The magnitude of the impulse delivered to the roof is -0.01134 kg·m/s, and the magnitude of the force of the impact is -30.70 N.
Explanation:The magnitude of impulse delivered to your roof can be calculated using the equation impulse = mass x change in velocity. In this case, the mass of the raindrop is given as 0.0014 g, which is equivalent to 0.0014 kg. The change in velocity is the final velocity (0 m/s) minus the initial velocity (8.1 m/s). Therefore, the magnitude of the impulse is 0.0014 kg x (-8.1 m/s) = -0.01134 kg·m/s.
To calculate the magnitude of the force of the impact, we can use the equation force = impulse/time. In this case, the impulse is the magnitude of the impulse calculated previously (-0.01134 kg·m/s) and the time is given as 0.37 ms, which is equivalent to 0.00037 s. Therefore, the magnitude of the force of the impact is -0.01134 kg·m/s / 0.00037 s = -30.70 N.
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In this physics problem, we calculate both the magnitude of the impulse delivered and the force of impact of a raindrop hitting a roof.
The magnitude of the impulse delivered to the roof can be calculated by multiplying the mass of the raindrop by its velocity.
The magnitude of the force of the impact can be determined using the formula for impulse, considering the time it takes for the raindrop to come to rest.
The potential difference between a storm cloud and the ground is 100 million V.If a charge of 1.2 C flashes in a bolt from cloud to Earth, what is the change of potential energy of the charge?Express your answer to two significant figures and include the appropriate units.
Answer:
Change in potential energy, [tex]U=1.2\times 10^8\ J[/tex]
Explanation:
Given that,
The potential difference between a storm cloud and the ground is 100 million V, [tex]V=100\ million V=100\times 10^6\ V=10^8\ V[/tex]
If a charge of 1.2 C flashes in a bolt from cloud to Earth, q = 1.2 C
We need to find the change of potential energy of the charge. The relation between the potential difference and the potential energy of the charge is given by :
[tex]U=qV[/tex]
U is the potential energy of the charge
[tex]U=1.2\times 10^8\ J[/tex]
So, the change of potential energy of the charge is [tex]U=1.2\times 10^8\ J[/tex]. Hence, this is the required solution.
According to the question,
Potential difference, V = 100 mV= 100×10⁶ V
= 10⁸ V
Charge, q = 1.2 CThe change in Potential energy:
→ [tex]U = qV[/tex]
By substituting the values, we get
[tex]= 1.2\times 10^{8} \ J[/tex]
Thus the above response is right.
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In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sectional area of 0.075 m2 and a large piston with a cross-sectional area of 0.237 m2 . An engine weighing 3400 N rests on the large piston. What force must be applied to the small piston in order to lift the engine? Answer in units of N.
Answer:
[tex]F_s=1075.9493\ N[/tex]
Explanation:
Given:
area of piston on the smaller side of hydraulic lift, [tex]a_s=0.075\ m^2[/tex]area of piston on the larger side of hydraulic lift, [tex]a_l=0.237\ m^2[/tex]Weight of the engine on the larger side, [tex]W_l=3400\ N[/tex]Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.
[tex]P_s=P_l[/tex]
[tex]\frac{F_s}{a_s}=\frac{W_l}{a_l}[/tex]
[tex]\frac{F_s}{0.075} =\frac{3400}{0.237}[/tex]
[tex]F_s=1075.9493\ N[/tex] is the required effort force.
Answer:
F = 1076 N
Explanation:
given,
small piston area, a = 0.075 m²
large piston area, A = 0.237 m²
weight on the large piston, W = 3400 N
force applied on the second piston, F = ?
using pascal law for the force calculation
[tex]\dfrac{F}{W}=\dfrac{a}{A}[/tex]
[tex]\dfrac{F}{3400}=\dfrac{0.075}{0.237}[/tex]
F = 0.3165 x 3400
F = 1076 N
The force applied to the small piston in order to lift the engine is equal to 1076 N.
Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of heat is transferred to the water and 5 kJ of heat is lost to the surrounding air. The paddle-wheel work amounts to 500 N·m. Determine the final energy of the system if its initial energy is 19.5 kJ.
Answer:
U₂=45 kJ
Explanation:
Given Data
Initial Energy U₁=19.5 kJ
Qin=30 kJ
Win=500 N.m
Qout= 5 kJ
To find
Final energy U₂
Solution
From The first law of thermodynamic
ΔE = Ein - Eout + ΔQ -ΔW
or
U₂=U₁+Qin+Win-Qout
where
U₂ is final energy
U₁ is initial energy
q = energy transferred as heat to a system
w = work done on a system
U₂=(19.5+30+500×10⁻³-5) kJ
U₂=45 kJ
Final answer:
To calculate the final energy of a system involving heat transfer and work, first, find the net heat transfer by considering heat added and lost, and then include the work done. The first law of thermodynamics is applied to combine these energies with the initial energy to find the final energy, which is 45 kJ.
