Answer:
[tex]a_{cm} = 9.64m/s^2[/tex]
[tex]Ff=6.42N[/tex]
Explanation:
The sum of torque on the sphere is:
[tex]m*g*sin\theta*R=I*\alpha[/tex]
[tex]m*g*sin\theta*R=2/3*m*R^2*\alpha[/tex]
[tex]m*g*sin\theta*R=2/3*m*R*a_{cm}[/tex]
Solving for a:
[tex]a_{cm}=9.64m/s^2[/tex]
Now, the sum of forces will be:
[tex]m*g*sin\theta-Ff=m*a_{cm}[/tex]
Solving for Ff:
[tex]Ff=m*g*sin\theta-m*a_{cm}[/tex]
Ff=-6.42N The negative sing tells us that it actually points downwards.
Answer:
a) a = 3.783 m/s^2
b) F_f = 5.045 N
Explanation:
Given:
- Mass of shell m = 2.0 kg
- Angle of slope Q = 40 degrees
- Moment of inertia of shell I = 2/3 *m*R^2
Find:
a) Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell.
b) Find the magnitude of the frictional force acting on the spherical shell.
Solution:
- Draw a Free body diagram for the shell. We see that the gravitational force F_g acting parallel to the plane of the inclined surface makes the sphere to roll down. The frictional force F_f between the inclined surface and the sphere gives the necessary torque for the sphere to roll down with out slipping. Under this conditions a sphere will roll down without slipping with some acceleration and the acceleration can be calculated from the equation of motion of the sphere:
m*g*sin(Q) - F_f = m*a
- Where, The frictional force produces the torque and due to this torque the sphere gets an angular acceleration.
- Then we can write the equation for the rotational motion as:
F_f*R = I*α
F_f = I*α / R
- Using moment mass inertia of the shell we have:
F_f = (2/3)*m*R^2*α/R
- Where the angular acceleration α is related to linear acceleration a with:
α = a / R
- combing the two equations we will have friction force F_f as:
F_f = (2/3)*m*R^2*a/R^2
F_f = (2/3)*m*a
- Now evaluate the equation of motion:
m*g*sin(Q) - (2/3)*m*a= m*a
- Simplify:
(5/3)*a = g*sin(Q)
a = (3/5)*g*sin(Q)
- Plug the values in: a = (3/5)*9.81*sin(40)
a = 3.783 m/s^2
- Now compute the Frictional force F_f from the expression derived above:
F_f = (2/3)*m*a
- Plug values in: F_f = (2/3)*2*3.783
F_f = 5.045 N
Briefly answer the following questions. a) A clock is mounted on the wall. As you look at it, what is the direction of the angular velocity vector of the second hand? What is the value of the angular acceleration of the second hand? b) If the angular acceleration of a rigid body is zero, what is the functional form of the angular velocity? c) What if another planet the same size as Earth were put into orbit around the Sun along with Earth. Would the moment of inertia of the system increase, decrease, or stay the same? d) Does increasing the number of blades on a propeller increase or decrease its moment of inertia, and why? e) Can you think of a body that has the same moment of inertia for all possible axes? If so, give an example, and if not, explain why this is not possible. Can you think of a body that has the same moment of inertia for all axes passing through a certain point? If so, give an example and indicate where the point is located. f) To maximize the moment of inertia of a flywheel while minimizing its weight, what shape and distribution of mass should it have? Explain. g) Is it possible to change the translational) kinetic energy of an object without changing its rotational kinetic energy? What about the reverse of it?
Answer:
Explanation:
a ) The direction of angular velocity vector of second hand will be along the line going into the plane of dial perpendicular to it.
b ) If the angular acceleration of a rigid body is zero, the angular velocity will remain constant.
c ) If another planet the same size as Earth were put into orbit around the Sun along with Earth the moment of inertia of the system will increase because the mass of the system increases. Moment of inertia depends upon mass and its distribution around the axis.
d ) Increasing the number of blades on a propeller increases the moment of inertia , because both mass and mass distribution around axis of rotation increases.
e ) It is not possible that a body has the same moment of inertia for all possible axes because a body can not remain symmetrical about all axes possible. Sphere has same moment of inertia about all axes passing through its centre.
f ) To maximize the moment of inertia of a flywheel while minimizing its weight, the shape and distribution of mass should be such that maximum mass of the body may be situated at far end of the body from axis of rotation . So flywheel must have thick outer boundaries and this should be
attached with axis with the help of thin rods .
g ) When the body is rotating at the same place , its translational kinetic energy is zero but its rotational energy can be increased
at the same place.
As a section of an workout program to toughen their pectoral muscles a person stretches a spring which has a spring constant k = 520 N/m. Write an equation for the work necessary to stretch the spring from the relaxed state to a distance x1.
To develop the problem it is necessary to apply the concepts related to Hooke's Law. For which it is understood that
[tex]F = kx[/tex]
Where,
k = Spring constant
x = Displacement of Spring
By the theorem of work we understand that it is the force realized when there is the path of an object therefore
[tex]dW =Fdx[/tex]
Our value of F is
[tex]dW = (kx)dx[/tex]
Integrating
[tex]W = k\int\limit_0^{x_1} xdx[/tex]
Therefore the expresion is
[tex]W = \frac{kx_1^2}{2}[/tex]
To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the north while your friend pulls with a force of 450 N to the northwest. The total force from the two ropes is:
a. more than 950 N.
b. exactly 950 N.
c. less than 950 N.
Answer:
C. less than 950 N.
Explanation:
Given that
Force in north direction F₁ = 500 N
Force in the northwest F₂ = 450 N
Lets take resultant force R
The angle between force = θ
θ = 45°
The resultant force R
[tex]R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}[/tex]
[tex]R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}[/tex]
R= 877.89 N
Therefore resultant force is less than 950 N.
