Answer:
Option B.
Explanation:
Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.
So, when we analyze the problem of "how long takes an object to hit the ground"
We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.
So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.
And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.
So, both objects will hit the ground at the same time.
(notice that here we are ignoring things like air resistance and other complex forces)
So here the correct option is b: Neither. Their vertical motion is the same, so they will reach the ground at the same time.
A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through its center.
Answer: [tex]2.4705\ kg.m^2[/tex]
Explanation:
Given
length of the rod is L=0.9 m
Mass of the rod m=3.8 kg
Point masses has mass of m=2.3 kg
Moment of Inertia of the rod about the center is
[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]
Moment of inertia of combined system is the sum of rod and two point masses.
[tex]\Rightarrow I=I_o+2mr^2[/tex]
[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]
Two balls of known masses hang from the ceiling on massless strings of equal length. They barely touch when both hang at rest. One ball is pulled back until its string is at 45 ∘, then released. It swings down, collides with the second ball, and they stick together.The problem can be divided into three parts: (1) from when the first ball is released and to just before it hits the stationary ball, (2) the two balls collide, and (3) the two balls swing up together just after the collision to their highest point. ..............conserved in parts (1) and (3) as the balls swing like pendulums. During the collision in part (2) ................. conserved as the collision is ................. Explain.Match the words in the left column to the appropriate blanks in the sentences on the rightboth energy and momentum areonly energy is only momentum is.........both energy and momentum are only energy is only momentum iselasticinelastic
Answer:
In parts 1 and 3 the energy
In part 2 moment. inelastic
conserved
Explanation:
In this exercise, we are asked to describe the conservation processes for each part of the exercise.
In parts 1 and 3 the energy is conserved since the bodies do not change
In part 2 the bodies change since they are united therefore the moment is conserved and part of the kinetic energy is converted into potential energy.
Energy
moment .inelastic
conserved
The two balls swing up together just after the collision to their highest point. energy is conserved.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
When the first ball is released and just before it hits the stationary ball, The two balls collide, The two balls swing up together just after the collision to their highest point. energy is conserved.
The balls swing like pendulums. During the collision in part (2) energy is conserved as the collision is inelastic.
We are requested to describe the conservation methods for each element of the activity in this exercise.
Because the bodies do not change in sections 1 and 3, energy is conserved.
Because the bodies change in part 2 is joined, the moment is conserved and some of the kinetic energy is transformed into potential energy.
Hence the two balls swing up together just after the collision to their highest point. energy is conserved.
To learn more about the law of conservation of momentum refer to;
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Un objeto de 0.5kg de masa se desplaza a lo largo de una trayectoria rectilínea con aceleración constante de 0.3m/s2. Si partió del reposo y la magnitud de su cantidad de movimiento en kg*m/s después de 8s es:
Answer:
p = 1.2 kg-m/s
Explanation:
The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".
Mass of the object, m = 0.5 kg
Acceleration of the object, a = 0.3 m/s²
We need to find the momentum after 8 seconds.
We know that,
[tex]p=F\times t[/tex]
i.e.
p = mat
So,
[tex]p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s[/tex]
So, the momentum of the object is 1.2 kg-m/s.
state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
A 1.64 kg mass on a spring oscillates horizontal frictionless surface. The motion of the mass is described by the equation: X = 0.33cos(3.17t). In the equation, x is measured in meters and t in seconds. What is the maximum energy stored in the spring during an oscillation?
Answer:
[tex]K.E_{max}=0.8973J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.64kg[/tex]
Equation of Mass
[tex]X=0.33cos(3.17t)[/tex]...1
Generally equation for distance X is
[tex]X=Acos(\omega t)[/tex]...2
Therefore comparing equation
Angular Velocity [tex]\omega=3.17rad/s[/tex]
Amplitude A=0.33
Generally the equation for Max speed is mathematically given by
[tex]V_{max}=A\omega[/tex]
[tex]V_{max}=0.33*3.17[/tex]
[tex]V_{max}=1.0461m/s[/tex]
Therefore
[tex]K.E_{max}=0.5mv^2[/tex]
[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]
[tex]K.E_{max}=0.8973J[/tex]
1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?
N = 52.0 N
Explanation:
Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]
[tex]m=6.00\:\text{kg}[/tex]
[tex]N = \text{normal force}[/tex]
The net force [tex]F_{net}[/tex] is given by
[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]
Solving for N, we get
[tex]N = mg - F_a\sin 20[/tex]
[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]
[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]
Calculate the volume of 10g of helium ( M= 4kg/kmol) at 25C and 600 mmHg
Answer:
T=273+25=298 K
n= m/M = 10/ 4 = 2.5
R=0.08206 L.atm /mol/k
760mmHg = 1 atm therefore
600mmHg = X atm
760 X = 600mmHg
X = 600/760 = 0.789 atm
P = 0.789 atm
V= ?
PV= nRT
0.789 V = 2.5 × 0.08206 × 298
V= 2.5 × 0.08206 ×298 / 0.789
V= 77.48 L
I hope I helped you ^_^
A rock, initially at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth. An observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth. Neglecting friction, the rock's speed when it hits the ground is _________ its speed at the top of the tower.
Answer:
the rock speed is increased
A rock is initially at rest concerning the earth, but the speed of the rock will increase when it hits the ground.
What is Friction?The resistance to something rolling or moving over another solid object is called friction. Even though frictional forces can be helpful, like the traction needed to walk without slipping, they also present a considerable amount of resistance to motion. About 20% of the engine power in a car is used to combat frictional forces in the moving parts.
