Answer:
high real rate of interest, depressing
Explanation:
If output is less than potential and the rate of policy has fall to the ground of zero, the subsequent decline in inflation results in high real rate of interest which further depressing output, causing inflation to even further decline.
If the actual output is less than potential output, demand or supply has fallen, resulting in job and output declines
The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear momentum. However, since the two pucks, mutual center of mass does not coincide with either of the pucks velocity vectors, they have angular momentum. This becomes evident after the collision when due to conservation of angular momentum the two pucks spin around their mutual center of mass.
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks above, this pair of pucks will rotate:
a. at the same rate.
b. one-fourth as fast.
c. four times as fast.
d. twice as fast.
e. one-half as fast.
Compared to the pucks given, the pair of pucks will rotate at the same rate.
Answer: Option A
Explanation:
The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.
The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.
By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since [tex]L=I \times \omega[/tex], the turning rate of the two-puck framework must stay unaltered.
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a spring is initially compressed 2.08 cm and fires a 12.3-gram dart straight up into the air. If the dart reaches a height of 3.25 m, determine spring constant.
The spring constant is 181.0 N/m
Explanation:
We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:
[tex]\frac{1}{2}kx^2 = mgh[/tex]
where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where
k is the spring constant of the spring
x = 2.08 cm = 0.0208 m is the compression of the spring
m = 12.3 g = 0.00123 kg is the mass of the dart
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h = 3.25 m is the maximum height of the dart
Solving for k, we find:
[tex]k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m[/tex]
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In 2005 a planet was found revolving about a distant sunlike star in 4.01 days. The radius of its orbit is 2 comma 600 comma 000 mi. What is its linear velocity (in mi/h) about its star?
Answer: linear velocity = 169,745.2mi/hr
Explanation:
Given: radius of orbit= 2,600,000mi
Time taken for orbit = 4.01 days = 4.01 × 24 hrs= 96.24hrs
Linear velocity = distance travelled/time
distance travelled = circumference of orbit = 2πr
= 2π × 2,600,000mi
= 16336281.8mi
Linear velocity = 16336281.8/ 96.24hrs
= 169745.2mi/hr
Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is switched off in 1.00 ms, inducing a 9.00 V emf in the other. What is their mutual inductance?
Answer:
Mutual inductance will be [tex]M=1.8\times 10^{-3}Hennry[/tex]
Explanation:
We have given induced current [tex]\Delta i=5A[/tex]
Time is given as [tex]t=1ms=10^{-3}sec[/tex]
We have to find the mutual inductance between the coils
Induced emf is given as e = 9 volt
We know that induced emf is given by
[tex]e=M\frac{\Delta i}{\Delta t}[/tex]
[tex]9=M\times \frac{5}{10^{-3}}[/tex]
[tex]M=1.8\times 10^{-3}Hennry[/tex]
A 0.150 kg toy is undergoing SHM on the end of a horizontal spring with force constant 300.0 N/m. When the object is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.200 m/s. Find(a) the total energy of the object at any point in its motion,(b) the amplitude of the motion, and(c) the maximum speed attained by the object during its motion.
Answer:
a)TE=0.0245 J
b)A = 0.0128 m
c)V=0.57 m/s
Explanation:
Given that
m = 0.150 kg
K= 300 N/m
x= 0.012 ,v= 0.2 m/s
The velocity of the toy at any point given as
[tex]v=\omega\sqrt{A^2-x^2}[/tex]
[tex]\omega=\sqrt{\dfrac{K}{m}}[/tex]
[tex]v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}[/tex]
[tex]0.2=\sqrt{\dfrac{300}{0.15}}\times \sqrt{A^2-0.012^2}[/tex]
2 x 10⁻⁵ = A² - 0.000144
A=0.0128 m
Amplitude ,A = 0.0128 m
The total energy TE
[tex]TE=\dfrac{1}{2}KA^2[/tex]
[tex]TE=\dfrac{1}{2}300\times 0.0128^2[/tex]
TE=0.0245 J
The maximum speed
[tex]V=\omega A[/tex]
[tex]V=\sqrt{\dfrac{K}{m}}\times A[/tex]
[tex]V=\sqrt{\dfrac{300}{0.15}}\times 0.0128[/tex]
V=0.57 m/s
(a) The total energy of the toy is 0.0246 J (b) The amplitude of its motion is 0.0128 m (c) The maximum speed it can attain is 0.573 m/s.
To solve the given problem involving a toy undergoing simple harmonic motion (SHM) on a horizontal spring, let's go step-by-step:
(a) The total energy (E) in SHM is the sum of kinetic energy (KE) and potential energy (PE) at any point.
Given the mass (m) is 0.150 kg, spring constant (k) is 300.0 N/m, displacement (x) is 0.0120 m, and speed (v) is 0.200 m/s, we calculate:
Kinetic Energy: [tex]KE = (\frac{1}{2})mv^2 = (\frac{1}{2})(0.150 kg)(0.200 m/s)^2[/tex] = 0.003 J
Potential Energy:[tex]PE = (\frac{1}{2})kx^2 = (\frac{1}{2})(300.0 N/m)(0.0120 m)^2[/tex] = 0.0216 J
Total Energy: E = KE + PE = 0.003 J + 0.0216 J = 0.0246 J
Amplitude: The total energy in SHM is also equal to the potential energy at maximum displacement (amplitude A). Therefore, using [tex]E = (\frac{1}{2})kA^2[/tex], we solve for A:
[tex]0.0246 J = (\frac{1}{2})(300.0 N/m)A^2[/tex]
[tex]A^2 = (\frac{0.0246 J}{150 N/m}) = 0.000164 m^2[/tex]
[tex]A = \sqrt{0.000164 m^2} = 0.0128 m[/tex]
(b) Maximum Speed: The maximum speed (vm)) occurs at the equilibrium position.
