Answer:
v = 3.6m / s , θ = 56º
Explanation:
This is a relative speed exercise, let's use the Pythagorean theorem
v = √ (v₁² + v₂²)
where v₁ is the speed of the sea still water and v₂ the speed of the current
let's calculate
v = √ (2² + 3²)
v = 3.6m / s
to find the direction we use trigonometry
tan θ = v₂ / v₁
θ = tan⁻¹ (v₂ / v₁)
let's calculate
θ = tan⁻¹ (3/2)
θ = 56º
Identify the following as combination, decomposition, replacement, or ion exchange reactions NaBr(aq) + Cl2(g) → 2 NaCl(aq) + Br2(g)
Answer:
Replacement
Explanation:
in replacements, like ions replace like. in this equation, we can see that Bromine replaced Chlorine. so, the answer is replacement.
Answer:
Single-replacement or replacement
Explanation:
The single-replacement reaction is a + bc -> ac + b, compare them.
NaBr + Cl2 -> 2 NACl + Br.
AB + C -> AC + B
As you can see they are the same ( even though the b is with the a and not with the c. The formula can be switched around a little with the order of b and c ) ((also like ions replace like ions in replacements, which they are in this))
A region of space contains a uniform electric field oriented along the y-axis. A frame of surface area A is placed perpendicular to the y-axis in the xz-plane. The magnitude of the electric flux through this frame is Φ0. A second frame is placed in the same electric field in such a way that the magnitude of the electric flux through it is Φ0/2. How is the plane of second frame oriented with respect to the plane of the first one?
Answer:
β = 30º
Explanation:
By definition, the vector flux across a surface, can be found integrating the dot product of the vector field (electric field E in this case) and the differential surface dA, across the entire surface.If the surface is placed perpendicular to the electric field, and this field is uniform, the total flux across the surface can be expressed as follows:Φ0 = E*A*cos 0º = E*A.
If the magnitude of the electric flux, is reduced to half of its original value, we can write the following equality:Φ0/2 = E*A*cos θ⇒ = Φ0 * cos θ (1)
where θ, is the angle between the electric field E and the vector
perpendicular to the plane traversed by E.
Rearranging terms in (1) we can solve for θ, as follows:
⇒ cos θ = 1/2 ⇒ θ = arc cos (1/2) = 60º
As this is the angle between the electric field, and the surface vector,
which is by definition, perpendicular to the plane, the angle between
the electric field and the plane can be found as follows:
β = 90º - θ = 90º - 60º = 30º
What is the length of the x-component of the vector shown below? A. 65.8 B. 90.6 C. 112 D. 33.2
Answer:
The correct answer is - option b. 90.6
Explanation:
The scalar x-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle
If you shine a light straight down onto that vector, then the length of its shadow on the x-axis is -
x-component = 112· cosine(36°)
x-component = 112 · (0.8090)
x-component = 90.60
Thus, The correct answer is - option b. 90.6
A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mass of 0.50 kg, determine each of the following values.
(a) the mechanical energy of the system
(b) the maximum speed of the object
m/s
(c) the maximum acceleration of the object
m/s2
Answer:
(a) the mechanical energy of the system, U = 0.1078 J
(b) the maximum speed of the object, Vmax = 0.657 m/s
(c) the maximum acceleration of the object, a_max = 15.4 m/s²
Explanation:
Given;
Amplitude of the spring, A = 2.8 cm = 0.028 m
Spring constant, K = 275 N/m
Mass of object, m = 0.5 kg
(a) the mechanical energy of the system
This is the potential energy of the system, U = ¹/₂KA²
U = ¹/₂ (275)(0.028)²
U = 0.1078 J
(b) the maximum speed of the object
[tex]V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s[/tex]
(c) the maximum acceleration of the object
[tex]a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2[/tex]
The characteristics of the simple harmonic motion allows to find the results for the questions of the oscillating mass are:
a) The total energy is: Em = 0.1078 J
b) The maximum speed is: v = 0.657 m / s
c) the maximum acceleration is: a = 15.4 m / s²
Given parameters
The amplitude A = 2.8 cm = 2.8 10⁻² m The spring constant k = 275 N / m Mass m = 0.50 kgTo find
a) Mechanical energy
b) Maximum speed
c) Maximum acceleration
the simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, it is described by the expression:
x = A cos (wt + Ф)
w² = k / m
Where x is the displacement, A the amplitude w the angular velocity, t the time, Ф a phase constant, k the spring constant and m the mass.
