Answer:
The maximum vertical displacement is 2.07 meters.
Explanation:
We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:
[tex]E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1[/tex]
Where [tex]U_e_0[/tex] is the elastic potential energy stored in the spring, [tex]K_1[/tex] is the kinetic energy of the block at point 1, [tex]U_g_1[/tex] is the gravitational potential energy of the block at point 1, and [tex]W_f_1[/tex] is the work done by friction at point 1.
Now, rearranging the equation we obtain:
[tex]\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1[/tex]
Where [tex]k[/tex] is the spring constant, [tex]x[/tex] is the compression of the spring, [tex]m[/tex] is the mass of the block, [tex]v_1[/tex] is the speed at point 1, [tex]g[/tex] is the acceleration due to gravity, [tex]h_1[/tex] is the vertical height of the block at point 1, [tex]\mu[/tex] is the coefficient of kinetic friction, [tex]N[/tex] is the magnitude of the normal force and [tex]s_1[/tex] is the displacement of the block along the ramp to point 1.
Since the force is in an inclined plane, the normal force is equal to:
[tex]N=mg\cos\theta[/tex]
Where [tex]\theta[/tex] is the angle of the ramp.
We can find the height [tex]h_1[/tex] using trigonometry:
[tex]h_1=s_1\sin\theta[/tex]
Then, our equation becomes:
[tex]\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}[/tex]
Plugging in the known values, we get:
[tex]v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s[/tex]
Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:
[tex]v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s[/tex]
Now, we have:
[tex]y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m[/tex]
Finally, the maximum vertical displacement [tex]h_2[/tex] is equal to the height [tex]h_1[/tex] plus the vertical displacement [tex]y[/tex]:
[tex]h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m[/tex]
It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.
Why does sound travel faster in water than in air?
A.
because water is a denser medium than air
B.
because air is a denser medium than water
C.
because water has a larger wavelength than air
D.
because air has a larger wavelength than water
Answer:
C.
Explanation:
Un móvil, que sale desde un punto situado 3 metros a la izquierda del origen y lleva un movimiento uniforme, se sitúa a 12 metros a la derecha del origen al cabo de 3 segundos. Tras esto invierte el sentido del movimiento, empleando 4 segundos más en llegar al origen. Realiza una gráfica que represente el movimiento descrito. Obtén el desplazamiento en cada tramo. (Sol: 15 m; −12 m) Determina la distancia total recorrida por el móvil. (Sol: 27 m) Determina la velocidad del móvil en cada tramo. (Sol: 5 m/s; −3 m/s) Representa gráficamente la velocidad del móvil frente al tiempo.
Answer:
x₁ = 15 m, x₂ = 12 m , x_total = 27m, v₁ = 5 m / s , v₂ = - 3 m / s
Explanation:
In this exercise we will use the kinematics of uniform motion
v = d / t
let's apply this equation for the first move
v₁ = Δx / t = (x₂-x₀) / t
v₁ = (12- (-3)) / 3
v₁ = 5 m / s
the distance traveled is x₁ = 15 m
Now let's analyze the return movement
v₂ = Δx / dt
v₂ = (0 - 12) / 4
v₂ = - 3 m / s
The negative sign indicates that the vehicle is moving to the left
the distance traveled is x₂ = 12 m
The total dystonia is
x_total = x₁ + x₂
x_total = 15 +12
x_total = 27m
In the attached we have the graphics of the movement
A migrating salmon heads in the direction N 45° E, swimming at 5 mi/h relative to the water. The prevailing ocean currents flow due east at 2 mi/h. Find the true velocity of the fish as a vector. (Assume that the i vector points east, and the j vector points north.)
Answer:
Explanation:
velocity of salmon with respect to water, v(s,w) = 5 mi/h at N 45° E
velocity of water with respect to ground, v(w,g) = 2 mi/h due east
Let the true velocity of salmon is velocity of salmon with respect to water is v(s,g)
First write the velocities in vector from
[tex]\overrightarrow{v}_{s,w}=5(Cos 45\widehat{i}+Sin 45\widehat{j})=3.54\widehat{i}+3.54\widehat{j}[/tex]
[tex]\overrightarrow{v}_{w,g}=2\widehat{i}[/tex]
Using the formula of relative speed,
[tex]\overrightarrow{v}_{s,w}=\overrightarrow{v}_{s,g}-\overrightarrow{v}_{w,g}[/tex]
[tex]3.54\widehat{i}+3.54\widehat{j}=\overrightarrow{v}_{s,g}-2\widehat{i}[/tex]
[tex]\overrightarrow{v}_{s,g}=5.54\widehat{i}+3.54\widehat{j}[/tex]
This i the true velocity of salmon.
