Answer:
from los angeles distance plan a = 1111.08 mi
from los angeles distance plan b = 888.92 mi
Explanation:
given data
new york to los angeles distance = 2000 miles
Plane b speed = 100 mph faster than plane a
Plane b takes off time = 1 hour after plane a
to find out
How far are they from los angeles when they pass
solution
we consider speed of plan a is = x mph
so speed of plan b will be = x + 100 mph
and we know plan b take here 1 hour less time than plan a so it mean time is distance divide speed i.e
[tex]\frac{2000}{x} - 1 =\frac{2000}{x+100}[/tex]
solve it we get x = 400 mph
it mean here
plan a speed is 400 mph
and plan b speed is 500 mph
and
now we consider they meet at time = t hour after a take off
then plan a travel = 400 t
and plan b travel = 500 ( t - 1 )
add both distance that is equal to 2000 mi
so 400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t-1) = 2000
solve we get
t = 2.777
so total distance travel plan a = 400 × 2.777 = 1111.08 mi
total distance travel plan b = 2000 - 1111.08
total distance travel plan b = 888.92 mi
Oppositely charged objects attract each other. This attraction holds electrons in atoms and holds atoms to one another in many compounds. However, Ernest Rutherford’s model of the atom failed to explain why electrons were not pulled into the atomic nucleus by this attraction. What change to the atomic model helped solve the problem seen in Rutherford’s model?
Answer:
A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.
Explanation:
The change to the atomic model that helped solve the problem seen in Rutherford's model was the discovery of the strong nuclear force.
Explanation:Rutherford's model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. However, Rutherford's model failed to explain why electrons were not pulled into the atomic nucleus by this attraction. The change to the atomic model that helped solve this problem was the discovery of the strong nuclear force, which is much stronger than electrostatic interactions and holds the protons and neutrons together in the nucleus.
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What is the kinetic energy of a 0.500 kg ball thrown at 30.0 m/s?
Answer: The answer is 225 joules
A high-wire walker always attempts to keep his center of mass (com) over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of 71.0 kg and a rotational inertia of 14.0 kg · m² about the wire. What is the magnitude of his angular acceleration about the wire if his com is 5.0 cm to the right of the wire under the following conditions?
(a) he carries no pole and
(b) the 14.0 kg pole he carries has its com 10 cm to the left of the wire?
Answer:
a)
2.5 rads⁻²
b)
1.5 rads⁻²
Explanation:
a)
[tex]R[/tex] = distance of the center of mass from the wire = 5 cm = 0.05 m
[tex]F_{g}[/tex] = Force of gravity on walker
[tex]M[/tex] = mass of walker = 71 kg
Force of gravity on walker is given as
[tex]F_{g} = Mg\\F_{g} = (71)(9.8)\\F_{g} = 695.8 N[/tex]
[tex]I[/tex] = Rotational inertia of walker = 14 kgm²
[tex]\alpha[/tex] = Angular acceleration about the wire
Torque equation is given as
[tex]F_{g} R = I \alpha \\(695.8) (0.05) = (14) \alpha \\\alpha = \frac{(695.8) (0.05)}{14} \\\alpha= 2.5 rads^{-2}[/tex]
b)
[tex]F_{p}[/tex] = Force of gravity on pole
[tex]m[/tex] = mass of walker = 14 kg
Force of gravity on pole is given as
[tex]F_{p} = mg\\F_{g} = (14)(9.8)\\F_{g} = 137.2 N[/tex]
[tex]r[/tex] = distance of the center of mass of pole from the wire = 10 cm = 0.10 m
Torque equation is given as
[tex]F_{g} R - F_{p} r = I \alpha \\(695.8) (0.05) - (137.2)(0.10) = (14) \alpha \\\alpha= \frac{21.07}{14} \\\alpha=1.5 rads^{-2}[/tex]
A bulb with a resistance of 4 Ohms is plugged into a 2V battery. It has a capacity of 3 amps-hours. How long will this bulb remain at a constant brightness? Show all your working.
Answer:
Bulb will remain at constant brightness for 6 hour
Explanation:
We have given resistance of the bulb R = 4 ohm
Potential difference of the battery V = 2 volt
So current [tex]i=\frac{V}{R}=\frac{2}{4}=0.5A[/tex]
Capacity of the battery is given as 3 amp hour
We have to find the time by which bulb will remain at constant brightness
So time will be [tex]=\frac{3}{0.5}=6hour[/tex]
So bulb will remain at constant brightness for 6 hour
A gust of wind blows an apple from a tree. As the apple falls, the gravita- tional force on the apple is 2.25 N downward, and the force of the wind on the apple is 1.05 N to the right.
