Answer:
422.36 N
Explanation:
given,
time of rotation = 4.30 s
T = 4.30 s
Assuming the diameter of the ring equal to 16 m
radius, R = 8 m
[tex]v = \dfrac{2\pi R}{T}[/tex]
[tex]v = \dfrac{2\pi\times 8}{4.30}[/tex]
v = 11.69 m/s
now, Force does the ring push on her at the top
[tex]- N - m g = \dfrac{-mv^2}{R}[/tex]
[tex] N + m g = \dfrac{mv^2}{R}[/tex]
[tex] N = \dfrac{mv^2}{R}- m g[/tex]
[tex] N = m(\dfrac{v^2}{R}- g)[/tex]
[tex] N = 58\times (\dfrac{11.69^2}{8}- 9.8)[/tex]
N = 422.36 N
The force exerted by the ring to push her is equal to 422.36 N.
The force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]
What will be the force does the ring push on her at the top of the ride?It is Given that
Time rotation T= 4.30 s
Mass m= 58 kg
Now the Velocity will be calculated as
[tex]V=\dfrac{2\pi r}{T} =\dfrac{2\pi 8}{4.30} =11.69 \frac{m}{s}[/tex]
Now by balancing the forces
[tex]N=\dfrac{mv^2}{R} -mg[/tex]
[tex]N=m(\dfrac{v^2}{R} -g)[/tex]
[tex]N=58\times (\dfrac{11.69^2}{8} -9.8)[/tex]
[tex]N=422.36 \ Newton[/tex]
Thus the force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]
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Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..
Answer:
Option c is correct
Explanation:
There are two types of collisions-elastic collision and inelastic collision.
In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.
Total energy is always conserved in both types. Thus, option a is incorrect.
In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.
Answer:
i believe its a?
Explanation:
In an inelastic collision, momentum is conserved
If Earth were twice as massive but it revolved at the same distance from the Sun, its orbital period would be
A) 1 year.
B) 3 years.
C) 2 years.
D) 4 years.
E) 6 months.
A) 1 Year, if the Earth is larger in mass, but still the same distance from the sun, it will orbit at the same rate.
The Earth will continue to orbit at the same rate even if its mass increases but its distance from the sun stays the same, or 1 year. Hence, option A is correct.
What are planets?A huge, spherical celestial object that is not a star nor a remnant is called a planet. The emission nebula hypothesis, which holds that a cosmic cloud collapses out of a supernova to produce a young primitive orbited by a planetary system, is the best theory currently available for explaining planet formation. The slow accumulation of matter propelled by gravity—a process known as accretion—leads to the formation of planets in this disk.
The planetary systems Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. Each of these planets revolves around an axis that is inclined with regard to its orbit pole.
Hence, option A is correct.
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A 2 kg rock is suspended by a massless
string from one end of a 7 m measuring stick.
What is the weight of the measuring stick
if it is balanced by a support force at the
1 m mark? The acceleration of gravity is
9.81 m/s^2
Answer in units of N.
Answer:
Weight = 7.848 N
Explanation:
Given:
Mass of the rock (m) = 2 kg
Length of the stick (L) = 7 m
Distance of support from the rock (x) = 1 m
The weight of the stick acts at the center, which is at a distance of 3.5 m from one end.
So, let 'M' be mass of the measuring stick.
Distance of 'W' from the supporting force (d) = 3.5 - 1 = 2.5 m
Now, for equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments.
Sum of clockwise moments = Sum of anticlockwise moments
[tex]m\times g\times x=M\times g\timesd\\\\M=\frac{mx}{d}[/tex]
Plug in the given values and solve for 'M'. This gives,
[tex]M=\frac{2\times 1}{2.5}=0.8\ kg[/tex]
Now, weight of the stick is given as:
[tex]W=Mg=0.8\times 9.81=7.848\ N[/tex]
Answer:
The weight at [tex]1[/tex]m mark is 7.848 N.
Explanation:
We have been given,
Mass of the rock= [tex]2[/tex] kg
Acceleration (gravity) = [tex]9.8[/tex] m/s^2
Length of the string = [tex]7[/tex] m
To find the weight at [tex]1[/tex] m mark let it be W.
So,
We know that torque will be balanced here.
Moment arm for the force (W) is [tex]3.5 - 1=2.5[/tex] m
As it is of [tex]7[/tex] m length so [tex]3.5[/tex] m is the COM (center of mass) of the string.
Let the moment be [tex]L_1\ and\ L_2[/tex] on clockwise and anti-clockwise direction.
[tex]L_1=L_2[/tex]
[tex]2.5*W = 2* 9.81*1[/tex]
Dividing both sides with [tex]2.5[/tex]
[tex]W=\frac{2*9.81*1}{2.5} =7.848\N[/tex]
So the weight is 7.848 N.