C. less than 950 N
Note- When these two force will act in the same direction then the resultant force will be 950 N.
The total force on the stump is less than the sum of the individual forces because the forces are vectors and not aligned in the same direction. The correct answer is (c) less than 950 N.
When you pull on a stump with two ropes at different angles, the total force is not simply the sum of the individual forces because the forces are vectors. To find the total force, you must consider both the magnitude and direction of each force. In this case, you pull with a force of 500 N to the north and your friend pulls with 450 N to the northwest. Since these forces are not in the same direction, you need to use vector addition to find the resultant force. Without calculating the exact value, we can already state that the resultant force will be less than 950 N because part of the forces cancel each other out due to the angle at which they are applied. Therefore, the correct answer is (c) less than 950 N.
Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity is 6.20 m/s2, and atmospheric pressure is 7.00 ✕ 104 Pa. The atmospheric temperature and pressure on this planet causes the density of the liquid methane ocean to be 415 kg/m3. (a) What force (in N) is exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of the ocean? N (b) What is the weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m? (Enter your answer in N.) N (c) What is the pressure (in Pa) at a depth of 10.0 m in the methane ocean?
Answer:
Explanation:
Atmospheric pressure = 7 x 10⁴ Pa
force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area
= 7 x 10⁴ x 3.14 x 2 x 2
= 87.92 x 10⁴ N
b )
weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m
Pressure x area
height x density x acceleration of gravity x π r²
= 10 x 415 x 6.2 x 3.14 x 2 x 2
=323168.8 N
c ) Pressure at a depth of 10 m
atmospheric pressure + pressure due to liquid column
= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)
= 7 x 10⁴ + 10 x 415 x 6.2
(7 + 2.57 )x 10⁴ Pa
9.57 x 10⁴ Pa
The Force exerted by the atmosphere on a disk-shaped area is 8.8 x 10^5 N. The weight of a 10 m cylindrical column of methane is 1.31 x 10^7 N. The pressure at a depth of 10 m in the methane ocean is 1.22 x 10^5 Pa.
Explanation:To answer this question, we need to apply concepts of fluid pressure and buoyancy.
(a) The force exerted by the atmosphere on a disk-shaped region is given by the atmospheric pressure multiplied by the area of the disk. This is calculated as Force = Pressure x Area. The Area of the disk is given by πr², where r is the radius. Therefore, the force = (7.00 x 10^4 Pa) x π(2 m)² = 8.8 x 10^5 N.
(b) The weight of the methane column can be found by using the formula for the weight of a liquid column which is Weight = density x gravity x height x Area. Plugging in the values from the question, it comes to (415 kg/m³) x (6.20 m/s²) x (10 m) x π(2 m)² = 1.31 x 10^7 N.
(c) The pressure at a depth in the methane ocean can be calculated using the equation P = Po + pgd, where Po is atmospheric pressure, p is the density of the fluid, g is the acceleration due to gravity and d is the depth. Substituting the given values, P = 7.00 x 10^4 Pa + (415 kg/m³)(6.20 m/s²)(10 m) = 1.22 x 10^5 Pa.
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A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on his back, but you have been hired to investigate the safety of this stunt. When you examine the mattress, you see that it effectively has a spring constant of 65144 N/m for the area likely to be impacted by the stuntman, but cannot depress more than 12.89 cm without injuring him. To approach this problem, consider a simplified version of the situation. A mass falls through a height of 3.32 m before landing on a spring of force constant 65144 N/m. Calculate the maximum mass that can fall on the mattress without exceeding the maximum compression distance.
=______________________ kg
Answer:
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg
Explanation:
Hi there!
Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:
PE = EPE
m · g · h = 1/2 k · x²
Where:
m = mass.
g = acceleration due to gravity.
h = height.
k = spring constant.
x = compression distance
The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:
EPE =1/2 k · x²
EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J
Then, using the equation of gravitational potential energy:
PE = m · g · h = 541.2 J
m = 541.2 J/ g · h
m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)
m = 16.6 kg
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.
When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.
Answer:
P = 1.99 10⁸ Pa
Explanation:
The definition of the bulk module is
B = - P / (ΔV / V)
The negative sign is included for which balk module is positive, P is the pressure and V that volume
They tell us that the variation in volume is 9.05%, that is
ΔV / V = 9.0Δ5 / 100 = 0.0905
P = - B DV / V
P = 2.2 10⁹ (0.0905)
P = 1.99 10⁸ Pa
Final answer:
The force per unit area water exerts on a container when it freezes is calculated using the bulk modulus of water, resulting in a pressure of 1.99 × 10⁴ N/cm². This explains why freezing water can cause significant damage to rigid structures.
Explanation:
When water freezes, its volume increases by 9.05% (ΔV/V = 9.05 × 10⁻²). The force per unit area that water can exert on a container when it freezes can be calculated using the bulk modulus of water, which is given as B = 2.2 × 10⁹ N/m². The change in volume fraction (ΔV/V) is directly related to the change in pressure (ΔP) and the bulk modulus (B) by the formula ΔP = -B(ΔV/V), where the negative sign indicates a reduction in volume leads to an increase in pressure, but in this case, we're dealing with an expansion, so ΔV/V is positive.
To find the pressure in N/cm², we first calculate the pressure in Pascals (Pa) using the given values:
ΔP = B(ΔV/V)
= (2.2 × 10⁹ N/m²)(9.05 × 10⁻²)
= 1.99 × 10⁸ Pa.
Since 1 Pa equals 1 N/m² and 1 N/m² equals 0.0001 N/cm²,
the pressure in N/cm² is 1.99 × 10⁴ N/cm².
A transverse harmonic wave travels on a rope according to the following expression:
y(x,t) = 0.14sin(2.1x + 17.7t)
The mass density of the rope is μ = 0.104 kg/m. x and y are measured in meters and t in seconds.