Using [tex]E = (\frac{1}{2})mv^2_m[/tex], we solve for vm:
[tex]0.0246 J = (\frac{1}{2})(0.150 kg)v^2_max[/tex]
[tex]v^2_m = (\frac{0.0246 J}{0.075 kg}) = 0.328 m^2/s^2[/tex]
[tex]v_m = \sqrt{0.328 m^2/s^2} = 0.573 m/s[/tex]
Therefore, the total energy of the system is 0.0246 J, the amplitude of oscillation is 0.0128 m, and the maximum speed is 0.573 m/s.
Sand falls from a conveyor belt at a rate of 14 m cubed divided by min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the height and the radius changing when the pile is 5 m high?
Answer:7.16 cm/min or 0.0716 min.
Explanation:
From the equation the parameters given are; rate= 14m^-3/min, height of the pile,h= 3/8, rate when the pile is 5m high= ??.
d= diameter, r= radius. From the definition of diameter, diameter, d=2× radius, r(that is, 2r)
h= 3/8×d = 3/8(2r)= 3/4 × r
r= 4/3 × h
V= 1/3×πr^2×h ---------------------(1).
V= 1/3×π×[4/3h]^2×h
V= 16/27×π×h^3
Differentiate dV with respect to time, t;
dV/dt= 3(16)/27 × π ×h^2 ×dh/dt.
dV/dt = 16/9×π×h^2 × dh/dt
Then we differentiate V implicitly with respect to time,t
14=16/9×π[5]^2×dh/dt
dh/dt = 10× 9/400π
dh/dt= 45/200π (m/min)
dh/dt= 0.0716 m/min.
Conversion to cm/min
= dh/dt= 4500/200π (cm/min)
dh/dt= 1125/50π
dh/dt= 7.16 cm/min.
To find the speed at the edge of the storm, we divide the distance by the time.
Explanation:To find how fast the height and radius are changing when the pile is 5 m high, we need to find the rates of change of both the height and radius at that height.
Let's start by finding the rate of change of the height. We know that the height is always three-eighths of the base diameter, so the base diameter is twice the height. Therefore, when the height is 5 m, the base diameter is 2 * 5 = 10 m.
Now, we can use the information about the rate of sand falling from the conveyor belt to find the rate of change of the radius. The volume of a cone is given by the formula V = (1/3) * π * r^2 * h, where V is the volume, π is a constant (approximately 3.14), r is the radius, and h is the height. The rate of change of the volume with respect to time can be found using the chain rule.
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Compute the flux of F⃗ =xi⃗ +yj⃗ +zk⃗ through just the curved surface of the cylinder x2+y2=9 bounded below by the plane x+y+z=2, above by the plane x+y+z=4, and oriented away from the z-axis
The flux of F through the curved surface of the cylinder is [tex]\( 18\pi \).[/tex]
The flux of the vector field F = xi + yj + zk through the curved surface of the cylinder [tex]\( x^2 + y^2 = 9 \),[/tex] bounded below by the plane [tex]\( x + y + z = 2 \)[/tex] and above by the plane [tex]\( x + y + z = 4 \),[/tex] and oriented away from the z-axis can be calculated using the surface integral formula.
The flux [tex]\( \Phi \)[/tex]is given by:
[tex]\[ \Phi = \iint_S \mathbf{F} \cdot d\mathbf{S} \][/tex]
Where [tex]\( d\mathbf{S} \)[/tex] is the outward-pointing normal vector to the surface [tex]\( S \).[/tex]
First, let's parameterize the surface. Since the cylinder is described by [tex]\( x^2 + y^2 = 9 \)[/tex], we can use cylindrical coordinates:
[tex]\[ x = 3\cos \theta \]\[ y = 3\sin \theta \]\[ z = z \][/tex]
The normal vector [tex]\( d\mathbf{S} \)[/tex] can be calculated using the cross product of the partial derivatives of [tex]\( \mathbf{r}(\theta, z) = (3\cos \theta, 3\sin \theta, z) \)[/tex] with respect to [tex]\( \theta \) and \( z \):[/tex]
[tex]\[ \frac{\partial \mathbf{r}}{\partial \theta} = (-3\sin \theta, 3\cos \theta, 0) \]\[ \frac{\partial \mathbf{r}}{\partial z} = (0, 0, 1) \][/tex]
Taking the cross product:
[tex]\[ d\mathbf{S} = \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial z} \right| d\theta dz = 3 \, d\theta dz \][/tex]
Now, calculate F.ds:
[tex]\[ \mathbf{F} \cdot d\mathbf{S} = (3\cos \theta, 3\sin \theta, z) \cdot (0, 0, 3) = 3z \, d\theta dz \][/tex]
The bounds of integration for [tex]\( \theta \) are \( 0 \) to \( 2\pi \)[/tex] since we want to cover the entire curved surface of the cylinder. For [tex]\( z \),[/tex] the bounds are from [tex]\( 2 \) to \( 4 \)[/tex] since the surface is bounded below by [tex]\( x + y + z = 2 \)[/tex] and above by [tex]\( x + y + z = 4 \).[/tex]
Now integrate:
[tex]\[ \Phi = \int_0^{2\pi} \int_2^4 3z \, dz \, d\theta \]\[ \Phi = \int_0^{2\pi} \left[ \frac{3}{2}z^2 \right]_{z=2}^{z=4} \, d\theta \]\[ \Phi = \int_0^{2\pi} \frac{3}{2} \cdot 12 \, d\theta \]\[ \Phi = 18\pi \][/tex]
So, the flux of F through the curved surface of the cylinder is [tex]\( 18\pi \).[/tex]
Complete Question:
Compute the flux of F⃗ =xi⃗ +yj⃗ +zk⃗ through just the curved surface of the cylinder x2+y2=9 bounded below by the plane x+y+z=2, above by the plane x+y+z=4, and oriented away from the z-axis
Two pipes are smoothly connected together without leaks. One has a diameter of 3 cm and the other has a diameter of 5 cm. Water flows through the pipes. In which pipe is the flow speed the greatest? in which pipe is the volume flow rate the greatest?
