Answer:
God is omnipresent.
Explanation:
This means God is everywhere and He works where ever we are in the world
Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1.0 cm. What is the flow rate R
Answer:
The flow rate is [tex]R =2.357 *10^{-4} \ m^3/s[/tex]
Explanation:
From the question we are told that
The velocity is [tex]v = 3.0 \ m/s[/tex]
The diameter of the pipe is [tex]d = 1.0 \ cm = 0.01 \ m[/tex]
The radius of the pipe is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.01}{2}[/tex]
[tex]r = 0.005 \ m[/tex]
The flow rate is mathematically represented as
[tex]R = v * A[/tex]
Where is the cross-sectional area of the pipe which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.005)^2[/tex]
[tex]A = 7.855 * 10^{-5} \ m^2[/tex]
So
[tex]R = 3.0 * 7.855 *10^{-5}[/tex]
[tex]R = 2.357*10^{-4} \ m^3 /s[/tex]
Current folw in which dirction
In practice, a good insulator In practice, a good insulator A. slows heat flow. B. speeds negative heat flow. C. stops heat flow. D. all of the above
Answer:
The answer is A. slows heat flow.Explanation:
An insulator is a material that impedes the movement of heat or electric current from flowing.
Theoretically good heat insulators stops the movement of heat, while practically this insulation can only be slowed down.
Hence from the options listed the correct answer practically is
A. slows heat flow.15. The
of a sine wave is the time it takes to complete one cycle of the wave.
O A. maximum amplitude
O B. minimum amplitude
O C. average value
O D. wavelength
That TIME is called the "period" of the wave.
(It's not one of the choices.)
When static equilibrium is established for a charged conductor, the electric field just inside the surface of the conductor is
Answer:
The electric field just inside the charged conductor is zero.
Explanation:
Electric field is defined as the region where electrical force is experienced by an electric charge usually as a result of the presence of another electric charge. A charged conductor is said to be in electrostatic equilibrium when it is in an electrostatically balanced state. This simply means a state in which the free electrical charges in the charged conductor have stopped moving.
For any charged conductor that has attained electrostatic equilibrium, the electric field at any point below the surface of the charged conductor falls to zero. Hence the electric field just inside the charged conductor is zero.
Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a stopwatch to find the time it takes the ball to roll each distance
Answer:
The Independent variable in this experiment is the time taken by the ball to roll down each distance.
The dependent variable is the distance through which the ball rolls
The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.
Explanation:
The complete question is
Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?
Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.
A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.
The relationship between the dependent and independent variables in an experiment is given as
y = f(x)
where y is the output or the dependent variable,
and x is the independent variable.
Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.
If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:
5.19 x 10³Hz
Explanation:The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;
[tex]X_C = \frac{1}{2\pi fC }[/tex]
Where;
f = frequency of the signal through the capacitor
C = capacitance of the capacitor.
Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;
V = I x [tex]X_{C}[/tex] [Substitute the value of
=> V = I x [tex]\frac{1}{2\pi fC}[/tex] [Make f the subject of the formula]
=> f = [tex]\frac{I}{2\pi VC}[/tex] ---------------------(i)
From the question;
I = 3.33mA = 0.00333A
C = 8.50nF = 8.50 x 10⁻⁹F
V = 12.0V
Substitute these values into equation (i) as follows;
f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex] [Taking [tex]\pi[/tex] = 3.142]
f = 5.19 x 10³Hz
Therefore, the frequency is closest to f = 5.19 x 10³Hz
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously.
Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:
"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."
Answer:
[tex]\mu_{sB}=0.126[/tex]
[tex]\mu_{sC}=0.168[/tex]
Explanation:
In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:
Sum of torques:
[tex]\sum \tau_{A}=0[/tex]
[tex]N(3m)-W(1.5m)=0[/tex]
When solving for N we get:
[tex]N=\frac{W(1.5m)}{3m}[/tex]
[tex]N=\frac{(1962N)(1.5m)}{3m}[/tex]
[tex]N=981N[/tex]
Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:
First, the forces in y.
[tex]\sum F_{y}=0[/tex]
[tex]-F_{By}+N_{c}=0[/tex]
[tex]F_{By}=N_{c}[/tex]
Next, the forces in x.