1)
What is the amplitude of the wave?
m
Your submissions:
2)
What is the frequency of oscillation of the wave?
Hz
Your submissions:
3)
What is the wavelength of the wave?
m
Your submissions:
4)
What is the speed of the wave?
m/s
Your submissions:
5)
What is the tension in the rope?
N
Your submissions:
6)
At x = 3.4 m and t = 0.48 s, what is the velocity of the rope? (watch your sign)
m/s
Your submissions:
7)
At x = 3.4 m and t = 0.48 s, what is the acceleration of the rope? (watch your sign)
m/s2
Your submissions:
8)
What is the average speed of the rope during one complete oscillation of the rope?
m/s
Your submissions:
9)
In what direction is the wave traveling?
+x direction
-x direction
+y direction
-y direction
+z direction
-z direction
Your submissions:
10)
On the same rope, how would increasing the wavelength of the wave change the period of oscillation?
the period would increase
the period would decrease
the period would not change
This answer clarifies various aspects related to a transverse harmonic wave, such as amplitude, frequency, and tension in the rope. The wave is traveling in the -x direction. And, the period of oscillation will increase if the wavelength increases.
Explanation:To answer such questions, we need to understand the equation of the wave given by y(x,t) = A sin (kx + ωt). Here, 'A' represents the amplitude of the wave, 'k' is the wave number, and 'ω' is the angular frequency. We get these values by comparing the given equation with the standard wave equation.
The amplitude of the wave is given by A and here A=0.14 m.The frequency of the wave is given by f=ω/2π=17.7/2π ≈ 2.82 Hz.The wavelength of the wave λ = 2π/k = 2π/2.1 ≈ 2.99 m.The speed of the wave is given by the relation v=λf ≈ 2.99 m * 2.82 Hz ≈ 8.43 m/s.The tension in the rope is given by T = μv^2 = 0.104 kg/m * (8.43 m/s)^2 ≈ 7.4 N.The velocity and acceleration at given values of x and t can be obtained from the time derivative of your wave equation. They require more advanced calculations and depend on the given values of x and t.The average speed of the rope during one complete oscillation is 0, as every point on the rope returns to its initial position during one oscillation.The wave is travelling in the -x direction since the sign before 't' in the wave equation is positive.If the wavelength of the wave is increased, the period of oscillation will increase as the period T=1/f and frequency f=v/λ and v is constant for a given medium.Learn more about Transverse harmonic wave here:https://brainly.com/question/14014748
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A rod of mass M = 4 kg, length L = 1.8 meters, and moment of inertia ML2/12 is free to move on a frictionless surface. The rod is at rest when a puck of mass m = 0.4 kg approaches with a speed vi = 20 m/s perpendicular to the rod's length and strikes the rod at a point d = 0.3 meters from its lower end. After the collision, the puck moves backward with a speed vf = 10 m/s. The rod's center of mass moves forward with speed v and rotates with angular speed ω. (a) What is the speed v of the rod's center of mass after the collision?
Answer:
The speed of the rod's center of mass after the collision is 6 m/s.
Explanation:
Given that,
Mass of rod = 4 kg
Length l = 1.8 m
Moment of inertia [tex]I=\dfrac{ML^2}{12}[/tex]
Mass of puck = 0.4 kg
Initial speed= 20 m/s
Distance = 0.3 m
Final speed = 10 m/s
(a). We need to calculate the speed v of the rod's center of mass after the collision
As there is no external force acting on the system so, linear and angular momentum of the system will be conserved.
Using conservation of momentum
[tex]m_{i}v_{i}=m_{f}v_{f}+Mv[/tex]
Put the value into the formula
[tex]0.4\times20=-0.4\times10+2v[/tex]
[tex]v=\dfrac{8+4}{2}[/tex]
[tex]v=6\ m/s[/tex]
Hence, The speed of the rod's center of mass after the collision is 6 m/s.
A 11 g object moving to the right at 29 cm/s overtakes and collides? elastically with a 24 g object moving in the same direction at 19 cm/s. Find the velocity of the slower object after the collision. AND find the velocity of the faster object after the collision. Answer in units of cm/s.
Answer:
vf₁ = 15.29 cm/s : to the right
vf₂ = 25.29 cm/s : to the right
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:
p=m*v
where
p:Linear momentum
m: mass
v:velocity
There are 3 cases of collisions : elastic, inelastic and plastic.
For the three cases the total linear momentum quantity is conserved:
P₀ = Pf Formula (1)
P₀ :Initial linear momentum quantity
Pf : Final linear momentum quantity
Data
m₁= 11 g : mass of object₁
m₂= 24 g : mass of object₂
v₀₁ = 29 cm/s , to the right : initial velocity of m₁
v₀₂= 19 cm/s, to the right i :initial velocity of m₂
Problem development
We appy the formula (1):
P₀ = Pf
m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂
We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:
( 11)*( 29) + (24 )*(19) = ( 11)*vf₁ +(24)*vf₂
775 = ( 11)*vf₁ +(24)*vf₂ Equation (1)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
[tex]e = \frac{v_{f2} -v_{f1}}{v_{o1} -v_{o2}}[/tex]
1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)
(29 - 19 ) = (vf₂ -vf₁)
10 = (vf₂ -vf₁)
vf₂ = 10 + vf₁ Equation (2)
We replace Equation (2) in the Equation (1)
775 = ( 11)*vf₁ +(24)*(10 + vf₁ )
775 = ( 11)*vf₁ +240+(24) vf₁
775 - 240= ( 35)*vf₁
535 = ( 35)*vf₁
vf₁ = 535 / 35
vf₁ = 15.29 cm/s : to the right : Final velocity of object₁
We replace vf₁ = 15.29 cm/s in the Equation (2)
vf₂ = 10 + vf₁
vf₂ =10 + 15.29
vf₂ = 25.29 cm/s, to the right: Final velocity of object₂
78. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
To solve this problem it is necessary to apply the concepts related to the condition of path difference for destructive interference between the two reflected waves from the top and bottom of a surface.