The true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].
The given parameters;
velocity of the salmon relative to the water, = 5 mi/hdirection of the velocity, θ = 45⁰ocean currents, = 2 mi/hThe true velocity of the fish as a vector is calculated as follows;
[tex]v_f = v\ cos(\theta)i \ + \ v\ sin(\theta)j \ + \ 2i\\\\v_f = 5cos(45) i \ + 5sin(45)j \ + \ 2i\\\\v_f = 3.54i \ + \ 3.54j \ + 2i\\\\v_f = 5.54i \ + \ 3.54 j[/tex]
Thus, the true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].
Learn more here:https://brainly.com/question/4945130
All particles made of quarks are called _____.
electrons
gluons
hadrons
nucleons
Answer:
All particles made of quarks are called hadrons.
All particles madeup of two or more quarks held together by a force are called ; ( C ) Hadrons
What are Hadrons ?Hadrons are subatomic particles which are made of two or more quarks which are held firmly by a force similar to electric force. An example of Hadron is Baryon which contains three quarks which are held closely together.
Hence we can conclude that All particles made of quarks are called Hadrons
Learn more about Hadrons : https://brainly.com/question/3734714
How does the eye and brain work together to give you perception of color
Answer:
The human eye and brain together translate light into color. Light receptors within the eye transmit messages to the brain, which produces the familiar sensations of color. ... Rather, the surface of an object reflects some colors and absorbs all the others. We perceive only the reflected colors.
Explanation:
Final answer:
The eye and brain work together to perceive color. Light enters the eye and is detected by cones in the retina, which send signals to the brain. The brain processes these signals and combines information from different cones to create the perception of color.
Explanation:
The eye and brain work together to give you perception of color through a complex process. When light enters the eye, it is detected by special cells called cones in the retina. These cones are sensitive to different wavelengths of light and send signals to the brain. The brain then processes these signals and combines the information from different cones to create the perception of color.
For example, if you see a red object, the red cones in your eye will be activated and send signals to the brain. The brain then interprets these signals as the color red. Similarly, the green cones are activated by green light and the blue cones by blue light. By combining the information from all three types of cones, the brain is able to perceive a wide range of colors.
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in the bearings causes the wheel to stop in just 13 s . If the moment of inertia of the wheel about its axle is 0.33 kg⋅m2 , what is the magnitude of the frictional torque?
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
Work is done on a ball when a soccer player kicks it. Is the player still doing work on the ball as it rolls across the ground? Explain
Answer:
Energy is conserved.
Explanation:
According to law of conservation of energy ,energy can neither be created nor be destroyed .
When the player kicks the soccer ball, the kinetic energy is transferred to the ball and it will roll for a while and then stops. The combination of force and distance is called as work.Greater the force faster the ball will go.
The force with which the soccer player player kicked the ball will be having it effects on the ball. As the ball rolls in the ground kinetic energy which the ball got from the player will be lost in the form of heat due to friction.
According to the Newton first law of motion,the object continues to remain in rest or move with constant speed unless acted upon by the external force.
Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the charge on B one-third the charge on A. How far apart would the two spheres then have had to be for A to have had the same deflection that it had before
Answer:
The new separation is [tex]\bf{(d/\sqrt{3})}[/tex].
Explanation:
The expression of the force between two spheres is given by
[tex]F = k\dfrac{q_{A}q_{B}}{d^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where, [tex]q_{A}[/tex] is the charge on sphere A, [tex]q_{B}[/tex] is the charge on sphere B, [tex]k[/tex] is constant and [tex]d[/tex] i the separation between two spheres.
The new value of charge on sphere B is [tex]q_{B}^{n} = \dfrac{q_{A}}{3}[/tex]. Consider the new separation between the spheres be [tex]d'[/tex]. Under the new configuration the force between the spheres is given by
[tex]F = k \dfrac{q_{A}(q_{A}/3)}{d'^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Equating equation (1) and equation (2), we have
[tex]~~~~&& \dfrac{1}{d'^{2}} = \dfrac{1}{3d^{2}}\\&or,& d' = \dfrac{d}{\sqrt{3}}[/tex]
So, the new separation is [tex](d/\sqrt{3})[/tex].
Now, remove the positive charge by dragging it back to the basket, and drag one negative charge toward the middle of the screen. Determine how the voltage is different from that of the positive charge. How does the voltage differ from that of the positive charge? The voltage distribution does not change. The voltages become negative instead of positive and keep the same magnitudes. The voltages are positive, but the magnitude increases with increasing distance.