Find the magnitude and direction of the net force on the apple.
Answer:
The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical
Explanation:
Consider the gravitational force as [tex]\vec{F_g}=2.25\ N[/tex]
Consider the wind force as [tex]\vec{F_w}=1.05\ N[/tex]
The angle between the gravitational force and wind force is 90°
From the parallelogram law we have
[tex]F=\sqrt{F_g^2+F_w^2+2F_gF_wcos\theta}\\\Rightarrow F=\sqrt{2.25^2+1.05^2+2\times 2.25\times 1.05\times cos90}\\\Rightarrow F=2.48294\ N[/tex]
The angle between the resultant and the gravitational force would be
[tex]tan^{-1}\frac{1.05}{2.25}=25.02^{\circ}[/tex]
The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical
Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? Ignore air resistance.
Answer:before throwing and after catching the ball
Explanation:
When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.
After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.
The point in time where the zero net force acted on the ball is before throwing and after catching the ball
The point at which Net force acted on the ball:At the time of basketball is in the hand of the player, the net force should be zero since holding force is canceled via gravity Force. At the time of overall motion, gravitational force should be acted on the ball which is acting downward. After catching the ball net force on it zero since holding force should be canceled via gravity force and ball should be continue in stationary motion.
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An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?A. 72.2 m/sB. 12.6 m/sC. 47.3 m/sD. 34.2 m/sE. 6.45 m/s
Answer:
The terminal speed of this object is 12.6 m/s
Explanation:
It is given that,
Mass of the object, m = 80 kg
The magnitude of drag force is,
[tex]F_{drag}=12v+4v^2[/tex]
The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.
[tex]F_{drag}=mg[/tex]
[tex]12v+4v^2=80\times 9.8[/tex]
[tex]4v^2+12v=784[/tex]
On solving the above quadratic equation, we get two values of v as :
v = 12.58 m/s
v = -15.58 m/s (not possible)
So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.
Thunderstorms are often produced ahead of this type of front
Answer:
Cold front
Explanation:
The occurrence of cold front takes place when colder air at the surface pushes up the warmer into the atmosphere.
The movement of these cold fronts takes place at a faster pace as compared to the other fronts and are these are linked with the most violent types of weather such as severe thunderstorms are produced.
A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 49.0-gram mass is attached at the 13.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
Answer: M = 118.9g
Explanation:
Using the principle of moment to solve the equilibrium question,
Sum of clockwise moments is equal to the sum of anticlockwise moments
*Check attachment for the diagram
Note that the mass of the body is always located at the center i.e 50cm from the end.
To get M, taking moment about the fulcrum,
Since moment is force × perpendicular distance and according to the above principle, we have;
49×26.2= M×10.8
M= 49×26.2/10.8
M = 118.9g
What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves in free fall?
The answer is 0N but could someone pls explain why?
Answer:
0 N
Explanation:
The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.
The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.
But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force[tex](N)[/tex] acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity[tex](g)[/tex]
So, we apply Newton's second law on the woman.
[tex]\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma[/tex]
Net force is equal to the difference of the downward force and upward force.
[tex]F_{net}=mg-N[/tex]
Now, [tex]F_{net}=ma[/tex]
[tex]mg-N=mg\\N=mg-mg=0[/tex]
Therefore, the reading on the spring scale is 0 N.
A 440.0-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of 3 Hz is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?
Answer:
The original frequency of the guitar string is 437.0 Hz
Explanation:
It is an interesting phenomenon that when two waves of slightly different frequencies (f1 and f2) interfere they produce beats, and the frequency (fb) of those beats is:
[tex] f_{b}=\mid f_{1}-f_{2}\mid [/tex] (1)
Note that because the equation takes the absolute value of the difference between f1 and f2 we have two possible results of [tex] f_{1}-f_{2} [/tex] that satisfy the equation, those are:
[tex] 440.0-f_{2}=3\Longrightarrow f_{2}=437.0\,Hz [/tex]
[tex] -(440.0-f_{2})=3\Longrightarrow f_{2}=443.0\,Hz [/tex]
Because the frequency of the beat is heard to decrease when the frequency of the guitar string is increased, that means the guitar string was at first under the pitch, so the original frequency of the guitar string is 437.0 Hz
Final answer:
The original frequency of the guitar string was 437 Hz, since the string was out of tune and produced a beat frequency of 3 Hz with a 440 Hz tuning fork, and the beat frequency decreased when the string was tightened.
