cladding have the same refractive indices.
a. True
b. False
Answer:
False
Explanation:
Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of 33 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
1. What is the ratio of the electric force on the bee to the bee's weight?
2. What electric field strength would allow the bee to hang suspended in the air?
3. What electric field direction would allow the bee to hang suspended in the air?
Answer:
A) 3.367 × 10^(-6)
B) 2.97 × 10^(7) N/C
C) Upwards
Explanation:
We are given;
Mass of bee; m = 100 mg = 100 × 10^(-6) kg
Charge on bee;q=33 pC = 33 × 10^(-12)C
Electric field strength; E = 100 N/C
A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N
Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N
ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)
B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;
mg = qE
100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E
E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))
E = 2.97 × 10^(7) N/C
C) From Newton's law, sum of forces = 0.
Thus;
F_n + F + W = 0
Where F is the normal force.
Thus;
F_n = -(F + W)
F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))
F_n = -9.8 × 10^(-4) N
Thus, applied electric field is;
E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C
This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.