Answer:
The speed will be the greatest in the smaller pipeThe volume flow rate is the same in both pipesExplanation:
This question can be answered using the equation of continuity. Which is
V = A1v1 = A2v2
where V is the volume flow rate
A is the cross sectional area of the pipe at that position
v is the velocity of the flow at that point.
From the equation we can see the volume flow rate does not change even as the area and flow speed change. Which is consistent with the assumptions used in this equation which is that
The tube must one entry and one exit.The fluid in question is non-viscous.The fluid is incompressible.The flow is steady.If the temperature does not change, meaning there is not expansion/contraction and it is incompressible then the volume flow rate does not change.
And the speed will be higher in the pipe with the smaller diameter because area and speed are inversely related.
The flow speed is greatest in the smaller pipe, while the volume flow rate is greatest in the larger pipe.
Explanation:The flow speed is the greatest in the pipe with the smaller diameter, which is the pipe with a diameter of 3 cm. This can be determined from the continuity equation, which states that the flow speed is inversely proportional to the cross-sectional area. Since the smaller pipe has a smaller cross-sectional area, the flow speed is greater.
On the other hand, the volume flow rate is determined by both the flow speed and the cross-sectional area. The volume flow rate is the greatest in the pipe with the larger diameter, which is the pipe with a diameter of 5 cm. This is because even though the flow speed is slower in the larger pipe, the larger cross-sectional area compensates for it and allows a greater volume of water to flow through.
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Use the work-energy theorem to solve each of these problems. You can use Newton’s laws to check your answers. (a) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?
Answer:
d = 5.8 m
Explanation:
Principle of work and energy
ΔE = Wf
ΔE = Ef-Ei
ΔE : mechanical energy change
Wf : Work done by kinetic friction force
Ef : final mechanical energy
Ei : initial mechanical energy
K =(1/2 )mv² : Kinetic energy
U= mgh :Potential energy
m: mass (kg)
v : speed (m/s)
h: high (m)
Data
vi= 5.00 m/s
vf=0
hi=0
hf=0
μk=0.220
Kinetic friction force
fk = μk* FN
FN = W : normal force (N)
W = m*g : weight (N)
FN= 9.8*m (N)
fk =0.220*9.8*m
fk = (2.156)*m Equation (1)
Work done by the kinetic friction force
Wf = -fk*d (J) Equation (1)
d: distance traveled by force
Principle of work and energy to the skier
ΔE = Ef-Ei
Ef = Kf + Uf = 0
Ei = Kf + Uf = (1/2)(m)(5)² + 0
ΔE = -(1/2)(m)(5)²
ΔE = Wf
0- (1/2)(m)(5)² = -fk*d
We replace fk = (2.156)*(m )of the equation (1)
-(1/2)(m)(5)²= -(2.156)*(m)*d
We divide by (-m ) both sides of the equation
(1/2)(5)²= (2.156)*d
12.5= ( 2.156)*d
d = (12.5) / ( 2.156)
d = 5.8 m
To calculate the distance a skier travels before stopping, we equate the skier's initial kinetic energy to the work done by friction. The force exerted by friction is calculated using the given coefficient of kinetic friction and the skier's weight. The distance traveled can be calculated using this force with the initial speed of the skier.
The problem you've posed can be solved using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. We can use the principle of conservation of energy here. Initially, the skier has kinetic energy and no potential energy.
When she stops, all her kinetic energy has been transferred into frictional work, hence her kinetic energy becomes 0. Kinetic energy can be defined as (1/2)mv^2 and the work done by friction is force times distance (Fd).
Force exerted by friction (F) can be calculated using the formula F = μN where μ corresponds to the coefficient of kinetic friction given as 0.220 and N is the normal force and in this case, equals to the weight of the skier (assuming the skier is on a horizontal plane).
To find the distance (d) travelled before she comes to a stop, we equate the skier's initial kinetic energy to the work done by friction, giving us (1/2)mv^2 = μmgd. The mass of the skier (m) cancels out, and the distance comes out to be d = v^2/(2μg).