Explanation:
The original frequency of the guitar string can be determined using the concept of beat frequency. The beat frequency is the absolution difference between the frequencies of two sounds. In this case, the beat frequency was 3 Hz when the guitar string was sounded with the 440.0 Hz tuning fork.
Since the beat frequency decreased when the string tension was increased, the original frequency of the string must have been lower than the tuning fork. Therefore, the original frequency of the guitar string could be either 437 Hz or 443 Hz. However, since tightening the string increases its frequency and resulted in a decrease in the beat frequency, this indicates that the original frequency was 437 Hz (440 Hz - 3 Hz).
Tia needs to produce a solenoid that has an inductance of 2.07 μ H . She constructs the solenoid by uniformly winding 1.19 m of thin wire around a tube. How long, in centimeters, should the tube be?
Answer:
[tex]l=0.068 m[/tex]
Explanation:
given,
inductance of solenoid = 2.07 μ H
winding of the wire = 1.19 m
Using formula of inductance
[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]
L is the inductance
N is number of turns of the coil
μ₀ is permeability of free space
L is length of winding
N (2π r) = 1.19
squaring both side
4π(N²(πr²))=1.19²
N² A = 0.113
now
[tex]2.07 \times 10^{-6}= \dfrac{4\pi\times 10^{-7}\times 0.113}{l}[/tex]
[tex]l= \dfrac{4\pi\times 10^{-7}\times 0.113}{2.07 \times 10^{-6}}[/tex]
[tex]l=0.068 m[/tex]
This question involves the concepts of the inductance of an inductor and the length of the winding.
The tube should be "6.84 cm" long.
The inductance of an inductor can be given by the following formula:
[tex]L=\frac{N^2\mu_oA}{l}[/tex]
where,
L = inductance = 2.07 μH = 2.07 x 10⁻⁶ H
N = No. of turns in coil = [tex]\frac{length\ of\ wire}{circumference\ of\ tube} = \frac{1..19\ m}{2\pi r}[/tex]
μ₀ = permeability of free space = 4π x 10⁻⁷ Wb/A.m
A = Area of tube = πr²
l = length of tube = ?
Therefore,
[tex]2.07\ x\ 10^{-6}\ H = \frac{(\frac{1.19\ m}{2\pi r})^2(\pi r^2)(4\pi\ x\ 10^{-7}\ Wb/A.m)}{l}\\\\l = \frac{1.4161\ x\ 10^{-7}}{2.07\ x\ 10^{-6}}\\\\[/tex]
l = 0.0684 m = 6.84 cm
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The attached picture shows the inductance formula.
Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC rock must be placed in 4.00 kg of 15.0ºC water to bring its temperature to 100ºC, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite. (answer in kg)
Answer:
m = 4.65 kg
Explanation:
As we know that the mass of the water that evaporated out is given as
[tex]m = 0.0250 kg[/tex]
so the energy released in form of vapor is given as
[tex]Q = mL[/tex]
[tex]Q = (0.0250)(2.25 \times 10^6)[/tex]
[tex]Q = 56511 J[/tex]
now the heat required by remaining water to bring it from 15 degree to 100 degree
[tex]Q_2 = ms\Delta T[/tex]
[tex]Q_2 = (4 - 0.025)(4186)(100 - 15)[/tex]
[tex]Q_2 = 1.41\times 10^6J[/tex]
total heat required for above conversion
[tex]Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J[/tex]
now by heat energy balance
heat given by granite = heat absorbed by water
[tex]m(790)(500 - 100) = 1.47 \times 10^6[/tex]
[tex]m = 4.65 kg[/tex]
The ability to move or change an object or what a wave carries
Waves can transfer energy and momentum from one place to another without a permanent shift in the mass. They propagate via different oscillation modes and can impact the propagation direction when they encounter an interface between two media.