Answer:
The correct option is;
The voltages become negative instead of positive and keep the same magnitudes
Explanation:
Here we have that there is a change in direction from positive to negative with the direction pointing to opposite of the initial direction whereby the field strength does not change due to the reversal of the. Therefore the voltage becomes negative instead of negative while keeping the the magnitude.
That is there is a electric field reversal or a change electric field direction.
The pressure at the bottom of a jug filled with water does not depend on
105. A cable with a linear density of μ=0.2kg/m is hung from telephone poles. The tension in the cable is 500.00 N. The distance between poles is 20 meters. The wind blows across the line, causing the cable resonate. A standing waves pattern is produced that has 4.5 wavelengths between the two poles. The speed of sound at the current temperature T=20°C is 343.00m/s . What are the frequency and wavelength of the hum?
Answer: 11.5 Hz, 29.83 m
Explanation:
Given
Linear density of the cable, μ = 0.2 kg/m
Tensión in the cables, T = 500 N
Wavelength of the wave, = 4.5 Waves
Distance between the poles, L = 20 m
Temperature of, t = 20° C
Speed of sound, v = 343 m/s
λ = length / number of waves =
λ = 20 / 4.5
λ = 4.44 m
Frequency of a standing wave is the same as frequency of a hum. Calculated using the formula
F = n/2L * √(T/μ)
F = 1/λ * √(T/μ)
F = 1/4.44 * √(500/0.2)
F = 0.23 * √2500
F = 0.23 * 50
F = 11.5 Hz
Wavelength of the hum,
λ = v/f
λ = 343 / 11.5
λ = 29.83 m
The wavelength of the hum produced by the cable is approximately 4.44 meters, and the frequency is approximately 11.26 Hz, calculated using the known linear density, tension, and distance between poles.
Explanation:The student is asking about the properties of a standing wave created when the wind blows across a cable hung between two poles. Given the linear density (µ) of 0.2 kg/m, the tension in the cable (500 N), and the distance between poles (20 meters), we are to find the frequency and wavelength of the hum produced by the cable.
Firstly, we can calculate the wavelength (λ) of the wave using the information that 4.5 wavelengths fit between the two poles that are 20 meters apart:
λ = 20 meters / 4.5 wavelengths = 4.44 meters (approx)
Next, we can find the speed (v) of the wave on the cable using the formula:
v = sqrt(T / µ)
Plugging in the given values, we get:
v = sqrt(500 N / 0.2 kg/m) = sqrt(2500 m2/s2)
v = 50 m/s
Now, using the wave speed (v) and the wavelength (λ), we can calculate the frequency (f) of the wave using the formula:
f = v / λ
f = 50 m/s / 4.44 m = 11.26 Hz (approx)
Therefore, the wavelength is approximately 4.44 meters, and the frequency of the hum is approximately 11.26 Hz.
If a truck driver is driving at 30 mph and she makes a U-turn then starts driving 30 mph in the opposite direction did the driver speed or velocity change after changing direction
The final velocity of the truck driver does not change after making a U-turn and driving in the opposite direction.
Explanation:The final velocity of the truck driver does not change after making a U-turn and driving in the opposite direction. While the truck's speed changes, the velocity remains the same because velocity takes into account both the speed and direction of motion. Since the speed remains constant at 30 mph and the direction changes, the velocity is still 30 mph, just in the opposite direction.
Several years from now you have graduated with an engineering/physics degree from OSU and have been hired by a nanoengineering firm as an intern. You have been assigned to work under a top engineer from the company. Their current project is to design a microscopic oscillator as a time keeping device. The engineering design involves placing a negative charge at the center of a very small positively charged metal ring. Your boss claims that the negative charge will undergo simple harmonic motion of displaced away from the center of the ring. Furthermore, they claim they can change the period (timing) of oscillation by adjusting the amount of charge on the ring. The first task they give you is to check the validity of their design.Consider a charge −???? located a small distance z above the center of a positively charged ring with total charge +Q and radius R. Write an expression for the net force exerted on the charge −???? due to the ring of charge. What is the magnitude of the force on the charge −???? if it is at the location z = 0?
Complete Question in order
See the first image attached
Answer and Explanation:
The three images attached discussed the solution to this question
second Image
Third Image
and Fourth Image
If the images are not clear enough right-click on it and open in a new tab
a force of 20N pushes an object of mass 5.0kg along a rough surface of 5.0N
Answer:
I'm sorry, I don't think there is any answer to give seeing as no question has been asked
Answer:
No question
Explanation: Sorry
A mass m = 1.2 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(Ït â Ï). The positive y-axis points upward.a. Find the angular frequency of oscillations in radians per second.b. Determine the value of A in meters.c. Determine the value of Ï in radians.d. Enter an expression for velocity along y axis as function of time in terms of A, Ï and t using the value of Ï from part c.e. What is the velocity of mass at time t = 0.25 s?f. What is the magnitude of mass's maximum acceleration?