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What is the mass of a large ship that has a momentum of 1.60×109kg·m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.
a) The mass of the ship is [tex]1.2\cdot 10^8 kg[/tex]
b) The ship has a larger momentum than the shell
Explanation:
a)
The momentum of an object is given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
For the ship in this problem, we have
[tex]p=1.60\cdot 10^9 kg m/s[/tex] is the momentum
[tex]v=48.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=13.3 m/s[/tex] is the velocity
Solving for m, we find the mass of the ship:
[tex]m=\frac{p}{v}=\frac{1.60\cdot 10^9}{13.3}=1.2\cdot 10^8 kg[/tex]
b)
The momentum of the artillery shell is given by
[tex]p=mv[/tex]
where
m is its mass
v is its velocity
For the shell in this problem,
m = 1100 kg
v = 1200 m/s
Substituting,
[tex]p=(1100)(1200)=1.32\cdot 10^6 kg m/s[/tex]
So, we see that the ship has a larger momentum.
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Momentum is the product of force and velocity. The mass of the ship and the momentum of the fired artillery is [tex] 1.2 \times 10^{8}[/tex] and [tex]1.32 \times 10^{6} kgm/s [/tex] respectively
Momentum = mass × velocity
momentum, p = [tex]1.60 \times 10^{9} [/tex] Speed, v = [tex] 48 km/hr [/tex] Speed in m/s = (48×1000)/3600 = 13.33 m/sMass of the ship = momentum / velocity
Mass of ship = [tex]\frac{1.60 \times 10^{9}}{13.33} = 1.2 \times 10^{8}[/tex]
2.)
Speed, v = 1200 m/s Mass, m = 1100 kgMomentum of fired artillery = 1100 × 1200 = [tex]1.32 \times 10^{6} kgm/s [/tex]
Therefore, the momentum of fired artillery is [tex]1.32 \times 10^{6} kgm/s [/tex]
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What is the relationship between the stoma and an air space
Answer:
Stoma represents the interface between the environment and the plant, helping to obtain the necessary CO2.
Explanation:
Stoma are groups of two or more specialized epidermal cells whose function is to regulate gas exchange and perspiration.
The frequency or density varies widely from a few tens to thousands per mm2, due to the influence of environmental factors, leaf morphology and genetic composition.
An area with excess electrons has a net _____ charge; an area with a deficit of electrons has a net _____ charge. Question 10 options: negative, positive positive, negative positive, neutral negative, neutral
Answer:
negative, positive
Explanation:
A we know that the electrons have a negative charge.
So, when a body has some excess electrons, it means it has negative charge.
When a body has an deficiency of eletrons, it means it gains a positive charge.
An area with excess electrons has a net _negative__ charge; an area with a deficit of electrons has a net _Positive____ charge.
Find an expression for the normal force n acting on the car when it is at the top of the arc. (Use any variable or symbol stated above along with the following as necessary: m and g.)
Answer:
[tex]F_n = mg - \frac{mv^2}{R}[/tex]
Explanation:
As we know that when an object moves in a circle with uniform speed then the force required by the object in moving the circular path is known as centripetal force.
This force is always towards the center of the circle and points towards it
This force is the sum of all forces towards the center
so we have
[tex]mg - F-n = F_c[/tex]
[tex]F_c = \frac{mv^2}{R}[/tex]
so we have
[tex]mg - F_n = \frac{mv^2}{R}[/tex]
[tex]F_n = mg - \frac{mv^2}{R}[/tex]
People who work around radioactivity wear film badges to monitor the amount of radiation that reaches their bodies. These badges consist of small pieces of photographic film enclosed in a light-proof wrapper. What kind of radiation do these devices monitor?
Answer:
the kind of radiations of the devices monitor are gamma, X-Rays and beta particles.
Explanation:
Film badges, which is usually worn in the front body, is a device to monitor the exposure of the radiation. This device measures the the exposure of X-Rays, Gamma Rays and beta Particles. This device is able to differentiate the energy of photon and is considerably accurate for the exposure greater than 100 millirem. However, under 20 milllirem, the device is less accurate for gamma rays.
A satellite explodes in outer space, far from any other body, sending thousands of pieces in all directions. How does the linear momentum of the satellite before the explosion compare with the total linear momentum of all the pieces after the explosion? Account for your answer.
Answer:
They would be the same
Explanation:
Since there's no external force, or the external net force acting on the satellite is 0. By the law of conservation of momentum, the momentum before and after the explosion must be the same.
Even if the satellite is stationary prior to the explosion, its momentum is 0. After explosion it sends many of its pieces in all direction with different non 0 momentum. The total momentum by account of their direction would sum up to be 0.
Final answer:
The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion.
Explanation:
The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion. This is because linear momentum is conserved in an explosion. Although the distribution of momentum among the individual pieces may change, the total momentum remains the same. In this case, when the satellite explodes, the momentum of the satellite is redistributed among the thousands of pieces that are scattered in all directions.
How do lone pairs of electrons affect the bond angle differently than electrons shared in a bond?
Answer:
Lone pairs cause bond angles to deviate away from the ideal bond angles
Explanation:
Bonded electrons are stabilized and clustered between the bonding electrons meaning they are much closer together. Non-bonding electrons however are not being shared between any atoms which allows them to roam a little further spreading the charge density over a larger space and therefore interfering with what would be an expected bond angle
5. A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest? (a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive? (b) How many revolutions did the wheel make during the time it was coming to rest?