If a rope loaded with a force of 100 lbf had a spring constant of 500 lbf/inch, how far would the rope stretch or elongate?
To solve this problem we will apply the concepts related to Hooke's law for which the force exerted on a spring is described as the product between the spring constant and its displacement, that is
[tex]F = kx[/tex]
Where k is the spring's constant and x is the elongation,
Rearranging to find the elongation we have
[tex]x = \frac{F}{k}[/tex]
Replacing,
[tex]x = \frac{100}{500}[/tex]
[tex]x = 0.2in[/tex]
[tex]x = 5.08mm[/tex]
Therefore the elongation produced in the rope from its original length is 0.2in or 5.08mm
At time t in seconds, a particle’s distance s(t), in centimeters, from a point is given by s(t)=13sin t+40. What is the average velocity of the particle from t= π3 to t= 13π3?
Answer:
Explanation:
Given
distance [tex]s(t)=13\sin (t)+40[/tex]
at [tex]t=\frac{\pi }{3}[/tex]
[tex]s(\frac{\pi }{3})=13\sin (\frac{\pi }{3})+40[/tex]
[tex]s(\frac{\pi }{3})=13\times \frac{\sqrt{3}}{2}+40[/tex]
at [tex]t=\frac{13\pi}{3}=4\pi+\frac{\pi}{3}[/tex]
[tex]s(4\pi+\frac{\pi}{3})=13\sin (4\pi+\frac{\pi}{3})+40[/tex]
Average velocity is change in position w.r.t time
[tex]v_{avg}=\frac{s(4\pi+\frac{\pi}{3})-s(\frac{\pi }{3})}{4\pi+\frac{\pi}{3}-\frac{\pi }{3}}[/tex]
[tex]v_{avg}=0[/tex]
A certain lightning bolt moves 78.3 C of charge. How many fundamental units of charge is this?
Answer:
There are [tex]4.89\times 10^{20}\ electrons[/tex] flowing in the lighting bolt.
Explanation:
Given that,
Charge on certain lighting bolt, q = 78.3 C
We need to find the number of charge flowing in that bolt. Let there are n number of electrons. It is a case of quantization of electric charge. It is given by :
q = ne
[tex]n=\dfrac{q}{e}[/tex]
e is the charge on an electron
[tex]n=\dfrac{78.3\ C}{1.6\times 10^{-19}\ C}[/tex]
[tex]n=4.89\times 10^{20}\ electrons[/tex]
So, there are [tex]4.89\times 10^{20}\ electrons[/tex] flowing in the lighting bolt.
For each of the statements below, decide which of the Maxwell equations (forstaticsituations) tells you that the statement is true. Briefly, justify your answers.
(a) There are no magnetic monopoles.
(b) There must be a scalar potential.
(c) There must be a vector potential.
(d) Charges create electric fields.2.
Answer:
(a)There are no magnetic monopoles. true
(c) There must be a vector potential. true
(d) Charges create electric fields.2. true
Explanation:
a) there are no magnetic monopoles because magnetic field is created by charges (electrons) and these electrons have dipole field so it is not possible to have magnetic dipoles, more ever Gauss's law always explained that there are not magnetic dipoles. furthermore magnetic monopoles aree caused by magnetic charges and we have electric charges.
c)vector potential is a vector field which serves as a potential for magnetic field so, the magnetic field B by Faraday and Gauss's law is also known as vector potential.
d) electric field is solely generated by charges be it static charges or moving charges if there are no charges it is not possible to have an electric field.
consider a perfectly inelastic head-on collision between a small car and a large truck traveling at the same speed. Which vehicle has a greater change in kinetic energy as a result of the collision
Answer: The truck has a greater kinetic energy
Explanation:
The initial velocities of both the truck and car are the same. Likewise, the final velocity is the same. The change in kinetic energy is dependent only on mass. The truck has a greater mass, therefore, the change in its kinetic energy is greater.
During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90 km/h during such a sneeze, how far does the car move during that time?
Answer:
12.5 m.
Explanation:
Speed: This can be defined as the rate of change of distance. The S.I unit of speed is m/s. Mathematically it can be expressed as,
Speed = distance/time
S = d/t......................... Equation 1
d = S×t ...................... Equation 2.
Where S = speed of the car, d = distance, t = time taken to shut the eye during sneezing
Given: S = 90 km/h, t = 0.50 s.
Conversion: 90 km/h ⇒ m/s = 90(1000/3600) = 25 m/s.
S = 25 m/s.
Substitute into equation 2.
d = 25×0.50
d = 12.5 m.
Hence the car will move 12.5 m during that time
Final answer:
To find out the distance a car covers during a sneeze while driving at 90 km/h, we first convert the speed to meters per second (25 m/s) and then multiply by the sneeze duration (0.50 s), resulting in a distance of 12.5 meters.