Mathematically this expression can be described under the equation
[tex]\delta = 2nt[/tex]
Where
n = Refractive index
t = Thickness
In terms of the wavelength the path difference of the reflected waves can be described as
[tex]\delta = \frac{\lambda}{4}[/tex]
Where
\lambda = Wavelenght
Equation the two equations we have that
[tex]2nt = \frac{\lambda}{4}[/tex]
[tex]t = \frac{\lambda}{8n}[/tex]
Our values are given as
[tex]\lambda = 380nm \rightarrow[/tex] Wavelength of light
[tex]n = 1.4[/tex]
[tex]t = \frac{380nm}{8*1.4}[/tex]
[tex]t = 33.93nm[/tex]
Therefore the minimum thickness of the oil for destructive interference to occur is approximately 34.0 nm
The maximum thickness of the oil film that will appear dark at all visible wavelengths is approximately 33.9 nm, ensuring destructive interference for all visible light.
Explanation:The oil film appears dark when its thickness is such that the path length difference is less than one-fourth the wavelength of the light. This is due to the destructive interference of light waves. Given that the thickest the oil can be and still appear dark when the path length difference is one-fourth the shortest visible wavelength in oil, we need to find this thickness value.
The shortest visible wavelength of light in air is typically about 380 nm, which is violet. However, the speed and consequently the wavelength of light will decrease in oil due to its refractive index being higher. Therefore, the wavelength of light in oil is shorter than in air and can be found by dividing the wavelength in the air by the index of refraction, which gives about 271 nm for violet light in oil.
Since the path length difference is twice the film thickness, the maximum thickness of the oil should be one-eighth the shortest visible wavelength in oil to appear dark, which is approximately 33.9 nm. This calculation ensures destructive interference for all visible wavelengths and hence, the film will appear black or dark at all visible wavelengths.
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A vertical spring with k=490N/m is standing on the ground. You are holding a 5.0kg block just above the spring, not quite touching it. Part A How far does the spring compress if you let go of the block suddenly? Express your answer to two significant figures and include the appropriate units.
Answer:
Compression in the spring, x = 0.20 m
Explanation:
Given that,
Spring constant of the spring, k = 490 N/m
Mass of the block, m = 5 kg
To find,
Compression in the spring.
Solution,
Since the block is suddenly dropped on the spring gravitational potential energy of block converts into elastic potential energy of spring. Its expression is given by :
[tex]mgx=\dfrac{1}{2}kx^2[/tex]
Where
x is the compression in the spring
[tex]x=\dfrac{2mg}{k}[/tex]
[tex]x=\dfrac{2\times 5\times 9.8}{490}[/tex]
x = 0.20 m
So, the compression in the spring due to block is 0.20 meters.
The vertical spring with spring constant of 490N/m will compress at a distance of 0.2m.
How to calculate distance of spring?According to this question, the following information were given:
Spring constant of the spring, k = 490 N/mMass of the block, m = 5 kgTo calculate the compression in the spring, we deduce that the gravitational potential energy of the spring converts into elastic potential energy. Its expression is given by:
mgx = ½kx²
Where;
x = compression in the springx = 2mg/k
x = 2 × 5 × 9.8/490
x = 98/490
x = 0.2m
Therefore, the spring will compress at a distance of 0.2m.
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Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.84 Hz?
Answer:
m = 3.91 kg
Explanation:
Given that,
Mass of the object, m = 3.74 kg
Stretching in the spring, x = 0.0161 m
The frequency of vibration, f = 3.84 Hz
When the object is suspended, the gravitational force is balanced by the spring force as :
[tex]mg=kx[/tex]
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{3.74\times 9.8}{0.0161}[/tex]
k = 2276.52 N/m
The frequency of vibration is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]m=\dfrac{k}{4\pi^2f^2}[/tex]
[tex]m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}[/tex]
m = 3.91 kg
So, the mass of the object is 3.91 kg. Hence, this is the required solution.
Suppose you are planning a trip in which a spacecraft is totravel at a constant velocity for exactly six months, as measuredby a clock on board the spacecraft, and then return home at thesame speed. Upon return, the people on earth will have advancedexactly 110 years into the future. According to special relativity,how fast must you travel? (Give your answer with at least sixsignificant digits as a multiple of c.)
Answer: 0.999959 c
Explanation:
According to the special relativity theory, time is measured differently by two observers moving one relative another, according to the Lorentz Transform Equation, as follows:
t = t’ / t=t^'/√(1-(v)2/c2 )
where t= time for the moving observer (relative to the spacecraft, fixed on Earth) = 110 years.
t’= time for the observer at rest respect from spacecraft = 1 year
v= spacecraft constant speed
c= speed of light
Solving for v, with a six decimals precision as a multiple of c, we get:
v = 0.999959 c
In designing a backyard water fountain, a gardener wants to stream of water to exit from the bottom of one tub and land in a second one. The top of the second tub is 0.5 m below the hole in the first tub, which has water in it to a depth of 0.15 m. How far to the right of the first tub must the second one be placed to catch the stream of water?
To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.