Answer:
a) -1.57 rad/s²
b) 800 revolutions
Explanation:
a)In here you can use the equations of velocity as if it were a linear movement. In this case:
wf = wo + at
wo is the innitial angular velocity, that we can get this value using the fact that a revolution is 2π so:
wo = 20 * 2π = 125.66 rad/s
We have the time of 80 seconds, and the final angular speed is zero, because it's going to a rest so:
0 = 125.66 + 80a
a = -125.66 / 80
a = -1.57 rad/s²
b) In this part, we will use the following expression:
Ф = Фo + wo*t + 1/2 at²
But as this it's coming to rest then:
Ф = 1/2at²
solving we have:
Ф = 0.5 * (-1.57)*(80)²
Ф = 5,024 rad
Ф = 5024 / 2π
Ф = 800 revolutions
(a) The angular acceleration of the grinding wheel is approximately [tex]-1.57 \, \text{rad/s}^2[/tex]. (b) The grinding wheel made about 800 revolutions during the time it was coming to rest.
To solve the problem of the grinding wheel, we need to find the angular acceleration and the number of revolutions during its deceleration.
Mass of the grinding wheel, [tex]m = 1.15 \, \text{kg}[/tex] Diameter of the grinding wheel, [tex]d = 22.0 \, \text{cm} = 0.22 \, \text{m}[/tex]Radius of the grinding wheel, [tex]r = \frac{d}{2} = 0.11 \, \text{m}[/tex]Initial angular velocity, [tex]\omega_0 = 20.0 \, \text{rev/s}[/tex]Time to come to rest, [tex]t = 80.0 \, \text{s}[/tex]First, we convert the initial angular velocity from revolutions per second to radians per second:
[tex]\omega_0 = 20.0 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 40\pi \, \text{rad/s} \approx 125.66 \, \text{rad/s}[/tex]
Angular acceleration [tex]\alpha[/tex] can be calculated using the formula:
[tex]\alpha = \frac{\omega_f - \omega_0}{t}[/tex]
where [tex]\omega_f = 0 \, \text{rad/s}[/tex] (final angular velocity, when the wheel comes to rest).
Substituting in the values:
[tex]\alpha = \frac{0 - 125.66}{80.0} = \frac{-125.66}{80.0} \approx -1.57 \, \text{rad/s}^2[/tex]
To find how many revolutions the wheel made while coming to rest, we can use the formula for angular displacement [tex]\theta[/tex]:
[tex]\theta = \omega_0 t + \frac{1}{2} \alpha t^2[/tex]
Substituting the values:
[tex]\theta = (125.66 \, \text{rad/s}) (80.0 \, ext{s}) + \frac{1}{2} (-1.57) (80.0)^2[/tex]
[tex]\theta = 10052.8 \, \text{rad} - \frac{1}{2} (1.57) (6400)\approx 10052.8 - 5024\approx 5028.8 \, \text{rad}[/tex]
Now convert radians to revolutions:
[tex]\text{Revolutions} = \frac{\theta}{2\pi} = \frac{5028.8}{2\pi} \approx 799.8 \, ext{rev} \approx 800 \, ext{rev}[/tex]
specific heat capacity of ice = 2,000 J/kg-°C
latent heat of fusion of water = 334,000 J/kg
specific heat capacity of water = 4,000 J/kg-°C
After the 1 kg block of ice melts, the amount of heat required to raise its temperature to 10 °C is
A. 2,000 J.
B. 4,000 J.
C. 20,000 J.
D. 40,000 J.
E. 334,000 J.
Answer: D. 40,000Joules
Explanation:
Heat capacity is defined as the heat required to change the state of a substance.
Simce the substance in question is a block of ice (solid), the heat needed to cause it to melt is Latent heat H = mL
m is the mass of the substance
L is the latent heat of fusion of water
H= 1 × 334,000
= 334,000Joules
Heat required to raise the temperature of the water to 10°C
Q = mCƦ
where C is specific heat capacity of water and Ʀ is change in temperature
Q = 1 × 4000 ×(10-0)
Q= 40,000Joules
Since the block already melts, the heat required will just be
40,000Joules (D)
Final answer:
To raise the temperature of 1 kg of melted ice water to 10°C, the heat energy required is calculated using the specific heat capacity of water (4,000 J/kg-°C) and the formula Q = mcΔT, resulting in 40,000 J.
Explanation:
The question is asking about the amount of heat energy required to raise the temperature of 1 kg of water (from melted ice) to 10 °C. After the ice melts, the water at 0 °C needs to be heated to 10 °C. Using the specific heat capacity of water, which is 4,000 J/kg-°C for this calculation, we can find the amount of energy needed with the formula:
Q = mcΔT
where Q is the heat energy, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values:
Q = (1 kg)(4,000 J/kg-°C)(10 °C - 0 °C)
Q = (1 kg)(4,000 J/kg-°C)(10 °C)
Q = 40,000 J
Therefore, the correct answer is D. 40,000 J.
A certain amount of a monatomic gas is maintained at constant volume as it is cooled by 50K. This feat is accomplished by removing 400 J of energy from the gas. How much work is done by the gas?
a) 400 J
b) -400 J
c) zero
d) none of the above
In an isochoric process, where volume is constant and the gas is cooled, no work is done by the gas. Therefore, the answer is zero.
Explanation:In the thermodynamics, if a given amount of a gas is maintained at a constant volume and is cooled, the whole process occurs in what is known as an isochoric process. In an isochoric process, the volume is held constant which means the gas does no work because there isn't any volume change. Therefore, work done by the gas in this case is zero, hence, the answer is (c) zero. Work = Change in Internal Energy (ΔU) - Heat (Q)
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The volume of the monatomic gas is kept constant. When the volume is constant, the work done by the gas is zero because work is done when there is a change in volume. Therefore, the correct answer to this physics question is c) zero.