Explanation:
To calculate the distance a car moves during a sneeze, we can use the formula for distance traveled at a constant speed, which is distance = speed imes time. Given that the car's speed is 90 km/h, we first need to convert this speed into meters per second. Since 1 km = 1000 meters and 1 hour = 3600 seconds, the conversion yields:
90 km/h imes (1000 meters/km) imes (1 hour/3600 seconds) = 25 m/s.
Now, using the time duration of the sneeze, which is 0.50 seconds, the distance traveled during the sneeze can be calculated as:
Distance = 25 m/s imes 0.50 s = 12.5 meters.
Therefore, the car travels 12.5 meters during a sneeze lasting 0.50 seconds.
A stretched string is fixed at both ends, 77.7 cm apart. If the density of the string is 0.014 g/cm and its tension is 600 N, what is the wavelength of the first harmonic?
Answer:
The wavelength of the first harmonic is 155.4 cm.
Explanation:
Given that,
The separation between a stretched string that is fixed at both ends, l = 77.7 cm
The density of the string, [tex]d=0.014\ g/cm^3[/tex]
Tension in the string, T = 600 N
We need to find the wavelength of the first harmonic. The wavelength of the nth harmonic on the string that is fixed at both ends is given by :
[tex]\lambda=\dfrac{2l}{n}[/tex]
Here, n = 1
[tex]\lambda=2l[/tex]
[tex]\lambda=2\times 77.7[/tex]
[tex]\lambda=155.4\ cm[/tex]
So, the wavelength of the first harmonic is 155.4 cm. Hence, this is the required solution.
The wavelength of the first harmonic will be "155.4 cm".
Wavelength and DensityAccording to the question,
Separation between stretched string, I = 77.7 cm
String's density, d = 0.014 g/cm³
String's tension, T = 600 N
We know the relation,
Wavelength, λ = [tex]\frac{2l}{n}[/tex]
here, n = 1
then, λ = 2l
By substituting the values,
= 2 × 77.7
= 155.4 cm
Thus the above answer is appropriate.
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The practical limit to an electric field in air is about 3.00 × 10^6 N/C . Above this strength, sparking takes place because air begins to ionize.
(a) At this electric field strength, how far would a proton travel before hitting the speed of light (100% speed of light) (ignore relativistic effects)?
(b) Is it practical to leave air in particle accelerators?
Answer:
(a) x=157 m
(b) No
Explanation:
Given Data
Mass of proton m=1.67×10⁻²⁷kg
Charge of proton e=1.6×10⁻¹⁹C
Electric field E=3.00×10⁶ N/C
Speed of light c=3×10⁸ m/s
For part (a) distance would proton travel
Apply the third equation of motion
[tex](v_{f})^{2} =(v_{i})^{2}+2ax[/tex]
In this case vi=0 m/s and vf=c
so
[tex]c^{2}=(0)^{2}+2ax\\ c^{2}=2ax\\x=\frac{c^{2} }{2a}[/tex]
[tex]x=\frac{c^{2}}{2a}--------Equation (i)[/tex]
From the electric force on proton
[tex]F=qE\\where\\ F=ma\\so\\ma=qE\\a=\frac{qE}{m}\\[/tex]
put this a(acceleration) in Equation (i)
So
[tex]x=\frac{c^{2} }{2(qE/m)}\\ x=\frac{mc^{2}}{2qE} \\x=\frac{(1.67*10^{-27})*(3*10^{8})^{2} }{2*(1.6*10^{-19})*(3*10^{6})}\\ x=157m[/tex]
For part (b)
No the proton would collide with air molecule
A silver dollar is dropped from the top of a building that is 1344 feet tall. Use the position function below for free-falling objects. s(t) = -16t^(2) + v0t + s0
(a) Determine the position and velocity functions for the coin. s(t)=_____________ v(t)=_____________
(b) Determine the average velocity on the interval [3,4]. ______________= ft/s
(c) Find the instantaneous velocities when t = 3 s and t = 4 s. v(3)=________ v(4)=________
(d) Find the time required for the coin to reach the ground level. t=___________s
(e) Find the velocity of the coin at impact. vf =___________ft/s
Answer:
a. position function of the coin:
[tex]s=-16t^2+1344[/tex]
Now the velocity function:
[tex]v=-32t[/tex]
b. [tex]v_{avg}=-112\ m.s^{-1}[/tex]
c. [tex]v_3=-96\ m.s^{-1}[/tex] & [tex]v_4=-128\ m.s^{-1}[/tex]
d. [tex]t=9.165\ s[/tex]
e. [tex]v_f=293.285\ m.s^{-1}[/tex]
Explanation:
Given:
height of dropping the silver dollar, [tex]h=1344\ ft[/tex]Given position function associated with free falling objects:
[tex]s=-16t^2+v_0t+s_0[/tex]
here:
[tex]s_0=[/tex] initial height