Explanation:
The question involves calculating the final energy of a system after heat transfer and work done on a closed pan of water on a stove. Given that 30 kJ of heat is transferred to the water, 5 kJ is lost to the surrounding air, and the paddle-wheel work on the system is 500 N·m (or 0.5 kJ since 1 N·m = 1 J), we can use the first law of thermodynamics to find the final energy of the system. The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
First, we calculate the net heat transfer (Q) to the system by subtracting the heat lost to the air from the heat supplied: Q = 30 kJ - 5 kJ = 25 kJ. Then, we add the work done on the system (which is positive because the work is done on the system, not by it): 25 kJ + 0.5 kJ = 25.5 kJ. This is the net energy input into the system.
Finally, to find the final energy of the system, we add this net input to the initial energy of the system: 19.5 kJ (initial energy) + 25.5 kJ (net input) = 45 kJ. Therefore, the final energy of the system is 45 kJ.
A 1720 kg car skidding due north on a level frictionless icy road at 239.44 km/h collides with a 2597.2 kg car skidding due east at 164 km/h in such a way that the two cars stick together. 2597.2 kg 164 km/h 239.44 km/h 1720 kg vf θ N At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at? Answer in units of ◦ .
Answer:
[tex]\theta=44.03^{o}[/tex]
Explanation:
Here we have an inelastic collision problem. We can use the momentum (p = mv) conservation law in each component of the displacement.
So, [tex]p_{i}=p_{f}[/tex]
X-component:
[tex]m_{1}v_{i1x}+m_{2}v_{i2x}=m_{1}v_{f1x}+m_{2}v_{f2x}[/tex] (1)
Now,
v(i1x) is 0, because the first car just moving in y-directionv(i2x) is 164 km/hv(f1x)=v(f2x), because both cars stick together after the collision, so they have the same x-component velocity.Then, using this information we can rewrite the equation (1).
[tex]m_{2}v_{i2x}=v_{fx}(m_{1}+m_{2})[/tex]
[tex]v_{fx}=\frac{m_{2}v_{i2x}}{m_{1}+m_{2}}=\frac{2597.2*164}{1720+2597.2}[/tex]
[tex]v_{fx}=98.66 km/h[/tex]
Y-component:
[tex]m_{1}v_{i1y}+m_{2}v_{i2y}=m_{1}v_{f1y}+m_{2}v_{f2y}[/tex] (2)
We can do the same but with the next conditions:
v(i1y) is 239.44 km/hv(i2y) is 0, because the second car just moving at the x-directionv(f1y)=v(f2y), because both cars stick together after the collision, so they have the same y-component velocity.Then, using this information we can rewrite the equation (2).
[tex]m_{1}v_{i1y}=v_{fy}(m_{1}+m_{2})[/tex]
[tex]v_{fy}=\frac{m_{1}v_{i1y}}{m_{1}+m_{2}}=\frac{1720*239.44}{1720+2597.2}[/tex]
[tex]v_{fy}=95.39 km/h[/tex]
Now, as we have both components of the final velocity, we can find the angle East of North. Using trigonometric functions, we have:
[tex]tan(\theta)=\frac{v_{y}}{v_{x}}[/tex]
[tex]\theta=arctan(\frac{v_{y}}{v_{x}})=arctan(\frac{95.39}{98.66})[/tex]
[tex]\theta=44.03^{o}[/tex]
I hope it helps you!
A small glass bead has been charged to 8.0 nC. What is the magnitude of the electric field 2.0 cm from the center of the bead? (k = 1/4ΔΉ0 = 8.99 × 109 N · m2/C2)A) 3.6 × 10-6 N/C B) 1.4 × 10-3 N/C C) 1.8 × 105 N/C D) 3.6 × 103 N/C
Answer:
C) 1.8×10⁵ N/C.
Explanation:
Electric field: This can be defined as the region where electric force is experienced.
Electric Field intensity: This is defined as the force per unit charge which it exert at that point.
The S.I unit of electric field is N/C.
Mathematically, Electric Field intensity can be represented as,
E = kq/r².................... Equation 1
Where E = electric field intensity, q = charge, r = distance. k = proportionality constant.
Given: q = 8.0 nC = 8×10⁻⁹ C, r = 2.0 cm = 0.02 m, k = 8.99×10⁹ Nm²/C²
Substituting into equation 1
E = (8.99×10⁹×8×10⁻⁹)/0.02²
E = 71.92/0.0004
E = 1.798×10⁵ N/C.
E ≈ 1.8×10⁵ N/C.
The right option is C) 1.8×10⁵ N/C.
in a lake, there is a patch of lily pads. every day, the patch doubles in size. if it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover quarter the lake?
Answer:
46 days
Explanation:
It is a very logical question which does not require any mathematical calculation rather requires thinking.
It is said in the question every day, the patch doubles in size. if it takes 48 days for the patch to cover the entire lake.
Let total area of the lake be 100 m^2. The at the end of 48th day, it would have covered 100 m^2. So, at the end of 47th day it would have covered only 50 m^2, since every day, the patch doubles in size. Similarly on 46th day it would have covered 25 m^2, which is quarter of 100 m^2.