Explanation:Considering the situation, we can note that the volume of the monatomic gas is kept constant. When the volume is constant, the work done by the gas is zero because work is done by a gas when it expands or contracts, which involves a change in volume. The only change occurring here is the cooling of the gas by 50K, that is achieved by removing 400J of energy from the gas, but this does not involve any work done by the gas itself.
Therefore, the correct answer to this physics question is c) zero. This answer can be found using the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat transferred to the system, minus the work done by the system. In this case, there is no work done by the system since the volume is constant.
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A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.
Answer:
a)
[tex]mv l[/tex]
b)
[tex]\frac{M }{(M + m)}[/tex]
Explanation:
Complete question statement is as follows :
A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.
(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)
(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)
a)
[tex]m[/tex] = mass of the bullet
[tex]v[/tex] = velocity of the bullet before collision
[tex]r[/tex] = distance of the line of motion of bullet from pivot = [tex]l[/tex]
[tex]L[/tex] = Angular momentum of the bullet-block system
Angular momentum of the bullet-block system is given as
[tex]L = m v r[/tex]
[tex]L = mv l[/tex]
b)
[tex]V[/tex] = final velocity of bullet block combination
Using conservation of momentum
Angular momentum of bullet block combination = Angular momentum of bullet
[tex](M + m) V l = m v l\\V =\frac{mv}{(M + m)}[/tex]
[tex]K_{o}[/tex] = Initial kinetic energy of the bullet
Initial kinetic energy of the bullet is given as
[tex]K_{o} = (0.5) m v^{2}[/tex]
[tex]K_{f}[/tex] = Final kinetic energy of bullet block combination
Final kinetic energy of bullet block combination is given as
[tex]K_{f} = (0.5) (M + m) V^{2}[/tex]
Fraction of original kinetic energylost is given as
Fraction = [tex]\frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}[/tex]
Fraction = [tex]\frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}[/tex]
Fraction = [tex]\frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}[/tex]
The question is about the physics principle of Conservation of Angular Momentum in a completely inelastic collision involving a bullet and a block attached to a rod. After the collision, their total angular momentum remains constant due to no external torques acting on the system. This allows the analysis of their post-collision movement.
Explanation:The subject of this question is Physics and it deals with the concept of Conservation of Angular Momentum during a collision. When the bullet travels horizontally and hits the wooden block, it becomes embedded in it. In this scenario, the system of the block and bullet then starts moving as a single unit. This event is typically referred to as a completely inelastic collision.
The conservation of angular momentum principle states that the total angular momentum of a system remains constant if no external torques act on it. Here, the bullet-block system doesn't experience any external torques. Therefore, their total angular momentum before the collision is equivalent to their total angular momentum after the collision. This principle allows us to analyze the movement of the bullet-block system after the collision.
It's worth noting that in this setup, the block, the rod, and the position at which the bullet hits the block together form a simple physical pendulum. Therefore, the post-collision dynamics of the system could also involve oscillatory movements, which relate to the aspects of pendulum physics.
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(a) Calculate the magnitude of the gravitational force exerted on a 1,320-kg satellite that is a distance of two earth radii from the center of the earth. N
(b) What is the magnitude of the gravitational force exerted on the earth by the satellite? N
(c) Determine the magnitude of the satellite's acceleration. m/s2
(d) What is the magnitude of the earth's acceleration? m/s2 Additional Materials
Answer:
(a) 3224.27 N
(b) 3224.27 N
(c) 2.443 m/s^2
(d) 5.37 x 10^-22 m/s^2
Explanation:
Mass of satellite, m = 1320 kg
mas of earth, M = 6 x 10^24 kg
Radius of earth, r = 6.4 x 10^6 m
(a) The force of gravitation between the earth and the satellite is given by
[tex]F=G\frac{Mm}{d^{2}}[/tex]
where, d is the distance between the two objects
[tex]F=6.67\times10^{-11}\frac{6\times10^{24}\times 1320}{\left (2\times 6.4\times 10^{6} \right )^{2}}[/tex]
F = 3224.27 N
(b) The force on earth is same as the force on satellite.
F = 3224.27 N
(c) Acceleration of satellite = Force on satellite / mass of satellite
Acceleration of satellite = 3224.27 / 1320 = 2.443 m/s^2
(d) Acceleration of earth = Force on earth / mass of earth
Acceleration of satellite = 3224.27 / (6 x 10^24) = 5.37 x 10^-22 m/s^2
Recrystallization involves any chemical, physical, or biological changes that take place after sediments are deposited and burial and lithification occur when unconsolidated sediments are transformed into sedimentary rocks. True or False?
The statement is correct. Recrystallization refers to any chemical, physical, or biological changes that occur after sediments are deposited, whereas burial and lithification occur when unconsolidated sediments become sedimentary rocks.
What is sedimentary rocks?Sedimentary rocks are formed by the accumulation or deposition of mineral or organic particles at the Earth's surface, followed by cementation. Sedimentation refers to the processes that cause these particles to settle in place. Sedimentary rocks form from previously existing rocks or fragments of once-living organisms. They form as a result of deposits accumulating on the Earth's surface. Layering or bedding in sedimentary rocks is common. Sedimentary rocks are formed by sediment accumulation. Sedimentary rocks are classified into three types. Clastic sedimentary rocks form as a result of mechanical weathering debris accumulation and lithification. Breccia, conglomerate, sandstone, siltstone, and shale are some examples.To learn more about sedimentary rocks, refer to:
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Gel electrophoresis sorts dna molecules on the basis of their
Answer:
Size and charge and speed
Explanation:
Gel electrophoresis is method designed for the separation of DNA molecules. The are also used to separate other micromolecules like RNA and protein, etc. This separation takes place based on their size and charge. The molecule travels through the gel in all different directions but their speed are different and thus they are separated.