[tex]v_0=[/tex]initial velocity
[tex]t=[/tex] time of observation
a)
position function of the coin:
[tex]s=-16t^2+1344[/tex]
∵the object is dropped it was initially at rest
Now the velocity function:
[tex]v=\frac{d}{dt} s[/tex]
[tex]v=-32t[/tex]
b)
we know average velocity is given as:
[tex]\rm v_{avg}=\frac{total\ displacement}{total\ time}[/tex]
Displacement in the given interval:
[tex]s_{_{3-4}}=s_4-s_3[/tex]
[tex]s_{_{3-4}}=(-16\times 4^2+1344)-(-16\times 3^2+1344)[/tex]
[tex]s_{_{3-4}}=-112\ ft[/tex]
Now,
[tex]v_{avg}=\frac{-112}{4-3}[/tex]
[tex]v_{avg}=-112\ m.s^{-1}[/tex]
c)
Instantaneous velocity at t = 3 s:
[tex]v_3=-32\times 3[/tex]
[tex]v_3=-96\ m.s^{-1}[/tex]
Instantaneous velocity at t = 4 s:
[tex]v_4=-32\times 4[/tex]
[tex]v_4=-128\ m.s^{-1}[/tex]
d)
At ground we have s=0:
Put this in position function:
[tex]0=-16t^2+1344[/tex]
[tex]t=9.165\ s[/tex]
e)
Velocity of the coin at impact:
[tex]v_f=-32\times 9.165[/tex]
[tex]v_f=293.285\ m.s^{-1}[/tex]
The position and velocity functions of the silver dollar are s(t) = -16t^2 + 1344 and v(t) = -32t, respectively. After a time of 9.6 seconds, the coin hits the ground with a velocity of -307.2 ft/s. The average speed between 3 and 4 seconds is -224 ft/s.
Explanation:
In this particular problem, we know that the silver dollar is dropped from rest from a height (s0) of 1344 feet. Thus, the initial velocity (v0) is 0.
(a) The position function with these initial conditions becomes s(t) = -16t^2 + 1344 and the velocity function is derived from the position function as v(t) = -32t.
(b) The average velocity on the interval [3,4] can be found by taking the change in position divided by the change in time in that interval and is calculated by [s(4)-s(3)]/[4-3] = -224 ft/s.
(c) The instantaneous velocities at t = 3 s and t = 4 s are found by substituting these times into the velocity function, yielding v(3) = -96 ft/s and v(4) = -128 ft/s.
(d) The time required for the coin to reach the ground level can be found by solving the position function s(t) = 0 for t: t = sqrt(1344/16) = 9.6 s.
(e) The velocity of the coin at impact is found by evaluating the velocity function at t = 9.6 s: vf = -32*9.6 = -307.2 ft/s.
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Determine the period of a 1.7-m-long pendulum on the Moon, where the free-fall acceleration is 1.624 m/s2. Express your answer with the appropriate units.
Answer:
6.42 s
Explanation:
Period: This can be defined as the time taken for a wave or an oscillating body to complete on oscillation. The S.I unit of period is Second (s).
The period of a simple pendulum is represented as,
T = 2π√(L/g)....................................... Equation 1.
Where T = period, L = length of the pendulum, g = acceleration due to gravity. π = pie.
Given: L = 1.7 m, g = 1.624 m/s² and π = 3.14
T = 2(3.14)√(1.7/1.624)
T = 6.28√(1.047)
T = 6.28×1.023
T = 6.42 s.
Thus, the period of the pendulum = 6.42 s
The period of a 1.7-m-long pendulum on the Moon, where the free-fall acceleration is 1.624 m/s² is 6.425 secs
The formula for calculating the period of a simple pendulum is expressed using the formula:
[tex]T=2 \pi\sqrt{\frac{l}{g} }[/tex]
l is the length of the pendulum
g is the acceleration due to gravity
Given the following parameters
g = 1.624 m/s²
l = 1.7m
Substitute the given values into the formula:
[tex]T=2(3.14)\sqrt{\frac{1.7}{1.624} } \\T=6.28\sqrt{1.0468}\\T=6.28(1.0231)\\T = 6.425secs[/tex]
Hence the period of a 1.7-m-long pendulum on the Moon, where the free-fall acceleration is 1.624 m/s² is 6.425 secs.
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Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 50 N/C .
Answer:
q₁ = q₂ = Q = 14.8 pC
Explanation:
Given that
q₁ = q₂ = Q = ?
Distance between charges = r =7.3 cm = 0.073 m
Combined electric field = E₁ + E₂ = E = 50 N/C
Using formula
[tex]E=2\frac{kQ}{r^2}[/tex]
Rearranging for Q
[tex]Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}[/tex]
[tex]Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC[/tex]
The end of Hubbard Glacier in Alaska advances by an average of 105 feet per year.