So ,it would take 46 days for the patch to cover quarter the lake.
When NASA's Skylab reentered the Earth's atmosphere on July 11, 1979, it broke into a myriad of pieces. One of the largest fragments was a 1770-kg lead-lined film vault, and it landed with an estimated speed of 120 m/s. What was the kinetic energy of the film vault when it landed?
Answer:
The kinetic energy when the film vault landed is 12744000J.
Explanation:
The kinetic energy is defined as:
[tex]k_{e} = \frac{1}{2}mv^{2}[/tex] (1)
Where m is the mass and v is the velocity.
By means of equation 1, the kinetic energy of the film vault when it landed can be determined
[tex]k_{e} = \frac{1}{2}(1770kg)(120m/s)^{2}[/tex]
But [tex]1 J = kg.m^{2}/s^{2}[/tex]
[tex]k_{e} = 12744000J[/tex]
Hence, the kinetic energy when the film vault landed is 12744000J.
The kinetic energy of the film vault when it landed was 1,919,200 Joules.
Explanation:The kinetic energy of the film vault can be calculated
The kinetic energy of the film vault can be calculated using the formula: KE = 0.5 * mass * velocity^2. Given that the mass of the film vault is 1770 kg and its velocity is 120 m/s, we can substitute these values into the formula:
KE = 0.5 * 1770 kg * (120 m/s)^2 = 0.5 * 1770 * 14400 = 1,919,200 J
Therefore, the kinetic energy of the film vault when it landed was 1,919,200 Joules. using the formula: b mass ity^2. Given that the mass of the film vault is 1770 kg and its velocity is 120 m/s, we can substitute these values into the formula:
KE = 0.5 * 1770 kg * (120 m/s)^2 = 0.5 * 1770 * 14400 = 1,919,200 J
Therefore, the kinetic energy of the film vault when it landed was 1,919,200 Joules.
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They each have a mass of 1.0 $ 1030 kg and a radius of 1.0 $ 105 m. They are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial value and (b) they are about to collide
Answer:
Explanation:
Here is the full question:
Two neutron stars are separated by a distance of 1.0 × 1010 m. They each have a mass of 1.0 × 1030 kg and a radius of 1.0 × 105 m. They are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial value and (b) they are about to collide?
solution:
G is the gravitational constant, the value is,
[tex]G=6.67\times 10^{-11}\frac{m^3}{kg.s^2}[/tex]
For half distance is,
[tex]U'=2U\\\\-dU=\frac{-2Gm^2}{R}\\\\dKE=-dU[/tex]
a)
If the sepperation confined to one-half its initial value,
The velocity is,
[tex]K=\frac{Gm^2}{2R}\\\\\frac{1}{2}mv^2=\frac{Gm^2}{2R}\\\\v=\sqrt{\frac{Gm}{R}}\\\\=\sqrt{\frac{(6.67\times 10^{-11}\frac{m^3}{kg.s^2})(10^{30}kg)}{(10^{10}m)}}\\\\=81,670m/s\\\\v=8.2\times 10^4m/s[/tex]
b)
[tex]dU=Gm^2(\frac{1}{R}-\frac{1}{2r})\\\\dKE=Gm^2(\frac{1}{2r}-\frac{1}{R})=mv^2\\\\\therefore dKE=-dU\\\\v=\sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}\\\\\sqrt{(6.67\times 10^{-11}\frac{m^3}{kg.s^2})(10^{30}kg))(\frac{1}{2(10^5m)}-\frac{1}{(10^{10}m)})}\\\\v=1.8\times 10^7m/s[/tex]
The pressure inside a hydrogen-filled container was 2.10 atm at 21 ∘C. What would the pressure be if the container was heated to 88 ∘C ?
Answer:
2.57832 atm
Explanation:
[tex]P_1[/tex] = Initial pressure = 2.1 atm
[tex]P_2[/tex] = Fianl pressure
[tex]T_1[/tex] = Initial Temperature = (21+273.15) K
[tex]T_2[/tex] = Final Temperature = (88+273.15) K
From Gay-Lussacs law we have
[tex]\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\Rightarrow P_2=\dfrac{P_1\times T_2}{T_1}\\\Rightarrow P_2=\dfrac{2.1\times (88+273.15)}{21+273.15}\\\Rightarrow P_2=2.57832\ atm[/tex]
The final pressure would be 2.57832 atm
Answer:
2.58 atm
Explanation:
P1 = 2.1 atm
T1 = 21 °C = 294 K
T2 = 88 °C = 361 K
P2 = ?
By using the gas laws
P / T = constant keeping the volume constant.
P1 / T1 = P2 / T2
2.10 / 294 = P2 / 361
P2 = 2.58 atm
Thus, the pressure becomes 2.58 atm.