Answer:
a. 3π/2 b. 0.36 m c. 0.234 m/s d. 42.55 m/s²
Explanation:
Here is the complete question
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y ( t ) = A c o s ( ω t − ϕ ) , with the positive y-axis pointing upward. At time t = 0 the mass is observed to be passing through its equilibrium height with an upward speed of v 0 = 3.9 m/s.
A. Find the smallest positive value of ϕ , in radians.
B. Calculate the value of A in meters.
C. What is the mass's velocity along the y-axis in meters per second, at the time t = 0.15 s?
D. What is the magnitude of the mass's maximum acceleration, in meters per second squared?
Solution
a. Since y ( t ) = A c o s ( ω t − ϕ ), the smallest possible value for ϕ is gotten when c o s ( ω t − ϕ ) = 0 ⇒ ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0, ω t − ϕ = ω 0 − ϕ = 0 − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
Since this is a negative angle, we add 2π to the right side.
So, ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
b. Since v = Aω = A√(k/m) where v = maximum velocity at time t = 0 = 3.9 m/s. A = amplitude, k = spring constant = 130 N/m and m = mass = 1.1 kg
A = v/√(k/m) = 3.9 m/s/√(130 N/m/1.1 kg) = 3.9/√118.18 = 3.9/10.87 = 0.36 m
c. To find its velocity, we differentiate y(t)
So, v = dy(t)/dt = dA c o s ( ω t − ϕ )/dt = -'ωAsin( ω t − ϕ ) = v₀sin( ω t − ϕ )
v = v₀sin( ω t − ϕ ) = v₀sin( ω t − ϕ)
Substituting the value of the variables,
v = 3.9sin( 10.87 t − 3π/2)
At t = 0.15 s,
v = 3.9sin( 10.87 × 0.15 − 3π/2)
v = 3.9sin( 1.6305 − 4.7124)
v = -'3.9sin( -3.0819)
v = -'3.9 × - 0.06
v = 0.234 m/s
d. The maximum acceleration, a
a = Aω² = Ak/m = 0.36 × 130/1.1 = 42.55 m/s²
Answer:
a) F = 10.4 rad/s
b) A = 0.375 m
c) ϕ = 3π/2
d) V(t) = -ωAsin( ω t - 3π/2 )
e) V = 0.144 m/s
f) a = 40.625 m/s²
Explanation: Given that
mass m = 1.2 kg
The spring constant k = 130 N/m Time t = 0
Distance d = 0.35 m
y( t ) = A c o s ( ω t − ϕ )
At time t = 0
Speed of Vo = 3.9 m/s.
a) Find the angular frequency of oscillations in radians per second
W = √(k/m)
2πF = √(k/m)
F = 1/2π√(k/m)
F = 1/2π √(130/1.2)
F = 1.66Hz
1 Henz = 6.28 rad/s therefore,
F = 1.66 × 6.28
F = 10.4 rad/s
b) Calculate the value of A in meters.
V = Aω = A√(k/m)
where V = 3.9 m/s the maximum velocity at time t = 0
A = amplitude
A = v/√(k/m)
A = 3.9/√(130/1.2)
A = 3.9/10.4
A = 0.375 m
c. Determine the value of ϕ in radians
If y( t ) = A c o s ( ω t − ϕ ) We can obtain the smallest possible value of ϕ when c o s ( ω t − ϕ ) = 0
ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0,
ω(0) − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
This is a negative angle, let us add 2π to the right side. So,
ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
d. Enter an expression for velocity along y axis as function of time in terms of A, ϕ and t using the value of ϕ from part c.
To find expression for velocity, we differentiate y(t) with respect to time t So,
V = dy/dt = dA c o s ( ω t − ϕ )/dt
V = -ωAsin( ω t − ϕ )
Therefore
V(t) = -ωAsin( ω t + π/2 )
Or
V(t) = -ωAsin( ω t - 3π/2 ) ...... (1)
e. What is the velocity of mass at time t = 0.25 s?
From equation (1)
V(t) = V₀sin( ω t − ϕ )
Substituting the value of the variables,
V = 3.9sin( 10.4t − 3π/2)
At t = 0.25 s,
V = 3.9sin( 10.4 × 0.25 − 3π/2)
V = 3.9sin( 2.6 − 4.7124)
V= -3.9sin( -2.1124)
V= -3.9 × - 0.037
V = 0.144 m/s
f. What is the magnitude of mass's maximum acceleration?