What is the speed of advance of the glacier in
m/s
?
Answer:
Speed of the glacier, [tex]v=1.0148\times 10^{-6}\ m/s[/tex]
Explanation:
Given that,
The average speed of Hubbard Glacier, [tex]v=105\ feet/year[/tex]
We need to find speed of advance of the glacier in m/s. As we know that,
1 meter = 3.28 feet
And
[tex]1\ year=3.154\times 10^7\ second[/tex]
Using the above conversions, we can write the value of average speed is :
[tex]v=1.0148\times 10^{-6}\ m/s[/tex]
So, the speed of advance of the glacier is [tex]1.0148\times 10^{-6}\ m/s[/tex]. Hence, this is the required solution.
Calculate the hoop stress at rupture if the burst (maximum internal) pressure is 136.4 MPa. The PV has a nominal thickness of 4 mm.
Answer:
[tex]\sigma_x = 1705 MPa[/tex]
Explanation:
Given data:
Internal pressure = 136.4 MPa
thickness is given as 4 mm
diameter is not given in the question hence we are assuming the diameter of given cylinder to be 100 mm
circumferential stress is given as
[tex]\sigma_x = \frac{P\times r}{t} [/tex]
[tex]\sigma_x = \frac{136.4 \times \frac{100}{2}}{4}[/tex]
[tex]\sigma_x = 1705 MPa[/tex]
An electronic package with a surface area of 1 m2 placed in an orbiting space station is exposed to space. The electronics in this package dissipate all 1850 W of its power to space through its exposed surface. The exposed surface has an emissivity of 1.0 and an absorptivity of 0.25. Given: σ = 5.67×10–8 W/m2·K4
Final answer:
The temperature of the patch is approximately 1387 Kelvin, and the rate of heat loss through the patch is approximately 0.07 Watts.
Explanation:
The temperature of the patch can be calculated using the Stefan-Boltzmann law of thermal radiation, which states that the power radiated by an object is proportional to the fourth power of its temperature. The equation to calculate the temperature is:
T = sqrt((P / (A * sigma * e))
Where T is the temperature, P is the power, A is the surface area, sigma is the Stefan-Boltzmann constant, and e is the emissivity. Substituting the given values:
T = sqrt((1850 W / (1 m2 * 5.67×10–8 W/m2·K4 * 1.0)))
T = sqrt((1850 W / 5.67×10–8 W/m2·K4))
Calculating the square root:
T = 1387 K
The temperature of the patch is approximately 1387 Kelvin.
The rate of heat loss through the patch can be calculated using the Stefan-Boltzmann law:
P = A * e * sigma * T4
Substituting the known values:
P = (0.05 m * 0.08 m * 5.67×10–8 W/m2·K4 * 0.300 * (1387 K)4)
Calculating the power:
P = 0.07 W
The rate of heat loss through the patch is approximately 0.07 Watts.
Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you directed a laser beam of wavelength 650 nm onto it at a rate of 1 mol of photons per second from a base on the moon. The spacecraft has a mass of 1.0 kg. Given that, after a period of acceleration from standstill, speed = (force/mass) x time, how many minutes would it take for the craft to accelerate to a speed of 1.0 m/s (about 2.2 mph)?
Answer:
(a) F = 6.14 *10⁻⁴ N
(b) P = 6.14* 10⁻¹⁰ Pa
(c) t = 27.2 min
Explanation:
Area of sail A = 1.0 km² = 1.0 * 10⁶m²
Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m
Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s
(a)
Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹
= 1.0192 * 10⁻²⁷ kg.m/s
Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.
Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.
dp/dt = R * h/ λ
This is the force acting on the sail.
F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N
F = 6.14 *10⁻⁴ N
b)
Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail
P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa
P = 6.14* 10⁻¹⁰ Pa
c)
Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²
As the spacecraft starts from rest, initial speed u=0,m/s ,
final speed is u = 1.0 m/s after time t
v = u+at
t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min
t = 27.2 min
Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 156.9 s ( 2 min and 36.9 s ) , whereas Jill, the faster swimmer, covers 10 lengths in the same time interval. Find the average velocity and average speed of each swimmer.