Nuclear fusion ________.
a. is the process in which a large nucleus spontaneously splits into two or more smaller nuclei.
b. is the process in which two smaller nuclei combine to form a larger nucleus.
c. is the process in which a large nucleus
d. is bombarded with a neutron and then splits into two or more smaller nuclei.
e. is the process used to generate electricity in nuclear power plant
Answer:
b. is the process in which two smaller nuclei combine to form a larger nucleus
Consider the hypothetical observation "a planet beyond Saturn rises in west, sets in east." This observation is not consistent with a Sun-centered model, because in this model __________.
a. there are no planets beyond Saturn
b. all objects in space must orbit the Sun in the same direction
c. planets beyond Saturn must orbit the Sun more slowly than closer-in planets
d. the rise and set of all objects depends only on Earth’s rotation
This observation is not consistent with a Sun-centered model, because in this model "the rise and set of all objects depends only on Earth’s rotation".
Answer: Option D
Explanation:
Copernican heliocentrism, the astronomical model's title which was created by Nicolaus Copernicus and released in 1543. This model placed the Sun in motionless position near the Universe's center, with Earth and the planets of other which orbiting it in a circular orbits, altered by epicycles, and also at constant speeds.
Out of several planets in a determined order the Earth orbits around a stationary sun having three motions like annual revolution, daily rotation, and its axis is tilted annually. The planets' retrograde motion is understood by motion from the Earth.
Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer the following questions.
(a) What new force will exist if qA is doubled?
(b) What new force will exist if qA and qB are cut in half?
(c) What new force will exist if d is tripled?
(d) What new force will exist if d is cut in half?
(e) What new force will exist if qA is tripled and d is doubled?
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
Coulomb's Law explains how changes in charge and distance affect the electric force between two charges.
Explanation:Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Let's analyze the given scenarios:
If qA is doubled, the new force will also be doubled because the force is directly proportional to the charges.If both qA and qB are halved, the new force will be 1/4th of the original force because the force is directly proportional to the product of the charges.If d is tripled, the new force will be 1/9th of the original force because the force is inversely proportional to the square of the distance between the charges.If d is halved, the new force will be 4 times the original force because the force is inversely proportional to the square of the distance between the charges.If qA is tripled and d is doubled, the new force can be calculated by multiplying the individual effects of the changes. The new force will be 6 times the original force because the force is directly proportional to the product of the charges and inversely proportional to the square of the distance.Learn more about Electric Force here:https://brainly.com/question/20935307
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When we look into the night sky, we notice that there are stars dotted across the landscape. We also notice that there are regions of darkness. Knowing that there are more stars in the universe than there are grains of sand on Earth, which of the following statements best explains the regions of darkness in our night sky?
Answer: these regions of darkness are due to no stars being in these regions of the night sky
A projectile is fired from ground level at time t = 0, at an angle \theta with respect to the horizontal. It has an initial speed v_{0}. In this problem, we are assuming that the ground is level.
Part A. Find the time t_{H} it takes the projectile to reach its maximum height. Express t_{H} in terms of v_{0}, \theta, and g (the magnitude of the acceleration due to gravity).
Part B. Find t_{R}, the time at which the projectile hits the ground. Express the time in terms of v_{0}, \theta, and g.
Part C. Find H, the maximum height attained by the projectile. Express the maximum height in terms of v_{0}, \theta, and g.
Part D. Find the total distance R (often called the range) traveled in the x-direction; in other words, find where the projectile lands. Express the range in terms of v_{o}, \theta, and g.
Answer:
(A) t_H = Vo×Sin (theta)/g
(B) t,_R = (2VoSin(theta))/g
(C) H = Vo²Sin²(theta) / 2g
(D) Range = Vo²Sin(2×theta) / g.
Explanation:
The detailed solution to the problem can be found in the attachment below
To find the time it takes for the projectile to reach its maximum height, use the equation t_H = v_0 sin(θ) / g. To find the time the projectile hits the ground, use the equation t_R = 2v_0 sin(θ) / g. The maximum height attained by the projectile can be found using the equation H = (v_0^2 sin^2(θ)) / (2g). The total distance traveled in the x-direction (range) can be found using the equation R = (v_0^2 sin(2θ)) / g.
Explanation:To find the time it takes for the projectile to reach its maximum height, we can use the equation for vertical motion:
t_H = v_0 sin(\theta) / g
To find the time the projectile hits the ground, we can use the equation for vertical motion:
t_R = 2v_0 sin(\theta) / g
The maximum height attained by the projectile can be found using the equation:
H = (v_0^2 sin^2(\theta)) / (2g)
The total distance traveled in the x-direction (range) can be found using the equation:
R = (v_0^2 sin(2\theta)) / g
Anthony sends a series of pulses of amplitude 43 cm down a string that is attached to a post at one end. Assume the pulses are reflected with no loss of amplitude.(a) What is the amplitude at a point where two pulses are crossing if the string is rigidly attached to the post?cm(b) What is the amplitude at a point where two pulses are crossing if the end at which reflection occurs is free to slide up and down?cm
The amplitude where two pulses cross on a string rigidly attached to a post is momentarily zero due to destructive interference. However, when the end is free to move and two pulses cross, they superimpose constructively, resulting in a combined amplitude of 86 cm.