Gel electrophoresis is a technique used to separate DNA fragments based on their size.
Explanation:Gel electrophoresis is a technique used to separate DNA fragments based on their size. It involves loading DNA samples onto a gel matrix and applying an electric current. Smaller DNA fragments move faster through the gel, resulting in bands at specific distances from the top of the gel. Molecular weight standard samples can be run alongside the DNA fragments to provide a size comparison. The separated DNA fragments can then be visualized by staining the gel with a DNA-specific dye.
Gel electrophoresis is a fundamental technique used in molecular biology, genetics, and forensics for various purposes, including DNA fingerprinting, DNA sequencing, and the analysis of PCR products. It provides a visual representation of DNA fragment sizes, enabling researchers to draw conclusions about genetic material and study genetic variations.
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Explain briefly why the intensity reflected off the back surface of the film (i.e., the right surface, where there is a liquid-to-air interface) is 2.78% of the intensity of the beam that was transmitted through the front surface.
Explanation:
Taking the incident light to be traveling in the + x-direction so that it is at normal incidence to the left side of the film(referred to as the "Front side"). This means the beam transmitted into the liquid is essentially as strong as the incident beam.
Almost all the light that is reflected off the back surface will get through the front surface. (But only 2.78% gets re-reflected off the the front surface back to the right) this means that there are two beams reflected to the - x-direction, one from the front surface and one from the back, and these beams are of almost equal intensity.
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Which of these types of mass movements is the fastest and least depended upon moist or saturated conditions?
a. solifluction
b. rock fall
c. rock slide
d. soil creep
e. talus creep
Answer:
Option (B)
Explanation:
A rockfall is defined as a type of mass movement process, where the small and large fragments of rocks are weathered and fall down from the top of the cliff under the influence of gravity. It is a process that takes place in a dry condition. This means that it does not depend on the moisture content that could be present in the rock. The rock particles fall down by bouncing and rolling. The base of this rockfall area forms a collection of smaller to larger size rock fragments. This process mainly takes place in the region where the temperature of the area is very high and experiences comparatively less amount of rainfall.
Thus, this type of mass movement is the fastest of all and is not dependent on the amount of moisture content.
Thus, the correct answer is option (B).
The option that gives the type of mass movement that is the fastest and depends upon moist or saturated conditions the least is;
Option B; Rockfalls
To answer this question, let us analyze each of the options.
Option A; Solifluction; This occurs when is the soil becomes saturated and then the wet soil starts to flow down a slope. This alongside creep are the slowest types of mass movement.
Option B; Rockfall; These occur when rock fragments that could be tiny or huge fall down from cliffs that are steep. Thus, this is the fastest type of mass movement as it is not dependent on moist or saturated conditions.
Option C; Rockslides; These are also known as landslides and they occur when numerous amounts of loose rocks get to join with soils and suddenly fall down from a slope.
Option D; Soil Creep; This means a slow movement of soil material down a slope under the Influence of gravity.
Option E; Talus Creep; This is a very slow movement of individual rock fragments individually down a slope.
Looking at the analysis of the options above, we can say that the fastest form of mass movement is Rockfalls.
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A uniform metal rod, with a mass of 3.7 kg and a length of 1.2 m, is attached to a wall by a hinge at its base. A horizontal wire bolted to the wall 0.51 m above the base of the rod holds the rod at an angle of 25 degrees above the horizontal. The wire is attached to the top of the rod.
Find the tension in the wire.
a.Find the horizontal component of the force exerted on the rod by the hinge.
b.Find the vertical component of the force exerted on the rod by the hinge.
Answer:
Explanation:
On the rod , three forces are acting .
1 ) its weight from the middle point .
2 ) Tension in the wire in the horizontal direction
3 ) reaction force by the hinge which balances the other two forces
Now taking moment of forces acting on the rod about the hinge
and equating opposite moment
3.7 x 9.8 x .6 x cos 25 = T x .51 ( T is tension in the horizontal wire)
T = 38.6 N
a )
The horizontal component of the force exerted on the rod by the hinge
will balance the tension in the wire acting in horizontal direction so it will be equal to
38.6 N
b ) The vertical component of the force exerted on the rod by the hinge
will balance the weight of rod so it will be equal to
3.7 x 9.8 N
= 36.26. N
=
(a) The horizontal component of the force exerted on the rod by the hi1ge is 38.67 N.
(b) The vertical component of the force exerted on the rod by the h1nge is 36.26N.
How to calculate the tension in the wire?The tension in the wire is calculated by applying the following formula as shown below;
We will apply the principle of moment, sum of clockwise moment is equal to sum of anti clockwise moment.
(a) The horizontal component of the force exerted on the rod by the hi1ge is calculated as;
taking moment of forces acting on the rod about the h1nge;
3.7 x 9.8 x .6 x cos 25 = T x .51
where;
T is the tension in the horizontal wire3.7 kg x 9.8 m/s² x 0.6 m x cos 25 = Tₓ x .51 m
19.72 = 0.51Tₓ
Tₓ = 19.72 / 0.51
Tₓ = 38.67 N
(b) The vertical component of the force exerted on the rod by the h1nge will balance the weight of rod so it will be equal to;
Ty = 3.7 x 9.8 N
Ty = 36.26N
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A potter is shaping a bowl on a potter's wheel rotating at constant angular speed. The friction force between her hands and the clay is 1.6 N total.(a) How large is her torque on the wheel, if the diameter of the bowl is 12 cm?(b) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hand? The initial angular velocity of the wheel is 1.6 rev/s, and the moment of inertia of the wheel and the bowl is .11 kg*m^2 Show all work and formulas for best rating.