[tex]\sum F_{x}=0[/tex]
[tex]-f_{sB}-f_{sC}+P_{x}=0[/tex]
We can find the x-component of force P like this:
[tex]P_{x}=360N(\frac{4}{5})=288N[/tex]
and finally the torques about C.
[tex]\sum \tau_{C}=0[/tex]
[tex]f_{sB}(1.75m)-P_{x}(0.75m)=0[/tex]
[tex]f_{sB}=\frac{288N(0.75m)}{1.75m}[/tex]
[tex]f_{sB}=123.43N[/tex]
With the static friction force in point B we can find the coefficient of static friction in B:
[tex]\mu_{sB}=\frac{f_{sB}}{N}[/tex]
[tex]\mu_{sB}=\frac{123.43N}{981N}[/tex]
[tex]\mu_{sB}=0.126[/tex]
And now we can find the friction force in C.
[tex]f_{sC}=P_{x}-f_{xB}[/tex]
[tex]f_{sC}=288N-123.43N=164.57N[/tex]
[tex]f_{sC}=N_{c}\mu_{sC}[/tex]
and now we can use this to find static friction coefficient in point C.
[tex]\mu_{sC}=\frac{f_{sC}}{N}[/tex]
[tex]\mu_{sC}=\frac{164.57N}{981N}[/tex]
[tex]\mu_{sB}=0.168[/tex]
A converging lens has the focal length of 25 cm. A 10-cm object is placed at 30 cm in front of the lens. How far is the image from the lens? What is the size of the image?
Explanation:
Given that,
Focal length of a converging lens, f = +25 cm
Size of the object, h = 10 cm
Object distance, u = -30 cm
We need to find the image distance and the size of the image.
Using lens formula, [tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex], v is image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{25}+\dfrac{1}{(-30)}\\\\v=150\ cm[/tex]
Magnification,
[tex]m=\dfrac{v}{u}=\dfrac{h'}{h}[/tex]
h' is the size of image
[tex]h'=\dfrac{vh}{u}\\\\h'=\dfrac{150\times 10}{(-30)}\\\\h'=-50\ cm[/tex]
So, the image distance is 150 cm the size of image is 50 cm.
Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answer:
The inductance is [tex]L = 40\mu H[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 163 \ turns[/tex]
The diameter is [tex]D = 6.13 \ mm = 6.13 *10^{-3} \ m[/tex]
The length is [tex]l = 2.49 \ cm = 0.0249 \ m[/tex]
The radius is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{6.13 *10^{-3}}{2}[/tex]
[tex]r = 3.065 *10^{-3} \ m[/tex]
The inductance of the Tarik's solenoid is mathematically represented as
[tex]L = \frac{\mu_o * N^2 * A }{l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
A is the area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 3.065*10^{-3}]^2[/tex]
[tex]A = 2.952*10^{-5} \ m^2[/tex]
substituting values into formula for L
[tex]L = \frac{ 4\pi *10^{-7} * [163]^2 * 2.952*10^{-5} }{0.0249 }[/tex]
[tex]L = 40\mu H[/tex]
23.15. Can an object carry a charge of 2.0 10-19 C?
Answer:
Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)
Where the charge of a single proton is:
C = 1.6x10^-19 C
Now, you need to remember that when we are working with charges, we are working with discrete math:
What means that?
If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)
So the consecutive charge will be:
C = 2*1.6x10^-19 C = 3.2x10^-19 C.
So, because we are working in discrete math, we can not have any object that has charge between 1.6x10^-19 C and 3.2x10^-19 C.
Particularly, 2.0x10^-19 C is in that range, so we can conclude that:
No, an object can not carry a charge of 2.0x10^-19 C.
You shine unpolarized light with intensity 54.0 W/m^2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 19.0 W/m^2. Find the angle between the polarizing axes of the two polarizers.°
Answer:
The angle between the polarizing axes of the two polarizers is 54°
Explanation:
Given;
intensity of unpolarized light, I₀ = 54.0 W/m²
intensity of light that emerges from second ideal polarizer, I₁ = 19.0 W/m²
The angle between the polarizing axes of the two polarizers is dtermined by applying Malus' law for intensity of a linearly polarized light passing through a polarizer.