When Anthony sends a series of pulses of amplitude 43 cm down a string tied to a post:
(a) If the string is rigidly attached to the post, the reflected pulse inverts. Therefore, when two pulses of the same amplitude cross, they interfere destructively at the instant they coincide and the amplitude at the point of crossing is momentarily zero. However, since they're not lasting interactions, the amplitude of each pulse remains 43 cm before and after crossing.
(b) If the end is free to slide up and down, the reflected pulse does not invert and remains in phase with the incident pulse. Now, as two pulses of the same amplitude (43 cm) cross, they temporarily superimpose constructively, resulting in a combined amplitude of 86 cm, which is the sum of their individual amplitudes.
The phenomenon described here is a demonstration of wave interference, specifically in the context of reflected transverse waves on a string. It shows the distinct difference between reflections from fixed end (inverting) and free end (non-inverting) boundaries.
For a rigidly attached end, the amplitude is 0 cm due to destructive interference, while for a free boundary, it is 86 cm due to constructive interference.
When dealing with wave pulses on a string, the amplitude of the resulting wave at a point where two pulses are crossing depends on the boundary conditions at the end of the string.
(a) Rigidly Attached EndWhen the string is rigidly attached, the reflected pulse inverts upon reflection. This means that the original pulse and the reflected pulse will be out of phase by 180 degrees when they meet. If the amplitude of each pulse is 43 cm, the resulting amplitude at the point of crossing will be zero (43 cm - 43 cm = 0 cm) due to destructive interference.
(b) Free to Slide Up and DownWhen the end at which reflection occurs is free to move up and down (a free boundary condition), the reflected pulse remains in phase with the incident pulse. Therefore, the amplitudes add constructively. Hence, at the crossing point, the resulting amplitude will be the sum of the individual amplitudes: 43 cm + 43 cm = 86 cm.
In summary, the conditions at the end of the string determine whether the interference is constructive or destructive, affecting the amplitude at the crossing point of two pulses.
A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at 100 rpm. What is its rotational kinetic energy?
Answer:
KE = 1.75 J
Explanation:
given,
mass of ball, m₁ = 300 g = 0.3 Kg
mass of ball 2, m₂ = 600 g = 0.6 Kg
length of the rod = 40 cm = 0.4 m
Angular speed = 100 rpm= [tex] 100\times \dfrac{2\pi}{60}[/tex]
=10.47\ rad/s
now, finding the position of center of mass of the system
r₁ + r₂ = 0.4 m.....(1)
equating momentum about center of mass
m₁r₁ = m₂ r₂
0.3 x r₁ = 0.6 r₂
r₁ = 2 r₂
Putting value in equation 1
2 r₂ + r₂ = 0.4
r₂ = 0.4/3
r₁ = 0.8/3
now, calculation of rotational energy
[tex]KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2[/tex]
[tex]KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)[/tex]
[tex]KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)[/tex]
[tex]KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)[/tex]
KE = 1.75 J
the rotational kinetic energy is equal to 1.75 J
The total rotational kinetic energy of the balls is 1.78 J.
The given parameters;
mass of the first ball, m₁ = 300 g = 0.3 kgmass of the second ball, m₂ = 600 g = 0.6 kgThe angular speed of the balls is calculated as follows;
[tex]\omega = 100 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s } \\\\\omega = 10.47 \ rad/s[/tex]
The radius of the balls is calculated as;
[tex]r_1 + r_2 = 0.4[/tex]
The torque on the rod due to each is calculated as;
[tex]F_1r_1 = F_2 r_2\\\\m_1 gr_1 = m_2 g r_2\\\\m_1 r_1 = m_ 2r_2\\\\r_2 = \frac{m_1r_1}{m_2} \\\\r_2 = \frac{0.3 r_1}{0.6} \\\\r_2 = 0.5r_1[/tex]
solve for the radius;
[tex]r_1 + 0.5r_1 = 0.4\\\\1.5r_1 = 0.4\\\\r_1 = \frac{0.4}{1.5} \\\\r_1 = 0.267 \ m\\\\r_2 = 0.5(0.267)\\\\r_2 = 0.133 \ m[/tex]
The moment of inertia of each ball is calculated as follows;
[tex]I_1 = m_1 r_1^2 = 0.3 \times (0.267)^2 = 0.0214 \ kgm^2\\\\I_2 = m_2 r_2^2= 0.6 \times (0.133)^2 = 0.011 \ kgm^2[/tex]
The total rotational kinetic energy of the balls is calculated as follows;
[tex]K.E = \frac{1}{2}I_1 \omega^2 \ + \ \frac{1}{2}I_2 \omega^2 \\\\K.E = \frac{1}{2} \omega^2(I_1 + I_ 2)\\\\K.E = 0.5 \times (10.47)^2 (0.0214 + 0.011)\\\\K.E = 1.78 \ J[/tex]
Thus, the total rotational kinetic energy of the balls is 1.78 J.
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Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the incoming ball has a speed of 200 m/s . Part APart complete What is the final speed of the incoming ball if it is much more massive than the stationary ball?