Explanation:The ability to move or change an object or what a wave carries, essentially describes the energy transfer capability of waves. According to Big Idea 6, waves can transfer energy and momentum from one place to another without permanently relocating the mass. This aspect characterizes waves and differentiates them from other phenomena.
Waves can propagate through various oscillation modes, such as transverse and longitudinal. For instance, when you shake one end of a rope, the perturbation moves in the form of a wave up to the other end of the rope, thus, facilitating energy transfer without a physical shift of the rope's parts.
Another important aspect to remember about waves is their ability to affect the path of propagation when they encounter an interface between two different media, as articulated by Enduring Understanding 6.E. This property explains various natural phenomena, such as the bending of light when it moves from air to water.
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A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charles’, or Boyle’s law, develop an expression that would solve for the new volume V2.
Answer:
Boyle's Law
[tex]\therefore P_1.V_1=P_2.V_2[/tex]
Explanation:
Given that:
initially:
pressure of gas, [tex]= P_1[/tex]
volume of gas, [tex]= V_1[/tex]
finally:
pressure of gas, [tex]= P_2[/tex]
volume of gas, [tex]= V_2[/tex]
To solve for final volume [tex]V_2[/tex]
According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.
According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.
Mathematically:
[tex]P_1\propto \frac{1}{V_1}[/tex]
[tex]\Rightarrow P_1.V_1=k\ \rm(constant)[/tex]
[tex]\therefore P_1.V_1=P_2.V_2[/tex]
Final answer:
To calculate the new volume V2 in Boyle's Law, use the formula V2 = P1V1/P2, ensuring that P1 and P2 are in the same pressure units and that V1 and V2 are in the same volume units.
Explanation:
To determine the new volume V2 after a change in pressure of an ideal gas, one can utilize Boyle's Law, which states that at a constant temperature, the pressure of a gas is inversely proportional to its volume. The formula representing Boyle's Law is P1V1 = P2V2. In this equation, P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. To solve for V2, you simply rearrange the equation to V2 = P1V1/P2.
Remember that in order for Boyle's Law to be applied correctly, both volumes should be in the same unit. If you solve for V2 and find it to be greater than V1, it indicates that the pressure has decreased, consistent with the inverse proportionality.
A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 245 Hz. A person on the platform right next to the speaker detects the sound waves reflected off the wall and those emitted by the speaker.
How fast should the platform move, vp, for the person to detect a beat frequency of 3.00 Hz?
Answer:
Explanation:
by Doppler effect
F apparent = F real x (Vair +- Vobserver) / (Vair +- Vsource)
case1 :if observer is approaching a source then put a + in numerator and - in denominator
case 2 : if observer is going away from source then do opposite.
it is case1 cos observer is moving toward wall . actually wall is acting as a source for observer
so
F apparent = 240 x (344 + vp ) / (344 - vp )
now F apparent - Freal = beat frequency of 6.00 Hz
so {240 x (344 + vp ) / (344 - vp )} - 230 = 6
240 x (344 + vp -344 + vp) / (344 - vp ) = 6
240 x 2 x vp = 6 x (344 - vp )
480 vp = 2064 - 6 x vp
486 x vp = 2064
vp = 4.24 m/s
A heavy-duty stapling gun uses a 0.147-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 32800 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 4.44 cm from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.53 cm when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.
Answer:
v = 13.75 m/s
Explanation:
mass of rod (m) = 0.147 kg
spring constant (k) = 32800 N/m
initial compression (Y1) = 4.44 cm = 0.0444 m
final compression (Y2) = 1.53 cm = 0.0153 m
find the speed at the instant of contact
kinetic energy of a spring = elastic potential energy
0.5m[tex]v^{2}[/tex] = 0.5k[tex]x^{2}[/tex]
[tex]v^{2}[/tex] = [tex]\frac{0.5k[tex]x^{2}[/tex]}{0.5m} \[/tex]
where x = Y1 - Y2
x = 0.0444 - 0.0153 = 0.0291 m
[tex]v^{2}[/tex] = [tex]\frac{0.5 x 32800 x [tex]0.0291^{2}[/tex]}{0.5 x 0.147} \[/tex]
[tex]v^{2}[/tex] = 188.95
v = 13.75 m/s
An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will separate when charged negatively. Over time, however, these leaves will cease separating, even without being touched by another object. Explain why this may occur.
Answer:
All these is caused by the repulsion force.
Explanation:
The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.
As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.
