Answer:
13330.86 J
Explanation:
mass of volume of 30.1 cc of water = density x volume
= 1 x 30.1 = 30.1 gm
= 30.1 x 10⁻³ kg
Heat to be withdrawn to cool water from 22.4 degree to 0 degree
= mass x specific heat x fall of temperature
= 30.1 x 10⁻³ x 4186 x 22.4
= 2822.36 J
Heat to be withdrawn to cool water 0 degree ice
mass x latent heat of freezing
30.1 x 10⁻³ x 334000
= 10053.4 J
Heat to be withdrawn to cool ice from 0 degree to -7.2 degree
mass x specific heat of ice x fall of temperature
= 30.1 x 10⁻³ x 2100 x 7.2
= 455.1 J
Total heat to be withdrawn
= 2822.36 + 10053.4 + 455.1
= 13330.86 J
A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^2, where k=0.25N*s^2/m^2. Determine the maximum speed of free fall for the sky diver and the speed reached after 100m of fall. Plot the speed of the sky diver as a function of time and as a function of distance fallen
Answer:
[tex]v_{max}=52.38\frac{m}{s}[/tex]
[tex]v_{100}=33.81[/tex]
Explanation:
the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:
[tex]\sum{F}=0=F_d-W[/tex]
[tex]F_d=W[/tex]
[tex]kv_{max}^2=m*g[/tex]
[tex]v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}[/tex]
To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:
[tex]\sum{F}=ma=W-F_d[/tex]
[tex]ma=W-F_d[/tex]
[tex]ma=mg-kv_{100}^2[/tex]
[tex]a=g-\frac{kv_{100}^2}{m}[/tex] (1)
consider the next equation of motion:
[tex]a = \frac{(v_{x}-v_0)^2}{2x}[/tex]
If assuming initial velocity=0:
[tex]a = \frac{v_{100}^2}{2x}[/tex] (2)
joining (1) and (2):
[tex]\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}[/tex]
[tex]\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g[/tex]
[tex]v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g[/tex]
[tex]v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}[/tex]
[tex]v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}[/tex] (3)
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}[/tex]
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}[/tex]
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}[/tex]
[tex]v_{100}=\sqrt{1,143.3}[/tex]
[tex]v_{100}=33.81[/tex]
To plot velocity as a function of distance, just plot equation (3).
To plot velocity as a function of time, you have to consider the next equation of motion:
[tex]v = v_0 +at[/tex]
as stated before, the initial velocity is 0:
[tex]v =at[/tex] (4)
joining (1) and (4) and reducing you will get:
[tex]\frac{kt}{m}v^2+v-gt=0[/tex]
solving for v:
[tex]v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }[/tex]
Plots:
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in this process?
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost = [tex]\dfrac{0.029158}{4.03176}[/tex]= 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J
Answer:
(a) 992.87 g
(b) [tex]6.419\times 10^{14} J[/tex]
Solution:
As per the question:
Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g
(a) To calculate mass of He produced:
We know that:
Atomic mass of hydrogen is 1.00784 u
Also,
4 Hydrogen atoms constitutes 1 Helium atom
Mass of Helium formed after conversion:
[tex]4\times 1.00784 = 4.03136 u[/tex]
Also, we know that:
Atomic mass of Helium is 4.002602 u
The loss of mass during conversion is:
4.03136 - 4.002602 = 0.028758 u
Now,
Fraction of lost mass, M' = [tex]\frac{0.028758}{4.03136} = 0.007133 u[/tex]
Now,
For the loss of mass of 1000g = [tex]0.007133\times 1000[/tex] = 7.133 g
Mass of He produced in the process:
[tex]M_{He} = 1000 - 7.133 = 992.87 g[/tex]
(b) To calculate the amount of energy released:
We use Eintein' relation of mass-enegy equivalence:
[tex]E = M'c^{2}[/tex]
[tex]E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J[/tex]
The fastest server in women's tennis is Sabine Lisicki, who recorded a serve of 131 mi/h (211 km/h) in 2014. Suppose that the acceleration of the ball was constant during the contact with the racket. Part A If her racket pushed on the ball for a distance of 0.15 m, what was the acceleration of the ball during her serve?
