Answer:
The mass of the hailstone is [tex]m_2 = 0.05[/tex]
Explanation:
From the question we are told that
The mass of the hockey puck is [tex]m_1 = 0.2 \ kg[/tex]
The initial speed of the puck is [tex]u_1 = 10 \ m/s[/tex]
The final speed of the puck is [tex]v_f = 8 \ m/s[/tex]
According to the principles of linear momentum
[tex]m_1u_1 + m_2u_2 = (m_1 + m_2)v_f[/tex]
At initial the velocity of the hailstone was zero so
[tex]u_2 = 0 \ m/s[/tex]
So
[tex]m_1 u_1 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex]0.2 * 10 = (0.2 + m_2 )* 8[/tex]
=> [tex]2 = (0.2 + m_2 )* 8[/tex]
=> [tex]0.25 = 0.2 + m_2[/tex]
=> [tex]m_2 = 0.05[/tex]
If you slide down a rope, it's possible to create enough thermal energy to burn your hands or your legs where they grip the rope. Suppose a 30 kg child slides down a rope at a playground, descending 2.5 m at a constant speed.
How much thermal energy is created as she slides down the rope?
Answer:
Q = 735 J
Explanation:
In this exercise we must assume that all the mechanical energy of the system transforms into cemite energy.
Initial energy
Em₀ = U = m g h
final energy
[tex]Em_{f}[/tex] = Q
Em₀ = Em_{f}
m g h = Q
let's calculate
Q = 30 9.8 2.5
Q = 735 J
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.
g You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 78 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
F=49.48 N
Explanation:
Given that
Diameter , d= 30 mm
Holding pressure = 70 % P
P=Atmospherics pressure
We know that
P= 1 atm = 10⁵ N/m²
The force per unit area is known as pressure.
[tex]P=\dfrac{F}{A}[/tex]
[tex]F=P\times A[/tex]
[tex]F=0.7\times 10^5\times \dfrac{\pi}{4}\times 0.03^2\ N[/tex]
Therefore the force will be 49.48 N.
F=49.48 N
An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron's probability of tunneling through the barrier increase
Answer:
factor that the electron's probability of tunneling through the barrier increase 2.02029
Explanation:
given data
kinetic energy = 10.1 eV
height = 18.2 eV
width = 1.00 nm
wavelength = 546 nm
solution
we know that probability of tunneling is express as
probability of tunneling = [tex]e^{-2CL}[/tex] .................1
here C is = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
here h is Planck's constant
c = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c = 2319130863.06
and proton have hf = [tex]\frac{hc}{\lambda } = {1240}{546}[/tex] = 2.27 ev
so electron K.E = 10.1 + 2.27
KE = 12.37 eV
so decay coefficient inside barrier is
c' = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
c' = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c' = 1967510340
so
the factor of incerease in transmisson probability is
probability = [tex]e^{2L(c-c')}[/tex]
probability = [tex]e^{2\times 1\times 10^{-9} \times (351620523.06)}[/tex]
factor probability = 2.02029
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:
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
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
Which statement describes an essential characteristic of data in an experiment?
Answer:
the data must be reliable
Explanation:
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 70.0 km directly north of his starting point.
a. What was the length of the third leg of his trip?b. What was the direction of the third leg of his trip?
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
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]
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.
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
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.
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 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 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.
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.
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 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
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 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 body moving with uniform acceleration has two points (5,15) and (20,60) on the (v t) graphs of its motion. calculate the acceleration
Answer:
3m/s²
Explanation:
The slope on a velocity time graph represents the acceleration, so if you simply use the slope formula, you can find the acceleration between those two points.
m=rise/run
m=(60-15)/(20-5)
m=45/15
m=3 m/s ² squared (therefore this is your constant acceleration from those two points).
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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what are the strengths and weaknesses of the four methods of waste management?
Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;
* It creates employment
* It keeps the environment clean
* The practice is highly lucrative
* It saves the earth and conserves energy
The weaknesses of the methods of waste management includes;
* The sites are often dangerous
* The process is mostly
* There is a need for global buy-in
* The resultant product had a short life
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.
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 8.1 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled x= 0 m. The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.
Answer:
v₀ = 0.5058 m/s
Explanation:
From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m
Now, the potential energy of the block at x = 0.08 m is ½kx²
where;
k is the spring constant given by; k = ω²m
ω is the angular velocity of the oscillation
m is the mass of the block.
Thus, potential energy of the spring at the bottle(x = 0.08 m) is;
U = ½ω²m(0.08m)²
Also, potential energy of the spring at the bottle(x = 0.05 m) is;
U = ½ω²m(0.05m)²
and the kinetic energy of the block at x = 0.05 m is;
K = ½mv₀²
Thus;
½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²
Inspecting this, ½m will cancel out to give;
ω²(0.08)² = ω²(0.05)² + v₀²
Making v₀ the subject, we have;
v₀ = ω√((0.08)² - (0.05)²)
So,
v₀ = 8.1√((0.08)² - (0.05)²)
v₀ = 0.5058 m/s
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 4.00 kg ball is moving at 4.00 m/s to the EAST and a 6.00 kg ball is moving at 3.00 m/s to the NORTH. The total momentum of the system is:___________.A. 14.2 kg m/s at an angle of 48.4 degrees SOUTH of EAST.B. 48.2 kg m/s at an angle of 24.2 degrees SOUTH of EAST.C. 48.2 kg m/s at an angle of 48.4 degrees NORTH of EAST.D. 24.1 kg m/s at an angle of 24.2 degrees SOUTH of EAST.
E. 24.1 kg m/s at an angles of 48.4 degrees NORTH of EAST.
Answer:
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EASTExplanation:
Momentum = mass*velocity of a body
For a 4.00 kg ball is moving at 4.00 m/s to the EAST, its momentum = 4*4 = 16kgm/s
For a 6.00 kg ball is moving at 3.00 m/s to the NORTH;
its momentum = 6*3 = 18kgm/s
Total momentum = The resultant of both momentum
Total momentum = √16²+18²
Total momentum = √580
total momentum = 24.1kgm/s
For the direction:
[tex]\theta = tan^{-1} \frac{y}{x}\\\theta = tan^{-1} \frac{18}{16}\\ \theta = tan^{-1} 1.125\\\theta = 48.4^{0}[/tex]
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST
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.