The maximum acceleration a = Aω²
a = Aω² =
a = Ak/m
a = 0.375 × 130/1.2
a = 40.625 m/s²
A diagnostic sonogram produces a picture of internal organs by passing ultrasound through the tissue. In one application, it is used to fid the size, location, and shape of the prostate in preparation for surgery or other treatment. The speed of sound in the prostate is 1540 m/s, and a diagnostic sonogram uses ultrasound of frequency 1.40 MHz. The density of the prostate is 1060 kg/m3.1) What is the wavelength of the sonogram ultrasound? 2)What is Youngâs modulus for the prostate gland?
Answer:
a) 1.1mm
b) 2.513kg/ms^2
Explanation:
You can use the formula for the calculation of the wavelength of a wave
f=1.40MHz=1.40*10^{6}Hz
1 )
[tex]\lambda=\frac{v}{f}=\frac{1540m/s}{1.40*10^{6}Hz}=1.1*10^{-3}m=1.1mm[/tex]
2)
The Young modulus can be computed by using the expression:
[tex]v=\sqrt{\frac{Y}{\rho}}\\\\Y=v^2\rho[/tex]
where Y is the Young modulus and p is the density of the material. Here, you have considered that the prostate gland can be taken as a vibrating membrane or string.
By replacing you obtain:
[tex]Y=(1540m/s)^2(1060kg/m^3)=2.513*10^9kg/ms^2[/tex]
hence, the Young modulus of the prostate glande is 2.513kg/ms^2
Answer:
1) 1.1 x [tex]10^{-3}[/tex] m
2)2.51 x[tex]10^{9}[/tex]Pa
Explanation:
Given:
Speed of sound in prostate 'V'= 1540m/s
frequency 'f' = 1.40 MHz = 1.40 x [tex]10^{6}[/tex]Hz
density of prostate'ρ' = 1060 kg/m3
1) As we know that the relationship of the speed of sound, its frequency, and wavelength is the same as for all waves
V= fλ
λ= V/f => 1540/1.40 x [tex]10^{6}[/tex]
λ= 1.1 x [tex]10^{-3}[/tex] m
Thus, the wavelength of the sonogram ultrasound is 1.1 x [tex]10^{-3}[/tex] m
2)The speed of sound in a solid the depends on the Young's modulus of the medium and the density
V=√Y/ρ.
V² = Y/ρ
Y= V² x ρ=> 1540² x 1060
Y= 2.51 x[tex]10^{9}[/tex]Pa
Thus, Young's modulus for the prostate gland is 2.51 x[tex]10^{9}[/tex]Pa
To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is 5.0×103kg/m35.0×10
3
kg/m
3
. What fraction of the specimen's apparent volume is solid?
Answer:
Fraction of the specimen's is 0.4.
Explanation:
We know,
Mass = volume × density
Weigh= mass × g
= volume × density× g
= density× g × volume
[tex]=\rho.g.V[/tex]
An object weighs less submerged due to buoyant force acting on it.
[tex]\therefore W_{wet}= W_{dry}-B[/tex]
[tex]B= W_{dry}-W_{wet}[/tex]
[tex]=W_{\textrm{fluid displaced}}[/tex]
[tex]=\rho_{fluid}. g.V_{submerged}[/tex]
Given that, the weighs of the specimen in dry is twice of the weighs in air.
[tex]W_{wet}=\frac 12W_{dry}[/tex]
Then ,
[tex]B= W_{dry}-W_{wet}[/tex]
[tex]= W_{dry}-\frac12W_{dry}[/tex]
[tex]=\frac12W_{dry}[/tex]
[tex]=\rho_{Rock}. g.V_{Rock}[/tex]
Therefore,
[tex]\rho_{fluid}. g.V_{submerged}=\frac12\rho_{Rock}. g.V_{Rock}[/tex]
[tex]\Rightarrow \rho_{Rock}. g.V_{Rock}=2\rho_{fluid}. g.V_{submerged}[/tex]
[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\rho_{fluid}. g}{\rho_{Rock}.g}[/tex]
[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\rho_{fluid}}{\rho_{Rock}}[/tex]
[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\times 1.0 \times 10^3\ kg /m^3}{5.0\times 10^3 \ kg/m^3}[/tex]
=0.4
Fraction of the specimen's is 0.4.
Two wave pulses with equal positive amplitudes pass each other on a string, one is traveling toward the right and the other toward the left. At the point that they occupy the same region of space at the same time
constructive interference occurs
destructive interference occurs.
a standing wave is produced.
a traveling wave is produced.
a wave pulse is produced.
Answer:
Constructive interference occurs
Explanation
This is because at the point they both occupy the same region of space at the same time, there is a superposition of the waves causing them to add up. Since their amplitudes are equal and positive, they are in phase. and thus there is increase the resultant amplitude..This increase in amplitude is due to constructive interference.