A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes the forces the objects exert on each other when the collide and provides the best explanation? (A) The force exerted by A on B will be twice as great as the force exerted by B on A, as great as the force exerted by B on A other are the same, because the product of because A has twice the mass of B (B) The force exerted by A on B will be half because A has half the speed of B (C) The forces exerted by each object on the mass and speed is the same for both objects. (D) The forces exerted by each object on the other are the same, because inter objects cannot exert forces of difi magnitude on each other.
Answer:D
Explanation:
Given
mass of A is twice the mass of B half the velocity of B
Suppose [tex]F_a[/tex] and [tex]F_b[/tex] be the average force exerted on A and B respectively
and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.
Thus [tex]F_a=-F_b[/tex] and option d is correct
According to Newton's third law of motion, the forces exerted by each object on the other are the same.
Explanation:The correct answer is (C) The forces exerted by each object on the other are the same because inter objects cannot exert forces of different magnitudes on each other.
According to Newton's third law of motion, whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force it exerts. This means that the forces exerted by object A on object B and by object B on object A are equal in magnitude.
Here, object A has twice the mass of object B, but because both objects have the same speed before the collision, the forces exerted on each other will be the same.
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A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. ????(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?
Answer:
[tex]v_f=8.17\frac{m}{s}[/tex]
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
[tex]W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m[/tex]
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]
The box is initially at rest, so [tex]v_i=0[/tex]. Solving for [tex]v_f[/tex]:
[tex]v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}[/tex]
An escalator carries you from one level to the next in an airport terminal. The upper level is 4.4m above the lower level, and the length of the escalator is 7.3m . Assume that your mass is 63kg .Part AHow much work does the up escalator do on you when you ride it from the lower level to the upper level?Express your answer to two significant figures and include the appropriate units.W =___________.Part BHow much work does the down escalator do on you when you ride it from the upper level to the lower level?Express your answer to two significant figures and include the appropriate units.W =___________.
Answer:
Part a)
[tex]W = 2716.5 J[/tex]
Part B)
[tex]W = -2716.5 J[/tex]
Explanation:
Part A)
While escalator is moving up work done to move the person upwards is given as
[tex]W = mgh[/tex]
here we know that
m = 63 kg
h = 4.4 m
now we have
[tex]W = 63 \times 9.8 \times 4.4[/tex]
[tex]W = 2716.5 J[/tex]
Part B)
Work done while we move from up to down
So we have
[tex]W = - mgh[/tex]
so we have
[tex]W = -2716.5 J[/tex]
An object is hanging from a rope. When it is held in air the tension in the rope is 8.86 N, and when it is submerged in water the tension is 7.84 N. What is the density of the the object
Answer:
8684.2 kg/m³
Explanation:
Tension in the rope as a result of the weight = 8.86 N
Tension in the rope when submerge in water = 7.84
upthrust = 8.86 - 7.84 =1.02 N = mass of water displaced × acceleration due to gravity
Mass of water displaced = 1.02 / 9.81 = 0.104 kg
density of water = mass of water / volume of water
make volume subject of the formula
volume of water displaced = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³
volume of the object = volume of water displaced
density of the object = mass of the object / volume of the object = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³
A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)
Final answer:
To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. Using the given equation, we can solve for the specific heat capacity of the metal, which is found to be 3.75 J/g°C.
Explanation:
To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. We can use the equation:m1c1(T1 - T) = m2c2(T - T2)
where:
m1 is the mass of the metal (34.5 g)
c1 is the specific heat capacity of the metal
T1 is the initial temperature of the metal (75°C)
T is the final temperature of the system (39°C)
m2 is the mass of the water (64.0 g)
c2 is the specific heat capacity of water (4.18 J/g°C)
T2 is the initial temperature of the water (25°C)
Plugging in the given values, we can solve for c1: 34.5g × c1 × (75°C - 39°C) = 64.0g × 4.18 J/g°C × (39°C - 25°C)
Simplifying the equation gives:34.5g × c1 × 36°C = 64.0g × 4.18 J/g°C × 14°C
Dividing both sides by 34.5g × 36°C gives:c1 = (64.0g × 4.18 J/g°C × 14°C) / (34.5g × 36°C)
Calculating the value gives:c1 = 3.75 J/g°C
The final temperature for both the aluminum and water at thermal equilibrium is 17.3 °C. This is calculated using the principle of conservation of energy. The heat lost by aluminum equals the heat gained by water.