Answer:
Jill average velocity is 0
Jack average velocity is 0.159337
Jill average speed = 1.593372
Jack average speed = 1.434034
Explanation:
given data
long swimming pool = 25.0 m
9 lengths of the pool = 156.9 s ( 2 min and 36.9 s )
10 lengths = same time interval
to find out
average velocity and average speed
solution
we know that average velocity that is express as
average velocity = [tex]\frac{displacement}{time}[/tex] .....................1
Jill come back where she start
so here velocity will be = 0
and
Jack ends up on the other end of pool
so average velocity = [tex]\frac{25}{156.9}[/tex]
average velocity = 0.159337
now we get here average speed that is express as
average speed = [tex]\frac{distance}{time}[/tex] .............2
jack speed = 9 × [tex]\frac{25}{156.9}[/tex]
jack speed = 1.434034
and
Jill speed = 10 × [tex]\frac{25}{156.9}[/tex]
Jill speed = 1.593372
An insulated beaker with negligible mass contains liquid water with a mass of 0.270 kg and a temperature of 82.5 ∘C . How much ice at a temperature of -22.3 C must be dropped into the water so that the final temperature of the system will be 34.0 C?
Answer:
Explanation:
Given
mass of water [tex]m_1=0.27\ kg[/tex]
Temperature of water [tex]T_{wi}=82.5^{\circ}C[/tex]
Initial Temperature of ice[tex]=-22.3^{\circ}C[/tex]
Final temperature of system [tex]T=34^{\circ}C[/tex]
specific heat of water [tex]c=4.18\ kJ/kg-K[/tex]
specific heat of ice [tex]c_i=2.108\ kJ/kg-K[/tex]
Latent heat of ice [tex]L=336\ kJ/kg[/tex]
Heat loss by Water is equal to heat gained by ice
Heat loss by water [tex]Q_1=m_w\times c\times \Delta T[/tex]
[tex]Q_1=0.27\times 4.18\times (82.5-34)=54.7371\ kJ[/tex]
Heat gained by ice [tex]Q_1=x\times c_i(0-(-22.3))+x\times L+x\times c\times (T-0)[/tex]
[tex]Q_2=x\times 2.108\times (22.3)+x\times 336+x\times 4.18\times 34[/tex]
[tex]Q_2=525.1284x\ kJ[/tex]
[tex]Q_1=Q_2[/tex]
[tex]x=\frac{54.73}{525.1284}[/tex]
[tex]x=0.104\ kg[/tex]
Joules are Nm, which is kg middot m^2/s^2 Use this relationship and 1 Btu = 1.0551 kJ to determine what Btu using primary English units (ft, lb_m, degree F, sec, etc.)
Answer:
[tex]23851.35840\ lbft^2/s^2[/tex]
Explanation:
[tex]1\ Btu=1.0051\ kJ\\\Rightarrow 1\ Btu=1.0051\times 1000 kgm^2/s^2[/tex]
The English units that are to be used here are pounds (lb) and feet (ft)
[tex]1\ kg=2.20462\ lb[/tex]
[tex]1\ m=3.28084\ ft[/tex]
[tex]1005.1\times 2.20462\times 3.28084^2=23851.35840\ lbft^2/s^2[/tex]
The answer is [tex]1\ Btu=23851.35840\ lbft^2/s^2[/tex]
How many photons does a green (532nm) 1mW laser pointer emit? Can we still treat the light as a continuous electromagnetic wave in this case? (2 pts)
Answer:
[tex]n=2.68\times 10^{15}[/tex]photons/s
Yes,we can treat the light as a continuous electromagnetic wave in this case.
Explanation:
We are given that
Wavelength of photon=[tex]\lambda=532 nm=532\times 10^{-9}m[/tex]
1nm=[tex]10^{-9} m[/tex]
Power of photon=1mW=[tex]0.0010J/s[/tex]
We have to find the number of photons emit from laser.
In 1 s, photon emit energy=0.0010 J
We know that
[tex]E=nh\nu=nh\times \frac{c}{\lambda}[/tex]
Where c[tex]\c=3\times 10^8 m/s[/tex]
[tex]E=[/tex]Energy of photon
[tex]\lambda[/tex]=Wavelength of light
[tex]h=6.626\times 10^{-34}Js[/tex]
Substitute the values then, we get
[tex]0.0010=n\times 6.626\times 10^{-34}\times \frac{3\times 10^8}{532\times 10^{-9}}[/tex]
[tex]n=\frac{0.0010\times 532\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^8}[/tex]
[tex]n=2.68\times 10^{15}[/tex]photons/s
Visible light is EM wave.
Range of wavelength of visible light =400nm- 700 nm
Green color lies in visible region because its wavelength lies in 400nm-700 nm.
Therefore, we can treat the light as a continuous electromagnetic wave in this case.
What is the speed v of a wave traveling down such a wire if the wire is stretched to its breaking point?
Answer:
v = 620.17 m/s
Explanation:
There are different formulas for calculating the speed of a wave. Based on the given parameters, the speed of the wave can be estimated as:
v = sqrt(breaking tensile strength/density)
Where:
The breaking tensile strength = 3*10^9 N/m^2
Density = 7800 kg/m^3
Therefore, we can estimate the speed of the wave as shown below:
v = sqrt(3*10^9/7800) = sqrt(384615.3846) = 620.17 m/s
The speed of a wave in a wire at its breaking point depends on the maximum tension the wire can sustain and its linear mass density, calculated with the formula v = (T/μ)¹⁄².