Answer:
Explanation:
Check the attachment for solution
Answer: The final speed of the incoming ball is approximately 200m/s
Explanation:
Using the law of conservation of momentum.
m1(u1) + m2u2 = m1v1 + m2v2
And also law of conservation of kinetic energy for elastic heads on collision we can derive the formula for elastic heads on collision which is given below:
For elastic heads on collision.
v1 = [( m1 - m2)/(m1+m2)] u1 ......1
v2 = [(2m1)/(m1+m2)]u1 ......2
Where,
m1 and m2 are the mass of the incoming and stationary ball respectively.
u1 and u2 are the initial speed of the incoming and stationary ball respectively.
v1 and v2 are the final speed of the incoming and stationary ball respectively.
a) to determine the final speed of the incoming ball using equation 1
v1 = [( m1 - m2)/(m1+m2)]u1
Since m1 >> m2
m1 - m2 ~= m1 and m1 +m2 ~= m1
So, equation 1 becomes
v1 ~= [m1/m1]u1
v1 ~= u1
Since u1 = 200m/s
v1 ~= 200m/s
Additional tips: using equation 2 we can derive the approximate final speed of the stationary ball following the same assumptions. If well solved v2 = 2u1 = 400m/s
Two pieces of plas- tic, a full ring and a half ring, have the same radius and charge density. Which electric field at the center has the greater magnitude? Discuss and defend your answer.
Answer: The magnitude of the Electric Field at the centre of both plastic would be equal to Zero {i.e E =0}
Explanation: A plastic either full circle or half is an insulator, and the charges in an insulator cannot move around.
Therefore, thier electric potential p(r) = 0.
Recall that,
Electric field intensity E = first derivative of p(r) with respect to r ( radius).since p(r) = 0, E = 0 inside both plastic.
Final answer:
The electric field at the center of the half ring has a greater magnitude than that of the full ring because the full ring's symmetric charge distribution cancels the field out, while the half ring's electric field remains unopposed.
Explanation:
To determine which electric field at the center has the greater magnitude between a full ring and a half ring with the same radius and charge density, we must analyze the symmetrical distribution of charges. In the case of a full ring, the charges are evenly distributed around the ring, resulting in a net electric field at the center that cancels out due to symmetry. However, for a half ring, the symmetry is broken, and there is a net electric field at the center directed away from the flat side of the half ring, because there are no opposing charges to cancel the field out. Therefore, the electric field at the center of the half ring has a greater magnitude than that at the center of the full ring, which is effectively zero.
A basketball is thrown up into the air. It is released with an initial velocity of 8.5 m/s. How long does it take to get to the top of its motion?
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 8.5 m/s
Final velocity, v = 0 m/s - At maximum height
Time, t = ?
Acceleration , a = -9.81 m/s²
Substituting
v = u + at
0 = 8.5 + -9.81 x t
t = 0.867 s
It takes 0.867 seconds to get to the top of its motion
Answer:
0.87 s
Explanation:
initial velocity, u = 8.5 m/s
Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0
Use first equation of motion
v = u - gt
where, g be the acceleration due to gravity
0 = 8.5 - 9.8 t
t = 0.87 s
Thus, the time taken to reach at top is 0.87 s.
Now find the components NxNxN_x and NyNyN_y of N⃗ N→N_vec in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector NNN and the angle θθtheta, with the components separated by a comma.
Answer:
[tex]N_{y} =Ncos[/tex]Ф,
[tex]N_{x} =-Nsin[/tex] Ф
Explanation:
Now find the components NxNxN_x and NyNyN_y of N⃗ N→N_vec in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector NNN and the angle θθtheta, with the components separated by a comma.
Vectors are quantities that have both magnitude and direction while scalar quantities have only magnitude but no direction.
This a vector quantity
from the diagram the horizontal component of the length of the vector will be
[tex]N_{y} =Ncos[/tex]Ф
the vertical component will be
[tex]N_{x} =-Nsin[/tex] Ф
this is in the opposite direction because the x can be extrapolated to the negative axis
To solve the problem we must know about vectors and coordinate geometry.
What is a Vector?We know that the vector quantities are those quantities that have magnitude as well as direction.
Each vector quantity can be divided into two parts a horizontal and vertical component, the vertical component is known as the sine component while the horizontal component is known as the cosine component.
Representation of Vector componentA vector component is represented as the product of its length and the component angle.
For example, F Sinθ represents the vertical component of force, with magnitude F, while the F Cosθ represents the horizontal component, and will be in the same line with the axis from which the angle is been measured.
Vector components of the tilted Coordinate,The coordinate in the tilted coordinate will be,
[tex]N_x = N Sin\theta[/tex]
[tex]N_y = N Cos\theta[/tex]
Where N is the vector magnitude and θ is the angle from the axis of measurement.
The sign of the vector will depend upon the coordinate in which it is.
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Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time?
Answer
given,
Force on car, F = 300 N
time, t = 0.250 s
distance of the force from the pivot, r= 0.3 m
Angular momentum = ?
Now,
torque acting is calculated by multiplying force with displacement.
