Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
b
[tex]k =722.2 \ N/m[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]
The period is [tex]T = 17 \ s[/tex]
The test weight is [tex]W = 13 \ N[/tex]
Generally the radial acceleration is mathematically represented as
[tex]a = w^2 r[/tex]
at maximum angular acceleration
[tex]r = A[/tex]
So
[tex]a_{max} = w^2 A[/tex]
Now [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2 * \pi }{T}[/tex]
Therefore
[tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]
substituting values
[tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
Generally this test weight is mathematically represented as
[tex]W = k * A[/tex]
Where k is the spring constant
Therefore
[tex]k = \frac{W}{A}[/tex]
substituting values
[tex]k = \frac{13}{0.018}[/tex]
[tex]k =722.2 \ N/m[/tex]
On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is
a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2
Answer:
C) 8 m/s²
Explanation:
Given:
v₀ = 40 m/s
v = 0 m/s
Δy = 100 m
Find: a
v² = v₀² + 2aΔy
(0 m/s)² = (40 m/s)² + 2a (100 m)
a = -8 m/s²
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus
Answer:
2.55m
Explanation:
Using 1/do+1/di= 1/f
di= (1/f-1/do)^-1
( 1/0.0500-1/0.0510)^-1
= 2.55m
In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?
Answer:
Explanation:
Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least and kinetic energy is maximum .
In this way , there is conservation of mechanical energy .
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
Find out more information about Force here:
https://brainly.com/question/25239010
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
Learn more about friction force:
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A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.
Answer:
27°
Explanation:
The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)
So theta = arcsin(0.45)
=27°
The angle between the wire and the magnetic field is 27°.
Calculation of the angle:Since The magnetic force per meter on a wire is measured to be only 45 %
So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field
Therefore,
theta = arcsin(0.45)
=27°
Hence, The angle between the wire and the magnetic field is 27°.
Learn more about wire here: https://brainly.com/question/24733137
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?
Answer:
39.94m/s.Explanation:
Kinetic energy is expressed as KE = 1/2 mv² where;
m is the mass of the body
v is the velocity of the body.
For the heavier shell;
m = 5kg
KE gained = 100J
Substituting this values into the formula above to get the velocity v;
100 = 1/2 * 5 * v²
5v² = 200
v² = 200/5
v² = 40
v = √40
v = 6.32 m/s
Note that after the explosion, both body fragments will possess the same velocity.
For the lighter shell;
mass = 2.0kg and v = 6.32m/s
KE of the lighter shell = 1/2 * 2 * 6.32²
KE of the lighter shell = 6.32²
KE of the lighter shell= 39.94m/s
Hence, the lighter one gains a kinetic energy of 39.94m/s.
The gain in the kinetic energy of the smaller fragment is 249.64 J.
The given parameters;
Mass of the shell, m = 7.0 kgMass of one fragment, m₁ = 2.0 kgMass of the second fragment, m₂ = 5.0 kgKinetic energy of heavier fragment, K.E₁ = 100 JThe velocity of the heavier fragment is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]
Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;
[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]
The gain in the kinetic energy of the smaller fragment is calculated as follows;
[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]
Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.
Learn more about conservation of linear momentum here: https://brainly.com/question/7538238
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
Which has more mass electron or ion?
Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is
Answer:
If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE
Explanation
This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.
Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra
Answer:
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Explanation:
Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:
[tex]v = v_{o} + g\cdot t[/tex]
[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.
[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.
Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:
[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]
[tex]v = -196.14\,\frac{m}{s}[/tex]
[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]
[tex]y-y_{o} = -1961.4\,m[/tex]
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda = 622 nm[/tex]
Explanation:
From the question we are told that
The distance of the slit to the screen is [tex]D = 5 \ m[/tex]
The order of the fringe is m = 6
The distance between the slit is [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]
The fringe distance is [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]
Generally the for a dark fringe the fringe distance is mathematically represented as
[tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]
=> [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]
substituting values
=> [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]
=> [tex]\lambda = 6.22 *10^{-7} \ m[/tex]
[tex]\lambda = 622 nm[/tex]