A) The mechanical energy is
Em = ½ k A²
Let's calculate.
Em = ½ 275 (2.8 10⁻²) ²
Em = 0.1078 J
b) Velocity is defined as the change of position with respect to time.
v = [tex]\frac{dx}{dt}[/tex] = - Aw sin ( wt + fi)
To obtain the maximum velocity, the sine function must be ±1
[tex]v_{max}[/tex] = w A
Let's calculate
w = [tex]\sqrt{\frac{275}{0.5} }[/tex]
w = 23.45 rad / s
[tex]v_{max}[/tex] = 23.45 2.8 10⁻²
[tex]v_{max}[/tex] = 0.657 m / s
c) maximum acceleration.
Acceleration is defined as the change in velocity withtrspect to time.
a = [tex]\frac{dv}{dt}[/tex] = - A w² cos (wt + fi)
To have the maximum value, the cosine function must be maximum, that is ±1
a = A w²
let's calculate
a = 2.8 10⁻² 23.45²
a = 15.4 m / s²
In conclusion using the characteristics of the simple harmonic motion we can find the results for the questions of the oscillating mass are:
a) The total energy is: Em = 0.1078 J
b) The maximum speed is: v = 0.657 m / s
c) the maximum acceleration is: a = 15.4 m / s²
Learn more here: brainly.com/question/17315536
It is always a good idea to get some sense of the "size" of units. For example, the mass of an apple is about 100 g , whereas a bicycle is about a hundred times more massive at around 10 kg . Estimate the mass of the objects.
Answer and Explanation:
The estimation of the mass of the objects is as follows
Given that
The mass of an apple is about 100 g
Just like that, we do some estimation of different objects
A dime is around 1 g
A smartphone is about 100 g
An adult male is approx 100 kg
A college physics textbook is around 1 kg
A ripe banana is around 100 g
A small car is around 1,000 kg
When an automobile moves with constant velocity, the power developed is used to overcome the frictional forces exerted by the air and the road. If the engine develops 40 hp, what total frictional force acts on the car at 140 mph
Answer:
476.82 N
Explanation:
In the given question, the speed of the car is given in mph,
therefore to convert mph in meter per second
= ( 140 x 1609) / 3600
= 62.58
Now, to find the frictional force, convert the power of the car that is hp into watt
we know, 1 hp = 746 W
therefore, 40 x 746
= 29840 watts.
Frictional force,
= power/ speed of car
= 29840/ 62.58
= 476.82 N
Thus, 476.82 N is the correct answer.
"A particle of dust lands 41.0 mm from the center of a compact disc (CD) that is 120 mm in diameter. The CD speeds up from rest, and the dust particle is ejected when the CD is rotating at 84.0 revolutions per minute. What is the coefficient of static friction between the particle and the surface of the CD?"