Explanation:The speed v of a wave traveling down a wire at its breaking point would be determined by the tension in the wire just before breaking and the wire's linear mass density. The formula for the speed of a wave on a stretched string is v = (T/μ)¹⁄², where T is the tension in the wire and μ is the linear mass density. For a wire stretched to its breaking point, the tension T would be at its maximum value that the wire can sustain without breaking.
7. A neutral aluminum rod is at rest on a foam insulating base. A negatively charged balloon is brought near one end of the rod but not in direct contact with it. In what way, if any, will the charges in the rod be affected?
Answer:
If a negatively charged balloon is brought near one end of the rod but not in direct contact, then the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.
A 75 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 620 m/s . Part A What is the recoil speed of the hunter?
Express your answer to two significant figures and include the appropriate units.
To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose we will define that the initial moment must be preserved and be equal to the final moment. Since the system was at a stationary moment, the moment before the bullet is fired will be 0, while the moment after the bullet is fired will be equivalent to the sum of the products of the mass and the velocity of each object. , this is,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
m = mass of each object
u = Initial velocity of each object
v = Final velocity of each object
Our values are,
[tex]\text{Mass of hunter} ) m_1 = 75 kg = 75000 g[/tex]
[tex]\text{Mass of bullet} = m_2 = 42 g[/tex]
[tex]\text{Speed of bullet} = v_2 = 6200 m/s[/tex]
[tex]\text{Recoil speed of hunter} =v_1 = ?[/tex]
Applying our considerations we have that the above formula would become,
[tex]m_1v_1 + m_2v_2 = 0[/tex]
Solving for speed 1,
[tex]v_1 = - \frac{m_2v_2}{m_1 }[/tex]
[tex]v_1 = - \frac{42 * 620}{75000 }[/tex]
[tex]v_1 = - 0.3472 m/s[/tex]
Therefore the recoil speed of the hunter is -0.3472m/s, said speed being opposite to the movement of the bullet.
The recoil speed of the hunter standing on the frictionless ice is 0.35m/s².
Given the data in the question;
Mass of hunter; [tex]m_1 = 75kg[/tex]Mass of bullet; [tex]m_2 = 42g = 0.042kg[/tex]Velocity of bullet; [tex]v_2 = 620m/s[/tex]Recoil speed of the hunter; [tex]v_1 =\ ?[/tex]
From the law of conservation of momentum:
[tex]MV = mv[/tex]
Where m is mass and v is velocity
So
[tex]m_1v_1 = m_2v_2\\\\v_1 = \frac{m_2v_2}{m_1}[/tex]
We substitute our given values into the equation
[tex]v_1 = \frac{ 0.042kg \ *\ 620m/s^2}{75kg} \\\\v_1 = \frac{26.04kg.m/s^2}{75kg} \\\\v_1 = 0.35m/s^2[/tex]
Therefore, the recoil speed of the hunter standing on the frictionless ice is 0.35m/s².
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Why is everything spinning? Moons, planets and stars all rotate on their axes, moons orbit planets, planets orbit stars, spiral galaxies rotate around super-massive black holes, etc. What force is causing all this spinning?
Answer:
Conservation of angular momentum
Explanation:
When the objects spread in universe after big bang, because of the tremendous force , they gained angular momentum and started to rotate. Since, then the object continue to rotate on their axis because of conservation of angular momentum. In vacuum of space there no other forces that can stop these rotation, therefore, they continue to rotate.
The sound from a trumpet radiates uniformly in all directions in air with a temperature of 20∘C. At a distance of 5.18 m from the trumpet the sound intensity level is 54.0 dB . The frequency is 561 Hz .
What is the pressure amplitude at this distance?
Answer:
[tex]P=1.44*10^{-2}Pa[/tex]
Explanation:
The pressure amplitude is given by:
[tex]P=\sqrt{2I\rho v[/tex]
Here, I is the sound intensity, [tex]\rho[/tex] the density of the air and v the speed of air.
The sound intensity can be calculated from:
[tex]I=10^\beta I_0[/tex]
Where [tex]\beta[/tex] is the sound intensity level in bels and [tex]I_0[/tex] is the reference sound intensity. So, replacing (2) in (1):
[tex]P=\sqrt{2(10^\beta) I_o \rho v}\\P=\sqrt{2(10^{5.4})(10^{-14}\frac{W}{m^2})(1.2\frac{kg}{m^3})(344\frac{m}{s})}\\P=1.44*10^{-2}Pa[/tex]
The pressure amplitude is found using the formulas for pressure amplitude and sound intensity. Once the sound intensity is found, the pressure amplitude can be calculated, given the speed and density of air. The pressure amplitude at the given distance is approximately 0.0143 Pa.