Answer:
Part A:
T=0.096 N m
Part B:
t=11.525≅11.53 seconds
Explanation:
Part A:
In order to find the torque we will use the following formula:
[tex]T=r*F[/tex]
Where T is the torue
r is the radius of bowel
F is the force
[tex]r=\frac{12*10^{-2} }{2} m\\[/tex]
r=0.06 m
[tex]T=0.06*1.6[/tex]
T=0.096 N m
Part B:
In order to find how long would it take for potter wheel to sto we will proceed the following way:
T=I*α
Where:
T is the Torque
I is the moment of inertia
α is the angular acceleration
α=T/I
we will take T from art A
α=0.096/0.11
α=0.872 [tex]rad/s^2[/tex]
α=ω/t
where:
α is the angular accceleration
ω is angular velocity in rad/s
ω=1.6*2ππ
ω=10.05 rad/s
t=ω/α
[tex]t=\frac{10.05}{0.872}[/tex]
t=11.525≅11.53 seconds
It will take 11.53 sec for wheel to stop
The torque exerted by the potter on the wheel is 0.192 Nm. If the only force acting on the wheel is the potter's hand, the wheel will stop in approximately 5.72 seconds.
Explanation:The question is asking about the concepts of torque, angular velocity, and moment of inertia in a real-life situation involving a potter's wheel. We can solve this problem in two parts.
Torque: Torque (τ) can be calculated using the formula τ = r*F, where r is radius and F is force. In this case, the radius, r = 12 cm = 0.12 m (since we need to convert cm to meters for the SI units.) and F = 1.6 N. So, the torque exerted by the potter can be calculated as τ = 0.12 m * 1.6 N = 0.192 Nm.
Time to stop: If the only torque acting on the wheel is due to the potter's hand, we can use the rotational analog of Newton's second law, τ = I*a, where a is the angular acceleration, I is the moment of inertia and τ is the torque. The angular acceleration can be calculated as a = τ / I = 0.192 Nm / 0.11 kg*m^2 = 1.75 rad/s^2. The time it takes to stop the wheel can then be calculated using the formula t = (final velocity - initial velocity) / angular acceleration. As the final angular velocity (when the wheel stops) is 0, and the initial angular velocity is 1.6 rev/s = 1.6 * 2π rad/s (as 1 rev/s = 2π rad/s), the time to stop the wheel, t = (0 - 1.6*2π rad/s) / 1.75 rad/s^2, t = approximately 5.72 seconds, assuming that the only force acting on the potter's wheel is the hand of the potter.
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A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are μs=0.5 and μk=0.4, the magnitude of the frictional force on the crate is:
The magnitude of the frictional force on the crate is 20 N.
Explanation:The magnitude of the frictional force on the crate can be determined using the equation fs = μsN, where fs is the frictional force, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is equal to the weight of the crate, which is 40 N. Therefore, the frictional force is fs = (0.5)(40 N) = 20 N.
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How many seconds does it take for sunlight to reach the earth
Answer:
498.67 seconds, wich is 8.31 minutes
Explanation:
We can calculate it because we know the distance between the sun and the earth, and the speed at which light travels.
The distance from the sun to the earth is:
[tex]d=149.6 x10^9m[/tex]
and the speed of light:
[tex]v=3x10^8m/s[/tex]
this the tie it takes for light of the sun to reach earth can be calculated by the next equation:
[tex]t=\frac{d}{v}[/tex]
Substituting the v and d values:
[tex]t=\frac{149.6 x10^9m}{3x10^8m/s}=498.67s[/tex]
This is the anwer: 498.67s
if we want it in minutes we just divide by 60:
[tex]498.67s/60=8.31 minutes[/tex]
You leave your 75 W portable color TV on for 6 hours each day and you do not pay attention to the cost of electricity. If the dorm (or your parents) charged you for your electricity use and the cost was $0.3 /kW · h, what would be your monthly (30 day) bill?Answer in units of dollars.
Final answer:
To calculate the monthly cost of electricity for a portable TV, convert the power to kilowatts, multiply by daily usage, then by the number of days in a month, and finally by the cost per kWh. The monthly bill for a 75 W TV used 6 hours a day at a rate of $0.3/kWh is $4.05.
Explanation:
Calculating Monthly Cost of Electricity for a Portable TV
First, we need to convert the power consumed by the TV from watts to kilowatts (kW):
75 W = 0.075 kW
Next, we calculate the energy used each day by multiplying the power consumption (in kW) by the time the TV is on (in hours):
Daily energy consumption = 0.075 kW × 6 h/day = 0.45 kWh/day
To find the monthly energy consumption, we multiply the daily consumption by 30 days:
Monthly energy consumption = 0.45 kWh/day × 30 days = 13.5 kWh/month
Finally, to calculate the monthly bill, we multiply the monthly energy consumption by the cost of electricity (in dollars per kWh):
Monthly bill = 13.5 kWh/month × $0.3/kWh = $4.05/month
The student's monthly electric bill for the portable TV would be $4.05.