To find the final temperature of a 32.5 g cube of aluminum and 105.3 g of water at thermal equilibrium, we use the principle of conservation of energy. The heat lost by the aluminum will be equal to the heat gained by the water.
Calculate the heat transfer equation:Let us simplify: (32.5 g) x (0.900 J/g°C) x (T_final - 45.8 °C) = - (105.3 g) x (4.186 J/g°C) x (T_final - 15.4 °C)
29.25 x (T_final - 45.8) = -440.8858 x (T_final - 15.4)19.25T_final - 1339.85 = -440.8858T_final + 6793.6422470.1358T_final = 8133.4922T_final = 17.3 °CTherefore, the final temperature of both substances at thermal equilibrium is **17.3 °C**.
A jet transport has a weight of 1.87 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel, and the plane's center of gravity is 12.0 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.
Answer:
(a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].
(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]
Explanation:
Given that,
Weight of jet [tex]W=1.87\times10^{6}\ N[/tex]
Distance = 16 m
Second distance = 12.0 m
We need to calculate the normal force exerted by the ground on the front wheel
Using formula of torque
[tex]\sum\tau=-Wl_{W}+F_{f}l_{f}=0[/tex]
Where, W = weight of jet
[tex]l_{f}[/tex]=lever arms for the forces [tex]F_{f}[/tex]
[tex]l_{w}[/tex]=lever arms for the forces W
Put the value into the formula
[tex]-(1.87\times10^{6})\times(16.0-12.0)+F_{f}\times16.0=0[/tex]
[tex]F_{f}=\dfrac{(1.87\times10^{6})\times(16.0-12.0)}{16.0}[/tex]
[tex]F_{f}=4.67\times10^{5}\ N[/tex]
(b). We need to calculate the normal force exerted by the ground on each of the two rear wheels
The sum of vertical forces equal to zero.
[tex]\sum F_{y}=F_{f}+2F_{r}-W=0[/tex]
We using 2 for two rear wheels
[tex]\sum F_{y}=0[/tex]
[tex]F_{f}+2F_{r}-W=0[/tex]
[tex]F_{r}=\dfrac{F_{f}-W}{2}[/tex]
Put the value into the formula
[tex]F_{r}=\dfrac{-4.67\times10^{5}+1.87\times10^{6}}{2}[/tex]
[tex]F_{r}=7.02\times10^{5}\ N[/tex]
Hence, (a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].
(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]
Dan is gliding on his skateboard at 4.00 m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.00 m/s . Dan's mass is 50.0 kg and the skateboard's mass is 7.00 kg .
Answer:
v₂ = 3.44 m/s
Explanation:
given,
Dan is gliding on skateboard at = 4 m/s
speed of skateboard = 8 m/s
mass of Dan = 50 Kg
mass of skateboard = 7 Kg
Speed of Dan when his feet hit the ground = ?
Using conservation of momentum
PE(initial) = PE(final)
(M+m)V = m v₁ + M v₂
(50+7)x 4 = 7 x 8 + 50 x v₂
50 x v₂ =172
v₂ = 3.44 m/s
speed of Dan when his feet hit the ground = v₂ = 3.44 m/s
A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface of length 2 m with a coefficient of friction 0.16, then compresses the second spring of constant 2 N/m. The acceleration of gravity is 9.8 m/s2.
How far will the second spring compress in order to bring the mass to a stop?
Answer:
[tex]X_2=25.27m[/tex]
Explanation:
Here we will call:
1. [tex]E_1[/tex]: The energy when the first spring is compress
2. [tex]E_2[/tex]: The energy after the mass is liberated by the spring
3. [tex]E_3[/tex]: The energy before the second string catch the mass
4. [tex]E_4[/tex]: The energy when the second sping compressed
so, the law of the conservations of energy says that:
1. [tex]E_1 = E_2[/tex]
2. [tex]E_2 -E_3= W_f[/tex]
3.[tex]E_3 = E_4[/tex]
where [tex]W_f[/tex] is the work of the friction.