Two wave pulses with equal positive amplitudes moving in opposite directions on a string will result in constructive interference where their displacements add up leading to an increased amplitude at the point of overlap. This is a momentary or transient event.
Explanation:When two wave pulses with equal positive amplitudes meet each other on a string, moving in opposite directions, what we experience is constructive interference. This happens because the displacement of the two waves at every point adds up. Essentially, they combine to produce a wave of increased amplitude. In this case, the resultant wave would have an amplitude that's twice that of the individual wave pulses. Please note: this phenomenon happens only at the instant when both waves overlap and not any time before or after that. In other words, it's a transient or temporary phenomenon, not a permanent one, causing a momentary increase in the height of the wave at the point of intersection.
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The cytoplasm is the watery fluid found within cells. The cytoplasm holds all of the organelles, except _______, in place within the cell.
A.
chloroplasts
B.
mitochondria
C.
vacuoles
D.
the nucleus
Answer:
The cytoplasm holds all the organelles, except the nucleus.
Explanation:
The cytoplasm holds all organelles, with the primary exception being the nucleus, which is separated by a nuclear envelope. Other organelles like chloroplasts, mitochondria, and vacuoles, are held within the cytoplasm.
Explanation:The cytoplasm, the watery fluid inside cells, houses almost all the organelles. The primary organelle that it does not encompass is the nucleus. The nucleus stays separated from the cytoplasm by a nuclear envelope.
Other organelles such as chloroplasts, mitochondria, and vacuoles are embedded within the cytoplasm. These organelles perform various roles within cells, while the nucleus controls the cell's activities and contains the cell's genetic information.
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1. The current through a light bulb connected across the terminals of a 120 V outlet is 0.50 A. At what rate does the bulb convert electric energy to light? 2. A 12.0 V battery causes a current of 2.0 A to flow through a lamp. What is the power used by the lamp? 3. What current flows through a 100. W light bulb connected to a 120. V outlet? 4. The current through a motor is 210 A. If a battery keeps a 12.0 V potential difference across the motor, what electric energy is delivered to the motor in 10.0 s?
Answer:
1. 60 W
2. 24 W
3. 0.83 A
4. 25200 J
Explanation:
1. What we are simply asked to look for is Electrical Power. It is the rate electrical energy is being transferred.
It is given as:
[tex]P = IV[/tex]
where I = current and V = voltage.
Therefore, Power is:
P = 0.50 * 120 = 60 W
2. Power, as given in the formula above is:
P = 2.0 * 12
P = 24.0 W
3. According to the formula of Power, current is given as:
[tex]I[/tex] = [tex]\frac{P}{V}[/tex]
Power is 100 W and voltage is 120 V, therefore, current is:
[tex]I = \frac{100}{120} \\\\\\I = 0.83 A[/tex]
4. Recall that power is the time rate of transfer of electrical energy. Mathematically:
[tex]P = \frac{E}{t}[/tex]
where t = time
This means that Electrical energy is:
[tex]E = Pt[/tex]
Recall that Power is:
[tex]P = IV[/tex]
Therefore, Electrical energy is:
[tex]E = IVt[/tex]
[tex]E = 210 * 12 * 10\\\\\\E = 25200 J[/tex]
1. At "60 W" the bulb convert electric energy to light
2. Power used by lamp will be "24.0 W".
3. The current flow will be "0.83 A".
4. The electric energy delivered to the motor will be "25200 J".
Potential difference and CurrentAccording to the question,
1. Current, I = 0.50 A
Voltage, V = 120 V
We know,
→ Power (P) = Current (I) × Voltage (V)
= 0.50 × 120
= 60 W
2. Current, I = 2.0 A
Voltage, V = 12.0 V
then,
→ P = 2.0 × 12
= 24.0 W
3. Power, P = 100 W
Voltage, V = 120 V
We know,
→ Current, I = [tex]\frac{P}{V}[/tex]
= [tex]\frac{100}{120}[/tex]
= 0.83 A
4. Current, I = 210 A
Potential difference, V = 12.0
We know,
→ E = Pt
P = IV
or,
The electrical energy be:
→ E = IVt
= 210 × 12 × 10
= 25200 J
Thus the above answer is correct.
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If the speed of a wave is 150 m/s and it’s frequency is 2 Hz , what is its wavelength ?
Answer:
75 m
Explanation:
wavelength = frequency/ speed = 75 m
A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.44 m long. The block is initially at rest. A bullet with mass 0.0134 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m , the tension in the cord is 4.76 N .
What was the initial speed v0 of the bullet?