Answer:
The coefficient of static friction is [tex]\mu = 0.474[/tex]
Explanation:
From the question we are told that
The position of the particle is [tex]x = 41.00 \ mm = 0.041 \ m[/tex]
The diameter of the CD is [tex]d =120 \ mm = 0.12 \ m[/tex]
The radius of the CD is evaluated as [tex]r = \frac{d}{2} = \frac{0.12}{2} = 0.06[/tex]
The angular velocity of the CD when particle was ejected [tex]w = 84.0 rpm = 84 .0 * \frac{2 * \pi}{60} = 8.7976 \ rad/s[/tex]
At the instant just before the particle is ejected from the CD
The frictional force of the particle = centrifugal force on the particle
So
[tex]\mu * m * g = mw^2 r[/tex]
=> [tex]\mu * g = w^2 r[/tex]
=> [tex]\mu = \frac{8.7976^2 * 0.06}{ 9.8}[/tex]
=> [tex]\mu = 0.474[/tex]
Suppose a disk drive has the following characteristics:
1. Four surfaces
2. 1,024 tracks per surface
3. 128 sectors per track
4. 512 bytes/sector
5. Track-to-track seek time of 5ms
6. Rotational speed of 5,000rpm 1.
Required:
a. What is the capacity of the drive?
b. What is the access time?
Answer:
a. 256MB
b. 11 ms
Explanation:
a. The capacity of the drive is
As we know that
[tex]= Number\ of\ surface \times tracks\ per\ surface \times sectors\ per\ surface \times bytes\ or\ sector[/tex]
[tex]= 4 \times 1024 \times 128 \times 512\ bytes[/tex]
= 256MB
b. Now the access time is
As we know that
Average access time= seek time + average rotational latency
where,
seek time = 5ms
And, average rotational latency is
[tex]= \frac{rotational latency}{2}[/tex]
[tex]= \frac{60}{RPM}[/tex]
[tex]= \frac{60}{5,000}[/tex]
= 12 ms
So, average rotational latency is 6 ms
So, average access time is
= 5 ms + 6 ms
= 11 ms
We simply applied the above formulas
To live a good life and be the sort of people we ought to be, we need to develop a virtuous character that
Answer:
I belive its THAT HELPS US BE A BETTER PERSON
Answer:
Understands the purpose of moral standards and how to best fulfill that.
Explanation:
A vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West B. 47 m/s C. 67 m/s D. 67 m/s West
Answer:
D
Explanation:
Speed =distance /time
=2345/35
=67m/s
A 50-cm-long spring is suspended from the ceiling. A 410 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 16 cm before coming to rest at its lowest point. It then continues to oscillate vertically. Part A What is the spring constant
Answer:
25.125 N/m
Explanation:
extension on the spring e = 16 cm 0.16 m
mass of hung mass m = 410 g = 0.41 kg
equation for the relationship between force and extension is given by
F = ke
where k is the spring constant
F = force = mg
where m is the hung mass,
and g is acceleration due to gravity = 9.81 m/s^2
imputing value, we have
0.41 x 9.81 = k x 0.16 = 0.16k
4.02 = 0.16k
spring constant k = 4.02/0.16 = 25.125 N/m
A transformer supplies 60 watts of power to a device that is rated at 20 volts (rms). The primary coil is connected to a 120-volt (rms) ac source. What is the current I1 in the primary coil?
Answer:
I = 0.5A
Explanation:
Hello,
Assuming the transformer is an ideal transformer, we can calculate the value of the current using the formula.
Data;
Power = 60 watt
Primary voltage = 120 volts (rms)
Power (P) = current (I) × voltage (V)
P = IV
This formula is the relationship between power, current and voltage.
Current = power / voltage
I = P / V
I = 60 / 120
I = 0.5A
The current in the primary coil is 0.5A
A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?
Answer:
(a) Fw = 101.01 N
(b) W = 282.82 J
(c) Fg = 382.2 N
(d) N = 368.61 N
(e) Net force = 0 N
Explanation:
(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:
[tex]F_g-F_w=0[/tex] (1)
Fg: gravitational force over the object
Fw: worker's force
However, in an incline you have that the gravitational force on the object, due to its weight, is given by:
[tex]F_g=Wsin\theta=Mg sin\theta[/tex] (2)
M: mass of the ice block = 39 kg
g: gravitational constant = 9.8m/s^2
θ: angle of the incline
You calculate the angle by using the information about the distance of the incline and its height, as follow:
[tex]sin\theta=\frac{0.74m}{2.8m}=0.264\\\\\theta=sin^{-1}(0.264)=15.32\°[/tex]
Finally, you solve the equation (1) for Fw and replace the values of all parameters:
[tex]F_w=F_g=Mgsin\theta\\\\F_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N[/tex]
The worker's force is 101.01N
(b) The work done by the worker is given by:
[tex]W=F_wd=(101.01N)(2.8m)=282.82J[/tex]
(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:
[tex]F_g=Mg=(39kg)(9.8m/s^2)=382.2N[/tex]
The gravitational force is 382.2N
(d) The normal force is:
[tex]N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N[/tex]
(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.