From the definition we know that the speed of a body can be described as a function of gravity and height
[tex]V = \sqrt{2gh}[/tex]
[tex]V = \sqrt{2*9.8*0.15}[/tex]
[tex]V = 1.714m/s[/tex]
Then applying the kinematic equation of displacement, the height can be written as
[tex]H = \frac{1}{2}gt^2[/tex]
Re-arrange to find t,
[tex]t = \sqrt{2\frac{h}{g}}[/tex]
[tex]t = \sqrt{2\frac{0.5}{9.8}}[/tex]
[tex]t = 0.3194s[/tex]
Thus the calculation of the displacement would be subject to
[tex]x = vt[/tex]
[tex]x =1.714*0.3194[/tex]
[tex]x = 0.547m[/tex]
Therefore the required distance must be 0.547m
Electricity is distributed from electrical substations to neighborhoods at 1.6×104 V . This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house. Part A How many turns does the primary coil on the transformer have if the secondary coil has 140 turns? Express your answer using two significant figures. n prim n p r i m = nothing turns SubmitRequest Answer Part B No energy is lost in an ideal transformer, so the output power P out from the secondary coil equals the input power P in to the primary coil. Suppose a neighborhood transformer delivers 230 A at 120 V . What is the current in the 1.6×104 V line from the substation? Express your answer using two significant figures. I i I i = nothing A SubmitRequest Answer Provide Feedback Next
Answer:
(A) 18667 turns
(B) 1.7 A
Solution:
As per the question:
Voltage at which the electricity is distributed, [tex]V_{p} = 1.6\times 10^{4}\ Hz[/tex]
Frequency of the oscillating voltage, f = 60 Hz
Step down voltage, [tex]V_{s} = 120\ V[/tex]
No. of turns in the secondary coil, [tex]N_{s} = 140\ turns[/tex]
Current in the secondary coil, [tex]I_{s} = 230\ A[/tex]
Now,
(A) To calculate the primary no. of turns, we use the relation:
[tex]\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}[/tex]
[tex]N_{p} = \frac{V_{p}}{V_{s}}\times N_{s}[/tex]
[tex]N_{p} = \frac{1.6\times 10^{4}}{120}\times 140 = 18,667\ turns[/tex]
(B) To calculate the current in the primary coil, [tex]I_{p}[/tex], we use the relation:
[tex]\frac{V_{p}}{V_{s}} = \frac{I_{s}}{I_{p}}[/tex]
[tex]I_{p} = \frac{V_{s}}{V_{p}} \times {I_{s}}[/tex]
[tex]I_{p} = \frac{120}{1.6\times 10^{4}} \times 230 = 1.7\ A[/tex]
A ball of mass 0.250 kg and a velocity of + 5.00 m/s collides head-on with a ball of mass 0.800 kg that is initially at rest. No external forces act on the balls. After the collision, the velocity of the ball which was originally at rest is + 2.38 m/s. What is the velocity of the 0.250 kg ball?
Answer:
[tex]v_1=-2.616\ m/s[/tex]
Explanation:
Given that,
Mass of ball 1, [tex]m_1=0.25\ kg[/tex]
Initial speed of ball 1, [tex]u_1=5\ m/s[/tex]
Mass of ball 2, [tex]m_2=0.8\ kg[/tex]
Initial speed of ball 2, [tex]u_2=0[/tex] (at rest)
After the collision,
Final speed of ball 2, [tex]v_2=2.38\ m/s[/tex]
Let [tex]v_2[/tex] is the final speed of ball 1.
Initial momentum of the system is :
[tex]p_i=m_1u_1+m_2u_2[/tex]
[tex]p_i=0.25\times 5+0[/tex]
[tex]p_i=1.25\ m/s[/tex]
Final momentum of the system is :
[tex]p_f=m_1v_1+m_2v_2[/tex]
[tex]p_f=0.25\times v_1+0.8\times 2.38[/tex]
[tex]p_f=0.25 v_1+1.904[/tex]
According the law of conservation of linear momentum :
initial momentum = final momentum
[tex]1.25=0.25 v_1+1.904[/tex]
[tex]v_1=-2.616\ m/s[/tex]
So, the final velocity of ball 1 is (-2.616)m/s.
Answer:
mass of ball 1= 0.25 kg
initial speed of ball 1= 5. m/s
mass ball 2= 0.8 kg
initial speed ball 2= 0 ( it was at rest)
final speed of ball 2= 2.38m/s
the formula is:
m1 . v 1 + m2 . v2 = m1 . v1f+ m2 . v2f
0.25 . 5 + 0.8 . 0 = o.25 . x + 0.8 . 2.38
-2.6288 m/s
Explanation:
i got it right on the flvs
⦁ A baseball is struck by a bat 46 cm from the axis of rotation when the angular velocity of the bat is 70 rad/s. If the ball is hit at a height of 1.2 m at an angle of 45 degrees, will the ball clear a 1.2 m fence 110 m away (assume the initial velocity of the ball is the same as the linear velocity of the bat at the point at which it is struck?
To determine if the ball will clear the fence, calculate the range and maximum height of the ball's trajectory using given initial velocity and angle. Compare the calculated range with the distance to the fence.
Explanation:To determine whether the ball will clear the fence, we need to calculate the range and maximum height of the ball's trajectory. Using the given initial velocity of the ball (which is equal to the linear velocity of the bat), we can decompose it into horizontal and vertical components. The horizontal component remains constant throughout the flight, while the vertical component is affected by gravity. We can use the equations of motion to find the range and maximum height.
First, let's calculate the horizontal distance (range) the ball will travel. We can use the formula: range = (initial horizontal velocity) * (time of flight). Since the initial velocity and angle are given, we can calculate the initial horizontal velocity using trigonometry:
initial horizontal velocity = initial velocity * cos(angle)Next, we need to find the time of flight. Since the vertical displacement is zero at the peak of the trajectory, we can use the equation: time of flight = 2 * (initial vertical velocity) / (acceleration due to gravity). From the given height and angle, we can calculate the initial vertical velocity using trigonometry:
initial vertical velocity = initial velocity * sin(angle)Now, we can substitute the calculated values into the formula for range and solve for the time of flight:
range = (initial velocity * cos(angle)) * (2 * (initial velocity * sin(angle)) / (acceleration due to gravity))Finally, we can compare the calculated range with the distance to the fence. If the range is greater than the distance to the fence, the ball will clear the fence.
Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of R = 1.22 m and a mass of 67.0 kg . To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3510 m/s2 .What is the maximum kinetic energy that can be stored in the fly-wheel?
The maximum kinetic energy that can be stored in the flywheel is 27,909.78 J.
Explanation:To calculate the maximum kinetic energy that can be stored in the flywheel, we need to find the moment of inertia. For a uniform solid disk, the moment of inertia is given by the formula I = 0.5 * m * r^2, where m is the mass of the disk and r is the radius. Plugging in the values, we get I = 0.5 * 67.0 kg * (1.22 m)^2 = 49.518 kg·m².
The maximum kinetic energy is equal to 0.5 * I * ω^2, where ω is the angular velocity. Since the radial acceleration of a point on the rim is given as 3510 m/s², we can calculate ω using the formula a = r * ω^2. Rearranging the formula, we get ω = √(a / r) = √(3510 m/s² / 1.22 m) = 33.24 rad/s.
Plugging in the values, the maximum kinetic energy is 0.5 * 49.518 kg·m² * (33.24 rad/s)^2 = 27,909.78 J.
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The maximum kinetic energy that can be stored in the flywheel is approximately 71,791 Joules.
To determine the maximum kinetic energy that can be stored in the flywheel, we need to use the concepts of rotational motion. The relevant formulas are:
Kinetic Energy of a rotating object: KE = 1/2 I ω², where I is the moment of inertia and ω is the angular velocity.
Moment of Inertia for a solid disk: I = 1/2 m R², where m is the mass and R is the radius.
Relation between radial acceleration and angular velocity: a = ω² R.
We are given:
Radius (R): 1.22 m
Mass (m): 67.0 kg
Maximum radial acceleration (a): 3510 m/s²
First, solve for angular velocity (ω) using the radial acceleration formula:
ω² = a / R
⇒ ω² = 3510 m/s² / 1.22 m
⇒ ω² ≈ 2877.05 s⁻²
⇒ ω ≈ 53.61 rad/s
Next, calculate the moment of inertia (I) for the flywheel:
I = 1/2 m R² => I = 0.5 × 67.0 kg × (1.22 m)²
⇒ I ≈ 49.917 kgm²
Finally, calculate the maximum kinetic energy:
KE = 1/2 I ω² => KE = 1/2 × 49.917 kgm² × (53.61 rad/s)²
⇒ KE ≈ 71,791 Joules
Thus, the maximum kinetic energy that can be stored in the flywheel is approximately 71,791 Joules.
A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L and cross-sectional area 0.520 cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 5.00×104 N at its ends. Find the length L of the nickel rod if the elongations of the two rods are equal. What is the stress in the brass rod? What is the stress in the nickel rod? What is the strain in the brass rod? What is the strain in the nickel rod?
Answer:
a) L₂ = 0.676 m
b) σ₁ = 2.28*10⁸ N/m²
σ₂ = 9.62*10⁸ N/m²
c) ε₁ = 0.00253678
ε₂ = 0.00457875
Explanation:
Given info
L₁ = 1.22 m
A₁ = 2.19 cm² = 2.19*10⁻⁴ m²
L₂ = ?
A₂ = 0.52 cm² = 0.52*10⁻⁴ m²
P = 5.00*10⁴ N
E₁ = 9*10¹⁰ N/m²
E₂ = 2.1*10¹¹ N/m²
In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that
ΔL₁ = ΔL₂ ⇒ P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)
⇒ L₂ = A₂*E₂*L₁ / (A₁*E₁)
⇒ L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)
⇒ L₂ = 0.676 m
The stress in the brass rod is obtained as follows
σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²
The stress in the niquel rod is obtained as follows
σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²
The strain in the brass rod is obtained as follows
σ₁ = E₁*ε₁ ⇒ ε₁ = σ₁ / E₁
⇒ ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678
The strain in the niquel rod is obtained as follows
σ₂ = E₂*ε₂ ⇒ ε₂ = σ₂ / E₂
⇒ ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875
A certain 45 Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 12 kV/m. show answer No Attempt 50% Part (a) What is the wavelength, in meters, of this very low frequency electromagnetic wave
To solve this problem we must rely on the concepts related to wavelength in terms of the speed of light and frequency.
Mathematically one of the ways to express the wavelength is,
[tex]\lambda = \frac{c}{f}[/tex]
Where,
c = Speed of light [tex](3*10^8m/s)[/tex]
f = Frequency
The value of the frequency given is
f = 45Hz
Therefore replacing we can find it,
[tex]\lambda = \frac{3*10^8}{45Hz}[/tex]
[tex]\lambda = 6.6667*10^6m[/tex]
Hence the wavelength of the very low frequency electromagnetic wave is [tex]6.6667*10^6m[/tex]
Final answer:
The wavelength of this very-low-frequency electromagnetic wave is approximately 6.67 x 10^6 meters.
Explanation:
The wavelength (λ) of an electromagnetic wave can be calculated using the formula:
λ = c/f
Where c is the speed of light, approximately 3 x 10^8 m/s, and f is the frequency of the wave. In this case, the frequency is 45 Hz. Substituting the values into the formula, we get:
λ = 3 x 10^8 m/s / 45 Hz = 6.67 x 10^6 m
Therefore, the wavelength of this very-low-frequency electromagnetic wave is approximately 6.67 x 10^6 meters.
Three 500-g point masses are at the corners of an equilateral triangle with 50-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the triangle and passing through one of the masses at a corner of the triangle?