I₁ = I₀Cos²θ
Cos²θ = I₁ / I₀
Cos²θ = 19 / 54
Cos²θ =0.3519
Cos θ = √0.3519
Cos θ = 0.5932
θ = Cos⁻¹(0.5932)
θ = 53.6°
θ = 54°
Therefore, the angle between the polarizing axes of the two polarizers is 54°
3. Identify the mathematical relationship that exists between pressure and volume, when temperature and quantity are held constant, as being directly proportional or inversely proportional. Explain your answer and write an equation that relates pressure and volume to a constant, using variables
Answer:
P = cte / V
therefore pressure and volume are inversely proportional
Explanation:
For this exercise we can join the ideal gases equation
PV = n R T
they indicate that the amount of matter and the temperature are constant, therefore
PV = cte
P = cte / V
therefore pressure and volume are inversely proportional
The driver of a stationary car hears a siren of an approaching police car at a frequency of 280Hz. If the actual frequency of the siren is 240Hz, find the speed of the police car (speed of sound is 343m/s).
Answer:
The speed of the police car is 294 m/s
Explanation:
Given;
frequency of the siren in air, f = 280 Hz
speed of sound in air, v = 343 m/s
Determine the wavelength of the sound in air to the stationary car:
v = fλ
where;
λ is wavelength of the sound
λ = v/f
λ = 343 / 280
λ = 1.225 m
Now, determine the speed at which the police car is approaching the stationary car;
The actual frequency of the police car, F = 240 Hz
V = Fλ
Where;
V is speed of the police car
λ is the distance between the police car and the stationary car, (wavelength)
V = 240 x 1.225
V = 294 m/s
Therefore, the speed of the police car is 294 m/s
When an object has a net force of zero, then it is said to be in ____________.
Answer:
Equilibrium
Explanation:
Physics terminology I guess? Equilibrium means that an object isn't moving.
Answer:
Balanced Forces
Explanation:
When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. A plane will fly at constant velocity if the acceleration is zero.
Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. The number of fringes per unit length on the screen can be doubled:______.
a. if the distance between the slits is doubled.
b. if the wavelength is changed to λ = λ/2.
c. if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ.
d. if any of the above occurs.
e. only if the width of the slits is changed to w = w/2
Answer:
d. if any of the above occurs
Explanation:
That is The number of fringes per unit length on the screen can be doubled if
if the distance between the slits is doubled.
if the wavelength is changed to λ = λ/2.
And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
The maximum velocity of a particle executing simple harmonic motion with amplitude 7.0 mm is 4.4 m/s. The period of oscillation is? A. 0.5 s B. 0.1 s C. 100 s D. 0.01 s E. 10 s Reset Selection
Answer:
D. 0.01 s
Explanation:
The maximum speed is the amplitude times the angular frequency.
v = Aω
4.4 m/s = (0.0070 m) ω
ω = 628.6 rad/s
The period is:
ω = 2π / T
T = 2π / ω
T = 2π / (628.6 rad/s)
T = 0.01 s
a large crane has a mass of 8500kg calculate the weight of the crane
Answer:
Weight is 83 385 N
Explanation:
Weight is calculated by multiplying the mass by the gravitational acceleration constant
Weight = mass* gravity
Assuming that the gravitational constant is 9.81 m/s^2
Weight = mass* gravity
Weight of crane = (8500 kg)*(9.81 m/s^2)
Weight = 83 385 kg*m/s^2 or 83 385 N
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .
Answer:
The weight of the rod is 32.87 N
Explanation:
Density of the rod = 7800 kg/m
length of the rod = 81.2 cm = 0.812 m
diameter of rod = 2.60 cm = 0.026 m
acceleration due to gravity = 9.80 m/s^2
The rod can be assumed to be a cylinder.
The volume of the rod can be calculated as that of a cylinder, and can be gotten as
V = [tex]\frac{\pi d^{2} l}{4}[/tex]
where d is the diameter of the rod
l is the length of the rod
V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3
We know that the mass of a substance is the density times the volume i.e
mass m = ρV
where ρ is the density of the rod
V is the volume of the rod
m = 4.3 x 10^-4 x 7800 = 3.354 kg
The weight of a substance is the mass times the acceleration due to gravity
W = mg
where g is the acceleration due to gravity g = 9.80 m/s^2
The weight of the rod W = 3.354 x 9.80 = 32.87 N
A 20 g bullet strikes a 1.5 kg wood block and becomes embedded in it (all horizontal motion). The coefficient of friction between the block and the surface is 0.3 and the impact makes the block move a distance of 7 m before it comes to rest. What was the muzzle speed (in m/s) of the bullet? Report your answer with two significant figures.