1. equation 1 is equal to:
[tex]\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2[/tex]
where K is the constant of the spring, x is the distance compressed, M is the mass and [tex]V_2[/tex] the velocity, so:
[tex]\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2[/tex]
Solving for velocity, we get:
[tex]V_2[/tex] = 65.319 m/s
2. Now, equation 2 is equal to:
[tex]\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd [/tex]
where M is the mass, [tex]V_2[/tex] the velocity in the situation 2, [tex]V_3[/tex] is the velocity in the situation 3, [tex]U_k[/tex] is the coefficient of the friction, N the normal force and d the distance, so:
[tex]\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2) [/tex]
Volving for [tex]V_3[/tex], we get:
[tex]V_3 = 65.27 m/s[/tex]
3. Finally, equation 3 is equal to:
[tex]\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2[/tex]
where [tex]K_2[/tex] is the constant of the second spring and [tex]X_2[/tex] is the compress of the second spring, so:
[tex]\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2[/tex]
solving for [tex]X_2[/tex], we get:
[tex]X_2=25.27m[/tex]
The second spring will compress by 25.27 meters to bring the mass to a stop.
To solve this problem, we will go step-by-step considering energy conservation and the work done by friction.
1) Calculate the potential energy stored in the first spring:
The spring constant k is 20 N/m and the compression x is 8 m. The potential energy (PE) in the spring is calculated using the formula:
PE = (1/2) * k * x²
PE = (1/2) * 20 N/m * (8 m)²= 640 J
When the spring is released, this potential energy will convert to kinetic energy (KE) of the mass:
KE = 640 J
Calculate the work done by friction (Wf):
The frictional force (f) is calculated by:
f = μ * m * g
f = 0.16 * 0.3 kg * 9.8 m/s² = 0.4704 N
The work done by friction over a distance (d) of 2 m is:
Wf = f * d
Wf = 0.4704 N * 2 m = 0.9408 J
The kinetic energy after overcoming friction is:
Remaining KE = 640 J - 0.9408 J = 639.0592 J
Calculate the compression of the second spring (x₂):
The remaining kinetic energy will convert into potential energy in the second spring with a constant of 2 N/m.
PE = (1/2) * k₂ * x₂²
We set the remaining KE equal to this potential energy:
639.0592 J = (1/2) * 2 N/m * x₂²
639.0592 J = x₂²
x₂ = √639.0592 = 25.27 m
Therefore, the second spring will compress by 25.27 meters.
A steel plate has a hole in it with a diameter of exactly 1.0 cm when the plate is at a temperature of 20 degrees C. A steel ring has an inner diameter of exactly 1.0 cm at 20 degrees C. Both the plate and the ring are heated to 100 degrees C. Which statement is true?
a. the hole in the plate gets smaller, and the opening in the ring gets larger.
b. The opening in the ring gets larger, but we need the relative size of the plate and the hole to know what happens to the hole.
c. the hole in the plate and the opening in the ring get larger.
d. The hole in the plate and the opening in the ring get smaller.
e.The hole in the plate and the opening in the ring gets smaller.
Answer:
c. the hole in the plate and the opening in the ring get larger.
Explanation:
The hole in the plate and the ring both of them get enlarged.
The thermal expansion is very much similar to the optical expansion of the shadow of an object.
The change in length of an object is given by the relation:
[tex]\Delta l=l.\alpha.\Delta T[/tex]
where:
l = original length of the object
[tex]\alpha=[/tex] coefficient of linear expansion
ΔT = change in temperature
When heated from 20 degrees C to 100 degrees C, the diameter of the hole in a steel plate and the inner diameter of a steel ring both increase due to thermal expansion. So the correct option is C.
Explanation:The correct answer to the question about what happens to the diameter of a hole in a steel plate and the inner diameter of a steel ring when both are heated is that both will get larger. This is due to the thermal expansion of materials, which states that objects expand in all dimensions as their temperature increases. When the steel plate and ring are heated from 20 degrees C to 100 degrees C, the particles within the steel vibrate more and push each other farther apart, causing both the solid material and the empty space within, like the hole and the opening of the ring, to increase in size.
At the top of a frictionless inclined plane, a 0.50-kilogram block of ice possesses 5.0 joules of gravitational poteRtial energy with respect to the bottom of the incline. After sliding halfway down the plane, the block's gravitational potential energy is _______.
The block of ice would have lost half of its gravitational potential energy after sliding halfway down the frictionless inclined plane. Therefore, its gravitational potential energy will be 2.5 Joules.