Answer:
0.25 kg m^2
Explanation:
mass of each , m = 500 g = 0.5 kg
distance, r = 50 cm = 0.5 m
Moment of inertia about the axis passing through one corner and perpendicular to the plane of triangle
I = mr^2 + mr^2
I = 2 mr^2
I = 2 x 0.5 x 0.5 x 0.5
I = 0.25 kgm^2
At time t = 0 an object is traveling to the right along the axis at a speed of 10 m/s with acceleration -2.0 m/s2. Which statement is true?
a. The object cannot have a negative acceleration and be moving to the right.
b. The object will continue to move to the right, slowing down but never coming to a complete stop.
c. The object will slow down, momentarily stopping, then pick up speed moving to the left.
d. The object will slow down, eventually coming to a complete stop.
Answer:
C. The object will slow down, momentarily stopping, then pick up speed moving to the left.
Explanation:
Given that
Speed of the object = 10 m/s ( right direction)
Acceleration ,a= - 2 m/s
A negative sign shows that acceleration is in the opposite direction to the speed.
We know that if speed and acceleration is in the opposite direction then the final speed of the objects becomes zero. for a moment and it will increase in the opposite direction.
Therefore the answer is C
The correct statement is c. The object will slow down, momentarily stop, then pick up speed moving to the left. This is due to the object's negative acceleration, which means it is slowing down.
Explanation:In this physics problem, an object is moving to the right with an initial velocity of 10 m/s and an acceleration of -2.0 m/s2. Negative acceleration, also known as deceleration, doesn't mean the object is moving in a negative direction, rather it is slowing down. The correct statement is c. The object will slow down, momentarily stopping, then pick up speed moving to the left. This is because as time progresses, the negative acceleration reduces the speed of the object until it comes to a stop, after which it starts moving in the opposite direction, towards the left.
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Compare the weight of a 5.00-kg object on Earth’s surface to the gravitational force between a 5.00-kg object that is one Earth radius from another object of mass equal to 5.98 × 10 24 kg . Use Newton’s universal law of gravitation for the second part of the question
Answer
given,
mass of an object = 5 Kg
mass of the other object = 5.98 x 10²⁴ Kg
radius of earth = 6371 Km = 6.371 x 10⁶ m
comparison between gravitational force between two object and weight of the object.
gravitational force between two objects
[tex]F = \dfrac{Gm_1m_2}{r^2}[/tex]
[tex]F = \dfrac{6.67 \times 10^{-11}\times 5 \times 5.98 \times 10^{24}}{(6.371 \times 10^6)^2}[/tex]
F = 49.134 N
weight of the object on the earth surface
F = m g
F = 5 x 9.81
F = 49.05 N
From the above two calculation we can conclude that gravitational force is approximately equal to weight of the body.
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at t = 0 from the position x = 0.0360 m. (a) Determine the location of the mass at t = 5.47 s? (b) Determine if the mass is moving in the positive or negative x direction at t = 5.47 s? positive x direction negative x direction
The location of the mass at t = 5.47s can be determined by the equation for the position associated with SHM. To ascertain the direction of motion at the same time, the velocity function (the derivative of the position function) can be used. If the velocity value is positive, the mass moves in the positive x direction; if negative, it moves in the negative x direction.
Explanation:The problem describes the motion of a mass attached to a spring - specifically, it's simple harmonic motion (SHM). This can be understood in terms of the mass moving back and forth around a central, or equilibrium, position.
a) To determine the location of the mass at a specific time, t = 5.47s, we need to use the equation for the position x(t) associated with SHM: x(t) = A * cos(2 * π * t / T), where A is the amplitude (0.0360m in this case), π is the mathematical constant Pi, t is the time, and T is the period (2.25s in this case). Substituting the given values into this equation will give the position of the mass at t = 5.47s.
b) To determine the direction of motion at that time, we use the derivative of the position function, which gives us the velocity function: v(t) = dx(t) / dt = -A * (2 * π / T) * sin(2 * π * t / T). If this value is positive, then the mass is moving in the positive x direction; if negative, then it's moving in the negative x direction.
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Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constant for Ar under these conditions is 1.5 × 10−3 mol/L·atm. Enter your answer in scientific notation.
Answer:
S= 1.40x10⁻⁵mol/L
Explanation:
The Henry's Law is given by the next expression:
[tex] S = k_{H} \cdot p [/tex] (1)
where S: is the solubility or concentration of Ar in water, [tex]k_{H}[/tex]: is Henry's law constant and p: is the pressure of the Ar
Since the argon is 0.93%, we need to multiply the equation (1) by this percent:
[tex] S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L} [/tex]
Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.
Have a nice day!
A cylindrical conductor with a circular cross section has a radius a and a resistivity p and carries a constant current I. (Take the current to be coming out of the page when the cross-sectional view of the conductor is in the plane of the page.) For the following questions, express your answer in terms of the variables I,a,p, and appropriate constants (u0,e0, and pi).a) what is the magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis?b) What is the magnitude of the magnetic field vector B at the same point?c) What is the magnitude of the Poynting vector S at the same point?d) What are the directions of these three vectors?
Answer:
Explanation:
a)
Using Ohms Law
[tex]R= \rho \frac{l}{\pi a^2 } \\V = IR = E l = I \rho \frac{l}{\pi a^2 } \\E = \frac{I \rho}{\pi a^2 } \\E = J \rho [/tex]
Where J is the current density [tex]J = \frac{I}{\pi a^2 } [/tex]
and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor [tex]E = \rhoJ[/tex] just inside at a radius a (and anywhere else).
b)
Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current
[tex]\oint B .dl = B 2 \pi a = \mu_{o} I [/tex]
such that
[tex]B = \frac{\mu_{o} I}{2 \pi a } [/tex]
and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.
c)
The Poynting vector is given by
[tex]S = \frac{1}{ \mu_o} |E \textrm{x}B| = \frac{\rho I^2}{2 \pi^2 a^3} [/tex]
and by the right hand rule it’s always pointing in towards the center of the conductor.
d)
Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c
a) The magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis is given by E = I / (2*pi*a*ε0). b) The magnitude of the magnetic field vector B at the same point is given by B = (μ0 * I)/(2*pi*a). c) The magnitude of the Poynting vector S at the same point is given by S = (I^2 * μ0) / (4 * pi * a * ε0). d) The electric field vector E is radially outward, the magnetic field vector B is tangent to circles centered on the conductor, and the Poynting vector S is in the same direction as the cross product of E and B.