Is it possible to do work on an object without changing the kinetic energy of the object? Now Why?
a) Yes, it is possible by raising the object to a greater height without acceleration.
b) Yes, it is possible by raising the object to a greater height with acceleration
c) Yes, it is possible by moving the object without acceleration at the same height.
d) Yes, it is possible by moving the object with acceleration at the same height.
Answer:
(a) Yes, it is possible by raising the object to a greater height without acceleration.
Explanation:
The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.
If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.
If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).
Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.
Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".
You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to be 2.40 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. What frequency is required?
Answer:
f = 1.69*10^5 Hz
Explanation:
In order to calculate the frequency of the sinusoidal voltage, you use the following formula:
[tex]V_L=\omega iL=2\pi f i L[/tex] (1)
V_L: voltage = 12.0V
i: current = 2.40mA = 2.40*10^-3 A
L: inductance = 4.70mH = 4.70*10^-3 H
f: frequency = ?
you solve the equation (1) for f and replace the values of the other parameters:
[tex]f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz[/tex]
The frequency of the sinusoidal voltage is f
A bus driver heads south with a steady speed ofv1 = 22.0 m/sfort1 = 3.00 min,then makes a right turn and travels atv2 = 25.0 m/sfort2 = 2.80 min,and then drives northwest atv3 = 30.0 m/sfort3 = 1.00 min.For this 6.80-min trip, calculate the following. Assume +x is in the eastward direction.(a)total vector displacement (Enter the magnitude in m and the direction in degrees south of west.)magnitude m direction ° south of west(b)average speed (in m/s)m/s(c)average velocity (Enter the magnitude in m/s and the direction in degrees south of west.)magnitude m/s direction ° south of west
Answer:
total displacement =6096.93m[tex]\alpha =[/tex] 26.2°(south of west)
average speed = 24.41m/saverage velocity = 14.94m/sExplanation:
Many cultures around the world still use a simple weapon called a blowgun, a tube with a dart that fits tightly inside. A sharp breath into the end of the tube launches the dart. When exhaling forcefully, a healthy person can supply air at a gauge pressure of 6.0 kPa.
Required:
What force does this pressure exert on a dart in a 1.5-cm-diameter tube?
Answer:
F = 1.06N
Explanation:
In order to calculate the force that produces the given pressure you take into account the following formula:
[tex]P=\frac{F}{A}[/tex] (1)
P: pressure = 6.0kPa = 6.0*10³ Pa
F: force = ?
A: cross-sectional area of the tube = π r^2
r: radius of the tube = 1.5cm/2 = 0.75cm = 0.75*10^-2 m
You solve the equation (1) for F, and replace the values for the area A and the pressure P:
[tex]F=PA=(6.0*10^{3}Pa)(\pi(0.75*10^{-2}m)^2)\\\\F=1.06N[/tex]
The force tha produces a pressure of 6.0kPa in a tube with a diameters of 1.5cm is 1.06N
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad/s2. Assume no slippage. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest
Answer:
a) The linear acceleration of the car is [tex]4.65\,\frac{m}{s^{2}}[/tex], b) The tires did 7.46 revolutions in 2.50 seconds from rest.
Explanation:
a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:
[tex]\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}[/tex]
Where:
[tex]a_{r}[/tex] - Magnitude of the radial acceleration, measured in meters per square second.