Using the formula for acceleration, a = (v^2) / 2d, and the given velocity and acceleration distance, the acceleration of the tennis ball during Lisicki's serve was approximately 11368 m/s². The force exerted by the racket is generally higher than the force due to gravity during this action.
Explanation:To calculate the acceleration of the tennis ball during Sabine Lisicki's serve, we can use the formula for acceleration: a = (v^2) / 2d, where 'v' corresponds to the final velocity and 'd' represents the distance over which the ball accelerated.
In this case, the final velocity 'v' is 211 km/h (converted to m/s gives us approximately 58.6 m/s), and Lisicki's racquet was in contact with the ball, causing it to accelerate over a distance 'd' of 0.15 m. Plugging these values into the formula gives us: a = (58.6 m/s)^2 / 2(0.15 m), which equals about 11368 m/s².
The average force exerted by the racket can be understood through the equation F = ma, a case of Newton's second law of motion. However, in this scenario, the force due to gravity is negligible as it's much smaller than the force exerted by the racket. The main focus here is the force exerted by the racket which made such a high acceleration possible.
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A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strikes the ground and (b) how long it takes for the ball to reach the ground. Repeat (a) and (b) for the ball thrown directly up from the same height and with the same initial speed.
Answer:
Explanation:
Ball is thrown downward:
initial velocity, u = - 20 m/s (downward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(a) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]
v = 39.69 m/s
(b) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = - 20 - 9.8 t
t = 2 second
Now the ball is thrown upwards:
initial velocity, u = 20 m/s (upward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(c) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]
v = 39.69 m/s
(d) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = + 20 - 9.8 t
t = 6.09 second
A 0.09 g honeybee acquires a charge of +23 pc while flying. The earth's electric field near the surface is typically (100 N/C, downward). What is the ratio of the electric force on the bee to the bee's weight? Multiply your answer by 10 before entering it below.
Answer:
Explanation:
Charge on honeybee q = 23 x 10⁻¹²
Force due to electric field E = E x q
= 100 x 23 x 10⁻¹²
= 23 x 10⁻¹⁰ N
Gravitational force on the honeybee
= m g = .09x 10⁻³ x 9.8
= .882x 10⁻³ N
Ratio of electric field and gravitational field
23 x 10⁻¹⁰ / .882x 10⁻³
26.07 x 10⁻⁷
= 26.07
The temperature of 1 m^3 of water is decreased by 10°C. If this thermal energy is used to lift the water vertically against gravity, what is the change in height of the center of mass?
Answer:
h = 4271.43 m
Explanation:
given,
Volume of the water = 1 m³
temperature decrease by = 10°C
heat removed from water
Q = m c ΔT
Q = ρ V c ΔT
= 1000 × 1 × 4186 × 10
= 4.186 × 10⁷ J
energy is used to do work to move the water against its weight
Q = force × displacement
4.186 × 10⁷ J = m g × h
4.186 × 10⁷ J = 1000 × 1 × 9.8 × h
h = 4271.43 m
hence, the change in height of is equal to h = 4271.43 m
Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving in a circular path perpendicular to a magnetic field of magnitude B. (Use any variable or symbol stated above along with the following as necessary: q and c.)
Explanation:
Let [tex]m_p[/tex] is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.
The magnetic force is balanced by the centripetal force acting on the proton as :
[tex]\dfrac{mv^2}{r}=qvB[/tex]
r is the radius of path,
[tex]r=\dfrac{mv}{qB}[/tex]
Time period is given by :
[tex]T=\dfrac{2\pi r}{v}[/tex]
[tex]T=\dfrac{2\pi m_p}{qB}[/tex]
Frequency of proton is given by :
[tex]f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}[/tex]
The wavelength of radiation is given by :
[tex]\lambda=\dfrac{c}{f}[/tex]
[tex]\lambda=\dfrac{2\pi m_pc}{qB}[/tex]
So, the wavelength of radiation produced by a proton is [tex]\dfrac{2\pi m_pc}{qB}[/tex]. Hence, this is the required solution.
Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a stopwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.43 m, and stop the watch when the top of the sun again disappears. If the elapsed time is it = 11.9 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.
Answer:
R=3818Km
Explanation:
Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.
Since you travel 2π radians in 24H, the angle can be calculated as:
[tex]\alpha =\frac{2*\pi *t}{24H}[/tex] t being expressed in hours.