Explanation:a) The magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis can be determined using Gauss's Law. Since the conductor is cylindrical and the current is constant, the electric field is uniform and given by E = I / (2*pi*a*ε0), where I is the current and ε0 is the permittivity of free space.
b) According to Ampère's Law, the magnitude of the magnetic field vector B at the same point is given by B = (μ0 * I) / (2 * pi * a), where μ0 is the permeability of free space.
c) The magnitude of the Poynting vector S at the same point is given by S = E x B = (I2 * μ0) / (4 * pi * a * ε0).
d) The direction of the electric field vector E is radially outward, away from the axis of the conductor. The magnetic field vector B is tangent to concentric circles centered on the conductor. The Poynting vector S is in the same direction as the cross product of E and B.
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If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)
a. so that it was also rotating?
b. It makes no difference.
c. so that it wasn't rotating?
To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as
[tex]KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]
Where,
m = Mass of the Vall
v = Velocity
I = Moment of inertia abouts its centre of mass
[tex]\omega =[/tex] Angular speed
Basically the two sums of energies is the consideration of translational and rotational kinetic energy.
a. so that it was also rotating?
The ball is rotating means that it has some angular speed:
[tex]KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]
[tex]1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]
When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.
[tex]1000J > \frac{1}{2}mv^2[/tex]
The ball will not go faster.
c. so that it wasn't rotating?
For the case where the angular velocity does not rotate it is zero therefore
[tex]KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]
[tex]1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2[/tex]
[tex]1000J = \frac{1}{2}mv^2[/tex]
All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.
b. It makes no difference.
Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.
The rotational movement of the ball doesn't contribute to its linear speed. If the same amount of energy is used, a ball, whether rotating or not, should travel at the same speed, ignoring air resistance.
Explanation:The rotational movement of the ball in this case would not alter the speed of the ball's travel when projected. One reason is that only the kinetic energy associated with linear motion (or translational motion) would affect the speed. Rotational energy doesn't contribute to the ball's translational or linear speed. So, regardless of whether the ball is rotating or not if the same amount of energy (1000 J in this case) is used, the ball would travel at the same speed. Therefore, the correct option would be "It makes no difference."
However, it's worth noting that in the real world, where we cannot ignore air resistance, a spinning ball might travel a different path due to the Magnus effect.
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A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.310 m along the surface before stopping. (a) What was teh initial speed of the bullet?
(b) What kind of collision took place between the bullet and the block? Explain your answer comparing the values of the kinetic energy before and after the collision.
(c) Calculate the impulse of the block from the moment just after the impact to his final position. Analyze the result.
Using conservation of momentum and the work-energy principle, the initial speed of the bullet can be calculated. The collision is inelastic because kinetic energy is not conserved. The impulse on the block is the change in momentum, calculated from the initial combined momentum to zero when it stops.
Initial Speed of the Bullet
To find the initial speed of the bullet, we use the principle of conservation of momentum for the collision and the work-energy principle for the block's movement after the collision.
For the conservation of momentum:
After the collision, the bullet and block move together, so the total final momentum is the combined mass times their velocity.
The normal force is equal to the weight of the block since the surface is horizontal.
Kinetic energy can then be equated to this work to find the velocity right after the collision.
Type of Collision
The initial momentum right after impact is the combined mass of the block and bullet times the velocity just after collision, which we obtained earlier.
Analyzing the result, we can discuss the magnitude of impulse and how it relates to the force of friction and the distance slid.
A car heading north collides at an intersection with a truck of the same mass as the car heading east. If they lock together and travel at 28 m/s at 27° north of east just after the collision, how fast was the car initially traveling? Assume that any other unbalanced forces are negligible.
A. 25 m/s
B. 13 m/s
C. 19 m/s
D. 50 m/s
Answer:
Initial speed of car = 25 m/s north direction
correct option is A. 25 m/s
Explanation:
given data
Final velocity of both = 28 m/s at 27 degrees north of east
to find out
how fast was the car initially traveling
solution
we consider both vehicle have same mass
mass of car m1 = mass of truck m2
so we will apply here conservation of momentum law
m1 × u1 j + m2 × u2 i = (m1 + m2) × ( v cosθ i + v sinθ j ) ...............1
here m is mass and u is initial velocity and v is final velocity
m × u1 j + m × u2 i = (2m) × ( 28 × cos27 i + 28 × sin27 j )
u2 i + u1 j = 2 ( 24.94 i + 12.71 j)
u2 i + u1 j = 49.88 i + 25.42 j
so compare both side we get
Initial speed of car u1 = 25.42 m/s north direction
Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generates electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.
Answer:
The power in this flow is [tex]9.56\times10^{8}\ W[/tex]
Explanation:
Given that,
Distance = 221 m
Power output = 680 MW
Height =150 m
Average flow rate = 650 m³/s
Suppose we need to calculate the power in this flow in watt
We need to calculate the pressure
Using formula of pressure
[tex]Pressure=\rho g h[/tex]
Where, [tex]\rho[/tex]= density
h = height
g = acceleration due to gravity
Put the value into the formula
[tex]Pressure=1000\times9.8\times150[/tex]
[tex]Pressure=1470000\ Nm^2[/tex]
We need to calculate the power
Using formula of power
[tex]P=Pressure\times flow\ rate[/tex]
Put the value into the formula
[tex]P=1470000\times650[/tex]
[tex]P=9.56\times10^{8}\ W[/tex]
Hence, The power in this flow is [tex]9.56\times10^{8}\ W[/tex]