[tex]a_{t}[/tex] - Magnitude of the tangent acceleration, measured in meters per square second.
Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:
[tex]\| \vec a \| = a_{t}[/tex]
[tex]\| \vec a \| = r \cdot \alpha[/tex]
Where:
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]r[/tex] - Radius of rotation (Radius of a tire), measured in meters.
Given that [tex]\alpha = 15\,\frac{rad}{s^{2}}[/tex] and [tex]r = 0.31\,m[/tex]. The linear acceleration experimented by the car is:
[tex]\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\| \vec a \| = 4.65\,\frac{m}{s^{2}}[/tex]
The linear acceleration of the car is [tex]4.65\,\frac{m}{s^{2}}[/tex].
b) Assuming that angular acceleration is constant, the following kinematic equation is used:
[tex]\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}[/tex]
Where:
[tex]\theta[/tex] - Final angular position, measured in radians.
[tex]\theta_{o}[/tex] - Initial angular position, measured in radians.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]t[/tex] - Time, measured in seconds.
If [tex]\theta_{o} = 0\,rad[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\alpha = 15\,\frac{rad}{s^{2}}[/tex], the final angular position is:
[tex]\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}[/tex]
[tex]\theta = 46.875\,rad[/tex]
Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)
[tex]\theta = 7.46\,rev[/tex]
The tires did 7.46 revolutions in 2.50 seconds from rest.
The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:
s = 4/t^2, where t is measured in seconds.
Required:
Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.
Answer:
At [tex]t = 1\; \rm s[/tex], the particle should have a velocity of [tex]-8\; \rm m \cdot s^{-1}[/tex].At [tex]t = 2\; \rm s[/tex], the particle should have a velocity of [tex]-1\; \rm m \cdot s^{-1}[/tex].At [tex]t = 3\; \rm s[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{27}\; \rm m \cdot s^{-1}[/tex].For [tex]a > 0[/tex], at [tex]t = a \; \text{second}[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}[/tex].
Explanation:
Differentiate the displacement of an object (with respect to time) to find the object's velocity.
Note that the in this question, the expression for displacement is undefined (and not differentiable) when [tex]t[/tex] is equal to zero. For [tex]t > 0[/tex]:
[tex]\begin{aligned}v &= \frac{\rm d}{{\rm d}t}\, [s] = \frac{\rm d}{{\rm d}t}\, \left[\frac{4}{t^2}\right] \\ &= \frac{\rm d}{{\rm d}t}\, \left[4\, t^{-2}\right] = 4\, \left((-2)\, t^{-3}\right) = -8\, t^{-3} =-\frac{8}{t^3}\end{aligned}[/tex].
This expression can then be evaluated at [tex]t = 1[/tex], [tex]t = 2[/tex], and [tex]t = 3[/tex] to obtain the required results.
A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C , directed downward. What is the ratio of the electric force on the bee to the bee's weight
Answer:
[tex]1.50\ *10^{-6} }[/tex]
Explanation:
Given
e=100 N/C
M=0.15 g
[tex]q=\ 22\ pC\\=\ 22\ *10^{-2}[/tex]
The ratio of the electric force on the bee to the bee's weight can be determined by the following formula
[tex]\frac{fe}{M*9.81}[/tex]
[tex]\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}[/tex]
[tex]=\ 1.50\ *10^{-6}[/tex]
A uniform, 4.5 kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.3 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 2.5 m/s in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved but not the linear momentum?
Answer:
a) Angular speed(w) = 2.02rad/sec
b) 73J ( It is Inelastic Collision)
Explanation:
Given:
Mass=45kg
Length on each side = 1.5m side which is hangs vertically from a frictionless pivot at the center of its upper edge.
We need to calculate
(a) What is the angular speed and
(b) To know why the angular momentum conserved but not the linear momentum
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -1.4 V during one revolution. Assuming the shuttle is a conducting sphere of radius 15 m, estimate the amount of charge it collects.