Answer:
v₀ = 240 m / s
Explanation:
This problem must be solved in two parts, first we must use the conservation of the moment, then the conservation of energy.
Let's start by applying moment conservation, to the system formed by the block and bullet, in this case the forces during the crash are internal and the moment is conserved
Instant starts. Before the crash
p₀ = m v₀
Final moment. Right after the crash
[tex]p_{f}[/tex] = (m + M) v
The moment is preserved
Po =p_{f}
M v₀ = (m + M) v
v = m / (m + M) v₀ (1)
This is the speed with which the bullet block system comes out, now we can use energy conservation
Starting point. Right after the crash
Em₀ = K = ½ (m + M) v²
Final point. Highest point of the path
[tex]Em_{f}[/tex] = U = (m + M) g y
Em₀ = Em_{f}
½ (m + M) v² = (m + M) g y
v = √2 g y (2)
We substitute 1 in 2
m / (m + M) v₀ = √ 2gy
v₀ = (m + M) / m √ 2gy
Let's calculate
v₀ = (0.0134 +0.800) /0.0134 √ (2 9.8 0.8)
v₀ = 240 m / s
Answer:
298.04 m/s
Explanation:
Let m = mass of bullet = 0.0134 kg and M = mass of block = 0.800 kg.
Since the bullet becomes embedded in the block and rises a vertical height,h = 0.800 m
The kinetic energy change of mass + block = potential energy of mass + block at height, h
ΔK = -ΔU
So, 1/2(m + M)(v² - V²) = -[(m + M)gh - 0] where v is the velocity of the bullet + block at height, h. Since the tension, T is the centripetal force at height, h, it follows that
T = (m + M)v²/r r = length of cord = 1.44 m
v = √(Tr/(m + M)) = √4.76 N × 1.44 m/(0.800 + 0.0134)kg = √(6.8544/0.8134) = √8.427 = 2.9 m/s
So. 1/2(v² - V²) = -gh
v² - V² = -2gh
V = √(v² + 2gh) = √((2.9 m/s)² + 2 × 9.8 m/s² × 0.8 m) = √(8.41 + 15.68) = √24.09 = 4.91 m/s
This is the velocity of the bullet plus block at collision.
From the law of conservation of momentum,
momentum of bullet = momentum of bullet plus block
mv₀ = (m + M)V where v₀ = initial speed of bullet
v₀ = (m +M)V/m = (0.0134 kg + 0.800 kg)4.91 m/s ÷ 0.0134 kg = 3.994 ÷ 0.0134 kg = 298.04 m/s
A charged particle is surrounded by an electric field and a magnetic field
Answer:
yeah
Explanation:
electric fields help the charged particles interact
and isn't magnetic field the same as electric field.
Think about multiplying the mass of each student by a factor to calculate each student's kinetic energy. Is there a
common factor that works for every student? If so, what's this factor?
Answer:
Multiplying the mass of any student by a factor of 4.5 gives the kinetic energy of the student.
Explanation:
because
Answer: the question was asked to think about how to multiply a mass speech student by a factor to include the student. Kinetic energy, are there any common factors that appeal to them? The table shows mass energy and constant velocity. Let's calculate a constant factor, multiply it by mass, get the energy and see if we can do it. It's very easy. You can do it just in case. Only kinetic energy divided by mass can be produced. That is, the first is 100 and 21.5 divided by 27. Now you have the next 4.5. This is one. Then it is 162 divided by 36. This is 4.5. January 3, 202
Divide 5 by 45 to get 4.5. And hopefully you can see the patterns that are appearing here. In the fourth one, 243 divided by 54 gives 4.5. The fifth 288 divided by 64. Since it is 4.5, you can guess what it will be like. And finally, in the sixth case, it's 3 to 8.5 divided by 73. This is also equivalent to 4. Well, this common factor 4.5 was found.
What is kinetic energy?Kinetic energy is a form of energy that an object or particle has due to its movement. When work is done on an object by applying a net force that transfers energy, the object accelerates, thereby gaining kinetic energy. In physics, the kinetic energy of an object is the energy that the object has due to its motion.
This is defined as the work required to accelerate an object of a given mass from a stationary state to a specified velocity. After the body gains this energy during acceleration, the body retains this kinetic energy unless the velocity changes.
All moving objects use kinetic energy. Examples of kinetic energy are walkers, thrown baseball, bread crumbs falling from a table, and charged particles in an electric field.
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A mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 30. cm3. If the combustion of this mixture releases 984 J of energy, to what volume (in L) will the gases expand against a constant pressure of 648 torr if all the energy of combustion is converted into work to push back the piston?