[tex]\tau = F \times r [/tex]
[tex]\tau = 300 \times 0.3 [/tex]
[tex]\tau = 90\ N.m[/tex]
we know,
torque is equal to change in angular momentum per unit time.
[tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]
Δ L = τ Δ t
initial angular momentum is zero
L = 90 x 0.25
L = 22.5 kg.m²/s
Angular momentum is equal to 22.5 kg.m²/s
Monochromatic light falls on two very narrow slits 0.040 mm apart. Successive fringes on a screen 4.70 m away are 5.9 cm apart near the center of the pattern. What is the wavelength of the light?
Answer:
λ = 502 n m
Explanation:
given,
slit separation, d = 0.040 mm
= 4 x 10⁻⁵ m
Distance between the fringes, D = 4.70 m
distance between the fringes, x = 5.9 cm
= 0.059 m
wavelength of the light = ?
the separation between the fringe is given by
[tex]x = \dfrac{n\lambda\ D}{d}[/tex]
[tex]\lambda = \dfrac{x d}{n\ D}[/tex]
where as n = 1
[tex]\lambda = \dfrac{0.059\times 4\times 10^{-5}}{1\times 4.70}[/tex]
λ = 5.02127 x 10⁻⁷ m
λ = 502 n m
hence, the wavelength of the light is equal to λ = 502 n m
If the final position vector of a moving object has a smaller magnitude than the initial position vector, then the change in the object's position vector has a positive magnitude.
Explanation:
The given statement is absolutely true. this is because magnitude of a vector is always non negative, it can not be zero unless its a zero vector. So, in the given question, final position vector of a moving object has a smaller magnitude than the initial position vector, so, magnitude is neither zero nor negative. Hence, it has a positive magnitude.
Final answer:
The displacement of an object is the final position vector minus the initial position vector and is a vector with both magnitude and direction. The magnitude of displacement does not have to be positive if the final position vector has smaller magnitude than the initial; it depends on the direction relative to the chosen coordinate system.
Explanation:
When discussing the displacement of an object, it is important to understand that it refers to the change in the object's position and is defined as the final position vector minus the initial position vector. Displacement, being a vector quantity, has both magnitude and direction. If the magnitude of the final position vector is smaller than the magnitude of the initial position vector, this does not necessitate that the magnitude of the displacement is positive. In fact, the magnitude of the displacement vector could be either positive or negative based on the assigned coordinate system.
In a one-dimensional coordinate system, the direction of motion can be designated as positive or negative. For example, if rightward motion is deemed positive, then a leftward motion would be negative. Similarly, in vertical motion, upward movement is often taken as positive and downward as negative, though these conventions can be reversed based on the scenario and convenience. The direction of the displacement vector essentially depends on the chosen orientation of the positive direction.
Therefore, if the final position vector has a smaller magnitude than the initial position vector, and if the positive direction is assigned towards the final position, the displacement vector magnitude could indeed be negative, not necessarily positive, as it signifies a change in position opposite to the assigned positive direction.
A toy car (0.50 kg) runs on a frictionless track and has an initial kinetic energy of 2.2 J, as the drawing shows. The numbers beneath each hill give the heights of the hills. Over which of the hills will the car coast?
Answer:
Height will be equal to 0.4489 m
Explanation:
We have given mass of the toy m = 0.50 kg
Initial kinetic energy K = 2.2 J
We have to fond the height of the hill over which car roast
When car will roast the hill its kinetic energy will be converted into potential energy and at maximum height all kinetic energy will be converted into potential energy
So at maximum height [tex]mgh=K[/tex]
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
So 0.50×9.8×h = 2.2
h = 0.4489 m
Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on the sled, what force is needed to pull the sled across the snow at constant speed?
Answer:
69.74 N
Explanation:
We are given that
Weight of sled=49 N
Coefficient of kinetic friction[tex],\mu_k=0.11[/tex]
Weight of person=585 N
Total weight==mg=49+585=634 N
We know that
Force needed to pull the sled across the snow at constant speed,F=Kinetic friction
[tex]F=\mu_k N[/tex]
Where N= Normal=mg
[tex]F=0.11\times 634=69.74 N[/tex]
Hence, the force is needed to pull the sled across the snow at constant speed=69.74 N
The vasa vasorum, the collection of small arteries and veins supplying blood to the smooth muscle and fibroblasts within blood vessels, is found in the ________________.
Answer:
The vasa vasorum, the collection of small arteries and veins supplying blood to the smooth muscle and fibroblasts within blood vessels, is found in the tunica externa.
Explanation:
In order to understand the answer above let explain the concept of tunica externa
This in Latin means outer coat and can also be called 'tunica adventitia' which also means "additional coat in Lain",it is the outermost layer of the a blood vessel covering the tunica media(This is a Latin word translated as middle coat in English which is the middle layer of an artery or a vein it lies between the tunica externa and the tunica initima which is the innermost part of the artery or vein).
An Artery
We can define an artery as a blood vessel that transports blood from the heart to other parts of the body.
An Vein
We can define a vein as a blood vessel that transports blood to the heart
and we can define blood vessel as a tube that carries blood in the circulatory system.
Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. How would you determine the magnitude and direction of the electric field if the object were (a) positively charged and (b) negatively charged? (a) The object has a charge of +20.0μC and the instrument indicates that the electric force exerted on it is
40.0μN due east. What are the magnitude and direction of the electric field? (b) What are the magnitude and direction of the electric field if the object has a charge of −10.0μC and the instrument indicates that the force is 20.0μN due west?
Answer:
2 N/C direction of the force
-2N/C opposite to the direction of the force
Explanation:
E = Electric field
q = Charge
Electrical force is given by
[tex]F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{40}{20}\\\Rightarrow E=2\ N/C[/tex]
The magnitude of the force is 2 N/C
The force acting on the charge is positive so the direction of the electric field is positive.
[tex]F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{20}{-10}\\\Rightarrow E=-2\ N/C[/tex]
The magnitude of the force is 2 N/C
The direction of the electric field is negative and opposite to the direction of the force as the charge is negative.
To determine the magnitude and direction of an electric field, use the formula E = F/Q. For a positively charged object, the electric field has the same direction as the force. For a negatively charged object, the electric field has the opposite direction of the force.
Explanation:To determine the magnitude and direction of an electric field given the charge of an object and the force exerted on it, you can use the formula: E = F/Q, where E is the electric field, F is the force, and Q is the charge.
For a positively charged object (+20.0μC) with a force of 40.0μN due east: The electric field E = F/Q = (40.0μN)/(20.0μC) = 2.0 N/C. The direction of the electric field is in the same direction as the force, due east.
For a negatively charged object (-10.0μC) with a force of 20.0μN due west: The electric field E = F/Q = (20.0μN)/(-10.0μC) = -2.0 N/C. The direction of the electric field is in the opposite direction of the force, due east, as per the convention that electric fields are directed away from positive charges or towards negative charges.
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A pendulum and a block hanging at the end of a spring are both carefully adjusted to make one vibration per second while on the Earth's surface. Will the period of vibration of the pendulum or of the block on a spring be affected if they are moved to the Moon?
Answer:
For the pendulum , yes and for the spring ,No
Explanation:
We will examine the two cases,
For pendulum;
[tex]T =2\pi \sqrt{\frac{l}{g} }[/tex]
we see that for a pendulum, the period depends on the gravitational acceleration of the region it is. so the period for earth will differ compared to that of the moon.
For a spring;
[tex]T=2\pi \sqrt{\frac{m}{k} }[/tex]
form the equation above is obvious that the period of a spring in vibration is independent of gravitational acceleration of the region it is.
The mass of the bob has no effect on the motion of a pendulum; the period of vibration is determined by the pendulum's length and the acceleration due to gravity.
The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity.
A ferry boat shaped just like a brick is 4.3 m wide and 8.5 m long. When a truck pulls onto it, the boat sinks 6.15 cm in the water. What is the weight of the truck? The acceleration of gravity is 9.81 m/s 2 . Answer in units of N.
Answer:
The weight of the truck is 22072.5 N.
Explanation:
Given that,
Width of the brick =4.3 m
Length of the brick = 8.5 m
Distance = 6.15 cm
We need to calculate the volume of the water displaced by the boat with truck
Using formula of volume
[tex]V=l\times w\times d[/tex]
Put the value into the formula
[tex]V=8.5\times4.3\times6.15\times10^{-2}[/tex]
[tex]V=2.25\ m^3[/tex]
We need to calculate the mass of the water displaced by the boat with truck
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
[tex]m=\rho\times V[/tex]
Put the value into the formula
[tex]m=1000\times2.25[/tex]
[tex]m=2250\ kg[/tex]
We need to calculate the weight of the truck
Using formula of weight
[tex]W= mg[/tex]
Put the value into the formula
[tex]W=2250\times9.81[/tex]
[tex]W=22072.5\ N[/tex]
Hence, The weight of the truck is 22072.5 N.
According to Erikson's theory, when __________, the psychological conflict of the first year is resolved on the positive side.
Answer: The balance of care is sympathetic and loving
Explanation: According to Erikson's theory, when the balance of care is sympathetic and loving the psychological conflict of the first year is resolved on the positive side.
According to Erikson's theory of psychosocial development, the psychological conflict of the first year is resolved on the positive side when the infant develops a sense of trust in their caregivers.
Explanation:According to Erik Erikson's theory of psychosocial development, when an infant develops a sense of trust in their caregivers during the first year of life, the psychological conflict of that stage is resolved on the positive side. Erickson named this first stage trust versus mistrust. In this stage, caregivers who are responsive and sensitive to their infant's needs help the child to develop a sense of trust; thus, the child will see the world as a safe, predictable place. If the caregivers are unresponsive and do not meet the infant's needs, feelings of anxiety, fear, and mistrust emerge; the child may see the world as unpredictable.
Successful completion of each developmental task at various stages results in a sense of competence and a healthy personality, according to Erikson. Failure to master these tasks may lead to feelings of inadequacy and affect the formation of a positive self-concept.
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