Explanation:The question is about the concept of energy conservation in physics, specifically the gravitational potential energy. The gravitational potential energy (U) of an object is given by the formula U = mgh, where 'm' is the mass of the object, 'g' is the acceleration due to gravity, and 'h' is the height from the reference point.
In this case, we know that the block of ice has a gravitational potential energy of 5.0 Joules when it's at the top of the incline. When the block of ice has slid halfway down the incline, it will have lost half of its height. Therefore, it would have lost half of its gravitational potential energy. Hence, after sliding halfway down the incline, the block's gravitational potential energy will be 2.5 Joules.
Learn more about Gravitational Potential Energy here:https://brainly.com/question/23134321
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If 12,000,000 kilocalories of chicken can be produced per hectare, how much land is needed to provide someone with 2,000 kilocalories/day for a year?a. 0.01 hab.0.02 hac. 0.06 ha
Answer:
it will need 0.06 hectare of land
So option (c) is correct
Explanation:
We have given that 12000000 kilo calories of chicken can be produced per hectare
It is given that someone with 2000 kilo calories / day for a year
Total number of days in a year = 365
So total kilo calories of chicken = 365×2000 = 730000 kilo calories
As 12000000 kilo calories of chicken can be produced per hectare
So 730000 kilo calories need a land of [tex]=\frac{730000}{12000000}=0.06hectare[/tex]
So it will need 0.06 hectare of land
Final answer:
To provide someone with 2,000 kilocalories/day for a year, 0.06 hectares of land used for chicken production is needed, as the closest correct answer.
Explanation:
The question asks how much land is needed to provide someone with 2,000 kilocalories/day for a year if 12,000,000 kilocalories of chicken can be produced per hectare. First, calculate the total kilocalories needed for a year by multiplying 2,000 kcal/day by the number of days in a year (365), which equals 730,000 kcal/year. Then, find how much land is needed by dividing the total annual kilocalorie requirement (730,000 kcal) by the amount of kilocalories produced per hectare (12,000,000 kcal).
This calculation gives us:
730,000 kcal / 12,000,000 kcal/ha = 0.0608333 ha
Therefore, the closest answer is c. 0.06 ha.
A 32.5 g iron rod, initially at 21.8°C, is submerged into an unknown mass of water at 63.3°C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.9°C. What is the mass of the water?
Final Answer:
The mass of the water is approximately 48.4 grams.
Explanation:
1. To solve this problem, we can use the principle of conservation of energy, specifically the principle of heat exchange.
2. We can calculate the heat gained by the iron rod and the heat lost by the water, and equate them since there is no heat exchange with the surroundings.
3. The formula for heat exchange is given by: [tex]\[ Q = mc\Delta T \][/tex]
where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.
4. We can set up the equation for heat exchange for the iron rod and the water:
[tex]\[ mc_{\text{iron}}\Delta T_{\text{iron}} = mc_{\text{water}}\Delta T_{\text{water}} \][/tex]
5. Rearranging the equation and solving for the mass of water:
[tex]\[ m_{\text{water}} = \frac{m_{\text{iron}}c_{\text{iron}}\Delta T_{\text{iron}}}{c_{\text{water}}\Delta T_{\text{water}}} \][/tex]
6. Substituting the given values, we can calculate the mass of water to be approximately 48.4 grams.
Therefore, the mass of the water is approximately 48.4 grams.
The mass of the water is approximately 199.8 grams.
Here's how to find the mass of the water:
1. Define the variables:
m_iron = mass of iron rod = 32.5 g
c_iron = specific heat capacity of iron = 0.49 J/g°C
T_iron_initial = initial temperature of iron rod = 21.8°C
T_water_initial = initial temperature of water = 63.3°C
T_final = final temperature of the mixture = 58.9°C
m_water = mass of water (unknown)
c_water = specific heat capacity of water = 4.186 J/g°C
2. Apply the principle of heat transfer:
Heat lost by iron = Heat gained by water
3. Set up the equation:
m_iron * c_iron * (T_iron_initial - T_final) = m_water * c_water * (T_final - T_water_initial)
4. Substitute the known values and solve for m_water:
32.5 g * 0.49 J/g°C * (21.8°C - 58.9°C) = m_water * 4.186 J/g°C * (58.9°C - 63.3°C)
5. Solve for m_water:
m_water ≈ 199.8 g
Therefore, the mass of the water is approximately 199.8 grams.