Answer:
-2.3 × 10^-9 Coulombs(C).
Explanation:
So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;
=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "
=> " Assuming the shuttle is a conducting sphere of radius 15 m".
So, in order to estimate the value for the charge we will be making use of the equation below:
Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.
Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.
So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.
= -2.3 × 10^-9 Coulombs(C).
Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?
Answer:
The ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil.
Explanation:
Oil is more less dense than water. Thus, the molecules that make up the oil are larger than those that that make up water, so they cannot pack as tightly together as the water molecules will do. Hence, they will take up more space per unit area and are we can say they are less dense.
So, we can conclude that the ball filled with water will reach the bottom of the container first this is because oil is less dense than water and so the glass ball filled with oil will be a lot less denser than the one which is filled with water.
50 μC of negative charge is placed on an insulating pith ball and lowered into a insulating plastic container, suspended from an insulating thread attached to the lid of the box. After the box is entirely sealed, the electric flux through the sides of the box is:_______
a. 5. 65 Times 10^6 N m^2/C.
b. 5. 65 Times 10^5 N m^5/C.
c. -5. 65 Times 10^6 N m^2/C.
d. 50 x 10^-6 N m^2/C.
e. -5.65 Times 10^5 N m^2/C.
f. Can't tell unless the dimensions of the box are given.
Answer:
c. [tex]-5. 65 \times 10^6 N m^2/C.[/tex]
Explanation:
The calculation of the electric flux through the sides of the box is shown below:-
Negative charge in insulating pitch ball, [tex]q = 50\times 10^{-6}[/tex]
[tex]Permittivity = 8.854 \times 10^{-12} F/m[/tex]
Now, we are placing the values into the formula which is here below:-
[tex]Flux = \frac{Negative\ charge}{Permittivity}[/tex]
[tex]= \frac{50\times 10^{-6}}{8.854 \times 10^{-12}}[/tex]
= [tex]-5. 65 \times 10^6 N m^2/C.[/tex]
Therefore we divided the negative charge by permittivity to reach out the electric flux through the sides of the box.
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and position. A motorcyclist travels around a curved path that has a radius of 250 ft . While traveling around the curved path, the motorcyclist increases speed by 1.05 ft/s2 . Part A - Finding the time interval for the motorcyclist to reach a given acceleration If the motorcyclist starts from rest, determine the time needed to reach an acceleration of 4.25ft/s2. Express your answer to three significant figures and include the appropriate units.
Answer:
a^2 = ar^2 + al^2 where ar is the radial acceleration and al is the
linear acceleration - since vectors ar and al are at right angles
ar^2 = a^2 - al^2 = 4.25^2 - 1.05^2
ar = 4.12 ft/s^2
ar = V^2 / R where ar is the radial acceleration
So V^2 = ar * R = 4.12 * 250 = 1030 ft^2/s^2
V = 32.1 m/s the linear speed of the cycle
Also, V = al t or t = V / al = 32.1 / 1.05 = 30.6 sec
A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:_________
a) 10 m/s; 9.8 m/s2.
b) 9.0 m/s; 9.0 m/s2.
c) 9.0 m/s; 9.8 m/s2.
d) 1.0 m/s; 9.8 m/s2.
e) 1.0 m/s; 9.0 m/s2.
Answer:
a) 10 m/s; 9.8 m/s²
Explanation:
After leaving the bench the boy undergoes a semi-projectile motion from top to bottom. Neglecting the effects of air friction, we can safely assume that the horizontal velocity remains constant throughout the motion. Thus, the magnitude of velocity is 10 m/s. The only acceleration in the boy is the acceleration due to gravity, due to free fall motion. Therefore, the magnitude of acceleration is 9.8 m/s².