Explanation:The pressure amplitude can be found with the formula p = √(2*ρ*v*I), where p is the pressure amplitude, ρ is the density of the medium (air), v is the speed of sound in the medium, and I is sound intensity. The sound intensity level (dB) can be converted to sound intensity (W/m^2) by using the formula I = I0 * 10^(L/10), where I0 is the reference intensity (10^-12 W/m^2) and L is the sound intensity level in decibels.
So, calculating the sound intensity I = (10^-12 W/m^2) * 10^(54/10) = 0.002 W/m^2. Once we have the sound intensity, we can calculate the pressure amplitude using the speeds of sound in air at 20°C (343 m/s), and the density of air at sea level (1.225 kg/m^3). Substituting these values into our formula results in a pressure amplitude of about 0.0143 Pa.
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The volume of a fixed mass of an ideal gas is doubled while the temperature is increased from 100 K to 400 K.
What is the final pressure in terms of its initial pressure?
a. 2 Pi
b. 3 Pi
c. 4 Pi
d. 1/2 Pi
Answer:
c. 4 Pi
Explanation:
The pressure law states that the pressure of a gas is directly proportional to its temperature provided volume and other physical quantities remain constant.
If t is the temperature and p is the pressure of a gas,
p ∝ t
p = kt where k s the constant of proportionality
k = p/t
If t1 and p1 are the initial temperature and pressure respectively and
t2, p2 are the final temperature and pressure
p1/t1 = p2/t2
t1 = 100k, t2 = 400k
p2 = p1 × 400/100
p2 = 4t1
The final pressure of the gas is 4 times the initial pressure.
The coefficient of performance of a refrigerator is 5.0.
A. If the compressor uses 10 J of energy, how much heat is exhausted to the hot reservoir?
B. if the hot reservoir temperature is 27 degrees Celsius, what is the lowest possible temperature in degrees Celsius of the cold reservoir?
Answer:
[tex]Q_x=60\ J[/tex]
[tex]T_L=-23\ ^{\circ}C[/tex] is the lowest possible temperature of the cold reservoir.
Explanation:
Given:
Coefficient of performance of refrigerator, [tex]COP=5[/tex]
A)
We know that for a refrigerator:
[tex]\rm COP=\frac{ desired\ effect }{\: energy\ supplied }[/tex]
Given that 10 J of energy is consumed by the compressor:
[tex]5=\frac{desired\ effect}{10}[/tex]
[tex]\rm Desired\ effect=50\ J[/tex]
Now the by the conservation of energy the heat exhausted :
[tex]Q_x=50+10[/tex]
[tex]Q_x=60\ J[/tex]
B)
Also
[tex]COP=\frac{T_L}{T_H-T_L}[/tex]
where:
[tex]T_H\ \&\ T_L[/tex] are the absolute temperatures of high and low temperature reservoirs respectively.
[tex]5=\frac{T_L}{(273+27)-T_L}[/tex]
[tex]1500-5\times T_L=T_L[/tex]
[tex]T_L=250\ K[/tex]
[tex]T_L=-23\ ^{\circ}C[/tex] is the lowest possible temperature of the cold reservoir.
Final answer:
The coefficient of performance (COP) of a refrigerator is the ratio of heat removed from the cold reservoir to the work done by the compressor. Using this definition, we can solve for the amount of heat exhausted to the hot reservoir and the temperature of the cold reservoir.
Explanation:
The coefficient of performance (COP) of a refrigerator is a measure of its efficiency. It is defined as the ratio of heat removed from the cold reservoir to the work done by the compressor:
COP = heat removed / work done
(A) To find the amount of heat exhausted to the hot reservoir, we can use the formula:
heat removed = COP x work done
Substituting the given values, we get:
heat removed = 5.0 x 10 J = 50 J
(B) To find the lowest possible temperature of the cold reservoir, we can use the Carnot efficiency formula:
COP = hot temperature / (hot temperature - cold temperature)
Substituting the given values and rearranging the formula, we get:
cold temperature = hot temperature - (hot temperature / COP)
cold temperature = 27°C - (27°C / 5.0) = 21.6°C
the number of electrons in a copper penny is approximately 10 10^23. How large would the force be on an object if it carried this charge and were repelled by an equal charge one meter away?
Answer:
0.000230144 N
Explanation:
n = Number of electrons = [tex]10\times 10^{23}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
r = Distance between obects = 1 m
The force would be
[tex]F=\dfrac{nkq^2}{r^2}\\\Rightarrow F=\dfrac{10\times 10^{23}\times 8.99\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\Rightarrow F=0.000230144[/tex]
The force would be 0.000230144 N