Answer:
Explanation:
Kinetic frictional force acting on the block = μ mg
where μ is coefficient of friction , m is mass of block.
.3 x 1.5 x 9.8 = 4.41 N .
Let v be the velocity of bullet + block after collision
kinetic energy of composite mass after the strike
= 1 /2 x 1.52 x v²
this will be equal to work done by friction .
.76 v² = 4.41 x 7
v² = 40.62
v = 6.37 m /s
Now we can obtain muzzle speed of bullet by applying conservation of momentum .
Let this speed be u
initial momentum of bullet
= .02 x u
final momentum of composite mass
= 1.52 x 6.37
.02 u = 1.52 x 6.37
u = 484.12 m /s .
= 480 m /s ( in two significant figures )
A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a location a perpendicular distance of 0.053 m from the wire due to the current in the wire
Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
a figures skater rotating at 5 rads with arms extended has a moment of inertia of 2.25 kg. if the arms are pulled in so the moment of inertia decrease to 1.8 what is the final angular speed
Answer:
The final angular speed is 6.25 rad/s
Explanation:
Given;
initial angular speed, ω₁ = 5 rad/s
initial moment of inertia, I₁ = 2.25 kg.m²
Final moment of inertia, I₂ = 1.8 kg.m²
final angular speed, ω₂ = ?
Based on conservation of angular momentum, we will have the following expression;
ω₁I₁ = ω₂I₂
ω₂ = (ω₁I₁ ) / I₂
ω₂ = (5 x 2.25) / 1.8
ω₂ = 6.25 rad/s
Therefore, the final angular speed is 6.25 rad/s
A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.
Answer:
Maximum acceleration is 800m/s^2
Explanation:
See attached file
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
The greater the frequency of the waves, the ____________ the pitch.
Answer:
Higher.
Explanation:
The greater the frequency the bigger the amplitude gets and the greater pitch gets.
Think - more energy, bigger waves, more waves, and higher sound
An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr,where α is a positive constant.
A). What is the magnitude of the electric field at a distance r from the center of the shell, where a
Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a
Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.
Answer:
E = α/2∈₀ [ 1 - a²/r² ]
Ф = α/2∈₀
Explanation:
Using Gauss Law:
ρ(r) = a/r, dA
= 4 π r²d r
Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA
Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA
= 4πα [tex]\int\limits^r_a[/tex] r'dr'
Ф[tex]_{encl}[/tex] = 4 π α 1/2(r²-a²)
E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀
= [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)
= α (r² - a²) / 2 ∈₀ (r²)
= α/2∈₀ [ r²/r² - a²/r² ]
E = α/2∈₀ [ 1 - a²/r² ]
Electric field of the point charge:
E[tex]_{q}[/tex] = q / 4π∈₀r²
[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²
For [tex]E_{total}[/tex] to be constant:
- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²
-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀
= α - αa² + αa² / 2 ∈₀
= α /2 ∈₀
Hence:
Ф = α/2∈₀
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.
Answer:
4.9x10^-6T
Explanation:
See attached file
A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s. m/s
Answer:
a)
6t i + t² j m
b)
6 i + 2 t j m/s
c)
x = 24 m and y = 16 m
d)
10 m/s
Explanation:
Explanation is given in the attached document.
A ballistic pendulum consists of a 4 kg wooden block originally at rest at θ = 0o . When a 2 g bullet strikes and becomes embedded it, it is observed that the block swings upward to a maximum angle of θ = 6o . Estimate the speed of the bullet just before the impact.
Answer:
733m/s
Explanation:
Assuming that Just after impact:
given that at lowest point Is
T2 + V2 = T3 + V3
1 /2(4 + 0.002) (vB)²2 + 0 = 0 + (4 + 0.002)(9.81)(1.25)(1 -cos 6°)
(vB)2 = 0.3665
m>s For the system of bullet and block: ( S+ ) Σmv1 = Σmv2 0.002(vB)1 = (4 + 0.002)(0.3665) (vB)1 = 733 m/s