Answer:
Explanation:
Given
Original volume V1=30cm^3 converting to L
=30/1000=0.03L
Constant pressure P= 648 tors
Converting to atm; 648 tors*1atm/760 torr=0.853 atm
Work=984J= 984**1L/101.33=9.7L.atm
Note before
W= -P(Vfinal-Vinitial)
-9.7/0.853+0.03L=11.68L
Answer:
Final expanded volume 11.43 L
Explanation:
The picture attached shows the solution
What r photons?
........................,........,..........
Answer:
in physics, a photon is a bundle of electromagnetic energy. It is the basic unit that makes up all light
Explanation:
Answer:
The quantum of light and other electromagnetic energy, regarded as a discrete particle having zero rest mass, no electric charge, and an indefinitely long lifetime. It is a gauge boson.
A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coasting. Describe each of the following events as an elastic, an inelastic, or a perfectly inelastic collision between the skater and the Frisbee.
Question:
(a) She catches the Frisbee and holds it.
Answer:
The correct option is;
A perfectly inelastic collision
Explanation:
A perfectly inelastic collision is one in which there is maximum amount of loss of kinetic energy in the system. In a perfectly inelastic collision, the colliding members lose their initial speed and they stick together resulting in a loss of kinetic energy.
Since she catches and holds on to the Frisbee, the kinetic energy of the Frisbee is lost as she holds on to it so as to combine her mass to that of the Frisbee.
Which of the following statements is FALSE?
a. After the Galveston Hurricane of 1900 the city built a 17-foot sea wall to protect Galveston from
future hurricanes.
b. One of the most devastating natural disasters to strike Texas was the Galveston Hurricane of
1900.
C. After the humcane of 1900, the city of Galveston removed 8 feet of soil to lower its elevation and
protect it from future hurricanes.
d. The rebuilding of Galveston after the hurricane of 1900 prompted the state to reorganize city
governments.
Answer:
C after the hurricane of 1900s
Explanation:
They didn’t remove 8 feet of soil they put 8 feet of soil
From the edge of a cliff, a 0.55 kg projectile is launched with an initial kinetic energy of 1550 J. The projectile's maximum upward displacement from the launch point is +140 m. What are the (a) horizontal and (b) vertical components of its launch velocity?
Answer:
(a) 38.5m/s
(b) 64.4m/s
Explanation:
First, we can obtain the launch speed from the definition of kinetic energy:
[tex]K=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2K}{m}}\\\\[/tex]
Plugging in the given values, we obtain:
[tex]v=\sqrt{\frac{2(1550J)}{0.55kg}}\\\\v=75.0m/s[/tex]
Now, from the conservation of mechanical energy, considering the instant of launch and the instant of maximum height, we get:
[tex]E_0=E_f\\\\K_0=U_g_f+K_f\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0_x^2\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0^2\cos^2\theta\\\\\implies \cos\theta=\sqrt{1-\frac{2gh_f}{v_0^2}}[/tex]
And with the known values, we compute:
[tex]\cos\theta=\sqrt{1-\frac{2(9.8m/s^2)(140m)}{(75.0m/s)^2}}\\\\\cos\theta=0.513\\\\\theta=59.12\°[/tex]
Finally, to know the components of the launch velocity, we use trigonometry:
[tex]v_0_x=v_0\cos\theta=(75.0m/s)\cos(59.12\°)=38.5m/s\\\\v_0_y=v_0\sin\theta=(75.0m/s)\sin(59.12\°)=64.4m/s[/tex]
It means that the horizontal component of the launch velocity is 38.5m/s (a) and the vertical component is 64.4m/s (b).
You find it takes 200 J of heat to take 4 kg of an unknown substance from 200 K to 240 K. It does not change phases during this interval. If you added 300 J of heat to the substance instead of 200 J, what would be its final temperature?
Answer:
300 K
Explanation:
First, we have find the specific heat capacity of the unknown substance.
The heat gained by the substance is given by the formula:
H = m*c*(T2 - T1)
Where m = mass of the substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
From the question:
H = 200J
m = 4 kg
T1 = 200K
T2 = 240 K
Therefore:
200 = 4 * c * (240 - 200)
200 = 4 * c * 40
200 = 160 * c
c = 200/160
c = 1.25 J/kgK
The heat capacity of the substance is 1.25 J/kgK.
If 300 J of heat is added, the new heat becomes 500 J.
Hence, we need to find the final temperature, T2, when heat is 500 J.
Using the same formula:
500 = 4 * 1.25 * (T2 - 200)
500 = 5 * (T2 - 200)
100 = T2 - 200
=> T2 = 100 + 200 = 300 K
The new final temperature of the unknown substance is 300K.