Therefore, One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:
a) 10 m/s; 9.8 m/s²
The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spring with a transverse wave showing the position of the mass on the spring will make a transverse wave shape if bouncing up and down and moving horizontally. The shortest spring has the mass at the top of each crest and are labeled A and E respectively. The position where the mass in in the trough of the transverse wave would be labeled C and stretches the spring the farthest. The middle length springs has masses where the equilibrium of the material of the medium containing the transverse wave would be at the equilibrium of the material would be and are labeled B and D. Which describes the system at point D? The velocity has the maximum upward value because the acceleration upward is at a maximum.
Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
The system at point D is defined by "The net force is zero, so the acceleration is zero"
What is Net force?When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.
Given the image which shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spring with a transverse wave showing the position of the mass on the spring will make a transverse wave shape if bouncing up and down and moving horizontally.
The shortest spring has the mass at the top of each crest and are labeled A and E respectively. The position where the mass in in the trough of the transverse wave would be labeled C and stretches the spring the farthest. The middle length springs has masses where the equilibrium of the material of the medium containing the transverse wave would be at the equilibrium of the material would be and are labeled B and D.
At D, the net force becomes zero. This makes the system to be in equilibrium or moving with constant velocity. The acceleration becomes zero.
Thus, the system at point D is defined as "The net force is zero, so the acceleration is zero"
Learn more about net force.
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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107 N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth? The bulk modulus of water is 2.3 × 109 Pa.
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
[tex]K = - V\cdot \frac{dP}{dV}[/tex]
Where:
[tex]K[/tex] - Bulk module, measured in pascals.
[tex]V[/tex] - Sample volume, measured in cubic meters.
[tex]P[/tex] - Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
[tex]-\frac{K \,dV}{V} = dP[/tex]
This resultant expression is solved by definite integration and algebraic handling:
[tex]-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP[/tex]
[tex]-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}[/tex]
[tex]\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}[/tex]
[tex]\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }[/tex]
The final volume is predicted by:
[tex]V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }[/tex]
If [tex]V_{o} = 1\,m^{3}[/tex], [tex]P_{o} - P_{f} = -10132500\,Pa[/tex] and [tex]K = 2.3\times 10^{9}\,Pa[/tex], then:
[tex]V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }[/tex]
[tex]V_{f} \approx 0.996\,m^{3}[/tex]
Change in volume due to increasure on pressure is:
[tex]\Delta V = V_{o} - V_{f}[/tex]
[tex]\Delta V = 1\,m^{3} - 0.996\,m^{3}[/tex]
[tex]\Delta V = 0.004\,m^{3}[/tex]
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
A charged particle moves from point A to point B in an external electric field, and in the process its kinetic energy decreases fro, 87.6 J at A to 57.3 J at B. The electric potential at A is -48.0 V, and the electric potential at B is 18.0 V. What is the charge of the particle, including sign
Answer:
-0.46 C
Explanation:
The relationship between total kinetic energy, KE, and total electric potential, V is:
ΔKE = ΔV * q
where ΔKE = change in kinetic
ΔV = change in voltage
q = charge
The kinetic energy changes from 87.6 J to 57.3 J while the electric potential changes from -48.0 V to 18.0 V.
Therefore:
57.3 - 87.6 = (18.0 - (-48.0)) * q
-30.3 = 66q
=> q = -30.3 / 66
q = -0.46 C
The charge of the particle is -0.46 C.
Suppose I have an infinite plane of charge surrounded by air. What is the maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs
Answer:
[tex]53.1\mu C/m^2[/tex]
Explanation:
We are given that
Electric field,E=[tex]3\times 10^6V/m[/tex]
We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.
We know that
[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]3\times 10^6=\frac{\sigma}{2\times 8.85\times 10^{-12}}[/tex]
[tex]\sigma=3\times 10^6\times 2\times 8.85\times 10^{-12}[/tex]
[tex]\sigma=5.31\times 10^{-5}C/m^2[/tex]
[tex]\sigma=53.1\times 10^{-6}C/m^2=53.1\mu C/m^2[/tex]
[tex]1\mu C=10^{-6} C[/tex]