The primary cause of friction between metals appears to be the forces of attraction, also known as adhesion, between the contact zones of the surfaces, which are always microscopically unequal. Because of friction caused by the imperfections of the tougher surface rubbing up against the softer surface, these "welded" connections are sheared.
To know more about Friction :
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a particle undergoes three consecutive displacement d1=(15i+30j+12k)cm,d2=(23i-14j-5.0k)cm and d3=(-13i+15j)cm find the component of the resultant displacement and magnitude?
Answer:
Explanation:
The density of pure water is 1 gram per 1 milliliter or one cubic cm. By knowing the density of water we can use it in dilution equations or to calculate the specific gravity of other solutions.
It can also help us determine what other substances are made of using the water displacement experiment. This is done by observing how much water is displaced when an object is submerged in the water. As long as you know the density of the water, the mass of the object being submerged and the volume of increase you can calculate the density of the object.
This was done by the great Archimedes in discovering what composed the kings crown.
What word chemical equation describes this chemical reaction?
Answer : sodium + chlorine → sodium chloride
convert 56km/h to m/s.
Explanation:
15.556 metres per second
true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields
Answer:
True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.30 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged
Answer:
The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.
Explanation:
Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:
First skater
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)
Second skater
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)
Where:
[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.
[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.
[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.
[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.
[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.
[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.
If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:
By (1):
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]
[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]
[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]
[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]
[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]
By (2):
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]
[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]
[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]
[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]
Images formed by a convex mirror are always
Answer:
Images formed by a convex mirror are always virtual
Explanation:
A virtual image is always created by a convex mirror, and it is always situated behind the mirror. The picture is vertical and situated at the focus point when the item is far away from the mirror. As the thing approaches the mirror, the image follows suit and increases until it reaches the same height as the object.
OAmalOHopeO
The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Explanation:
Let T is the period of a pendulum. The SI unit of time is seconds (s).
It depends on the acceleration of gravity, g, and the length of the pendulum, l.
The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.
If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.
So,
[tex]T\propto \sqrt{\dfrac{l}{g}}[/tex]
Hence, this is the required solution.
Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
a concrete has a height of 5m and has unit area 3m² supports a mass of 30000kg.
Determine the stress, strain and change in height
Answer:
stress = 98000 N/m^2
strain = 3.92 x 10^-6
change in height = 0.0196 mm
Explanation:
Height, h = 5 m
Area, A = 3 m²
mass, m = 30000 kg
Stress is defined as the force per unit area.
[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]
Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2
Young's modulus is defined as the ratio of stress to the strain.
[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]
let the change in height is h'.
Strain is defined as the ratio of change in height to the original height.
[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]
It takes 20 Joules of Work to push 4 coulombs of charges Across the filament of a bulb.'find the potential difference Across the filament
Answer:
V = 5 Volts
Explanation:
Given the following data;
Work done = 20 Joules
Charge = 4 Coulombs
To find the potential difference;
Mathematically, the work done in moving a charge is given by the formula;
W = qv
Where;
W is the work done
q is the quantity of charge
v is the potential difference
Substituting we have;
20 = 4 * v
V = 20/4
V = 5 Volts
The value found for the universal gravitational constant, G, will vary depending on the materials used for the balls of a Cavendish balance. Question 11 options: True False
Answer:
false
Explanation:
took the test
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
Complete question:
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
a) 60 ⁰
b) 90 ⁰
c) 120 ⁰
Answer:
(a) When the angle, θ = 60 ⁰, force = 4.07 N
(b) When the angle, θ = 90 ⁰, force = 4.7 N
(c) When the angle, θ = 120 ⁰, force = 4.07 N
Explanation:
Given;
length of the wire, L = 2.8 m
current carried by the wire, I = 5.6 A
magnitude of the magnetic force, F = 0.3 T
The magnitude of the magnetic force is calculated as follows;
[tex]F = BIl \ sin(\theta)[/tex]
(a) When the angle, θ = 60 ⁰
[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N[/tex]
(b) When the angle, θ = 90 ⁰
[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N[/tex]
(c) When the angle, θ = 120 ⁰
[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N[/tex]
Please assist with solving this problem and showing the steps
Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 412 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.
Answer:
ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Explanation:
Given the data in the question;
T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE[tex]_{int[/tex],₁ + W₁
Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112
Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J
Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Find the refractive index of a medium
having a velocity of 1.5 x 10^8*
Explanation:
someone to check if the answer is correct
An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 19 m/s and measures a time of 24.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Answer:
1.56 m/s²
Explanation:
Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.
The total time (time of flight) of an object is given by:
T = 2usinθ / g
where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity
Since the astronaut throws a rock straight up, hence θ = 90°, u = 19 m/s, T = 24.4 s.
T = 2usinθ / g
Substituting:
24.4 = 2(19)(sin90)/g
g = 2(19)(sin90) / 24.4
g = 1.56 m/s²
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.
A
cook
holds a 3.2 kg carton of milk at arm's length.
75.9
w
25,5 cm
What force FB must be exerted by the bi-
ceps muscle? The acceleration of gravity is
9.8 m/s2. (Ignore the weight of the forearm.)
Answer in units of N.
Answer:
Explanation:
From the given information:
From the rotational axis, the distance of the force of gravity is:
d_g = 25+5.0 cm
d_g = 30.0 cm
d_g = 30.0 × 10⁻² m
However, the relative distance of FB cos 75.9° from the axis is computed as:
d_B = 5.0 cm
d_B = 5.0 × 10⁻² m
The net torque rotational equilibrium = zero (0)
i.e.
[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]
= 772.4 N
Thus, the force exerted = 1772.4 N
What is (a) the x component and (b) the y component of the net electric field at the square's center
Answer:
What is (a) the x component and (b) the y component of the net electric field at the square's center