[tex]\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad[/tex]
From the triangle formed by A,B and the center of the planet, we know that:
[tex]cos(\alpha )=\frac{r}{r+H}[/tex] Solving for r, we get:
[tex]r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km[/tex]
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.4 m/s^2 until he catches his friend. How much time does it take until he catches his friend (after his friend passes him)?
Final answer:
It will take approximately 1.61 seconds for the bicyclist to catch up to his friend after his friend passes him.
Explanation:
To determine the time it takes for the bicyclist to catch up to his friend, we can use the equation:
distance = initial velocity * time + 0.5 * acceleration * time^2
Since the friend is traveling at a constant speed of 3.1 m/s, the distance traveled by the bicyclist during the 2-second delay is 6.2 m. Using the equation above:
6.2 m = 0 m/s * t + 0.5 * 2.4 m/s^2 * t^2
Simplifying the equation:
2.4 m/s^2 * t^2 = 6.2 m
t^2 = 6.2 m / 2.4 m/s^2
t^2 = 2.5833 s^2
t ∼ 1.61 s
Therefore, it will take approximately 1.61 seconds for the bicyclist to catch up to his friend.
Light travels at a speed of close to 3 x 10^5 km/s in vacuum. Given that it takes light 8 min and 19 s to travel the distance from the center of the Earth to the center of the Sun, how far away is the Sun from the Earth? (Astronomers use this as a "distance unit" called 1 Astronomical Unit or 1 au)
Answer:
1497×10⁵ km
Explanation:
Speed of light in vacuum = 3×10⁵ km/s
Time taken by the light of the Sun to reach the Earth = 8 min and 19 s
Converting to seconds we get
8×60+19 = 499 seconds
Distance = Speed × Time
[tex]\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km[/tex]
1 AU = 1497×10⁵ km
The Sun is 1497×10⁵ km from Earth
which graph shows the variation with amplitude a of the intensity i for a wavelength for this spectrum
Answer: the right answer is C
Explanation:
Transverse waves travel with a speed of 20 m/s on a string under a tension 0f 6.00 N. What tension is required for a wave speed of 30.0 m/s on the same string?
Answer:
[tex]T_2=13.5\ N[/tex]
Explanation:
Given that,
Speed of transverse wave, v₁ = 20 m/s
Tension in the string, T₁ = 6 N
Let T₂ is the tension required for a wave speed of 30 m/s on the same string. The speed of a transverse wave in a string is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]........(1)
T is the tension in the string
[tex]\mu[/tex] is mass per unit length
It is clear from equation (1) that :
[tex]v\propto\sqrt{T}[/tex]
[tex]\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}}[/tex]
[tex]T_2=T_1\times (\dfrac{v_2}{v_1})^2[/tex]
[tex]T_2=6\times (\dfrac{30}{20})^2[/tex]
[tex]T_2=13.5\ N[/tex]
So, the tension of 13.5 N is required for a wave speed of 30 m/s. Hence, this is the required solution.
Final answer:
The tension required for a wave speed of 30.0 m/s on a string with an initial tension of 6.00 N is 900 N.
Explanation:
To find the tension required for a wave speed of 30.0 m/s, we can use the equation: wave speed = square root of (tension/linear mass density).
Given that the initial wave speed is 20.0 m/s and tension is 6.00 N, we can rearrange the equation to solve for tension using the new wave speed.
Substituting the values, we have: 30.0 m/s = sqrt(tension/linear mass density). After squaring both sides of the equation, we get: 900 = tension/linear mass density
Since the linear mass density remains constant, the tension required for a wave speed of 30.0 m/s would be 900 N.
A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including its uncertainty
Answer:
(177.94 ± 3.81) cm^2
Explanation:
l + Δl = 21.7 ± 0.2 cm
b + Δb = 8.2 ± 0.1 cm
Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2
Now use error propagation
[tex]\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}[/tex]
[tex]\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}[/tex]
[tex]\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81[/tex]
So, the area with the error limits is written as
A + ΔA = (177.94 ± 3.81) cm^2
Which of the following does not have the appropriate SI unit? work - Joule
acceleration - m/s2
power - Watt
momentum - kg.m/s
force - Pound
Answer:
Explanation:
The SI system of units is the system which is Standard International system. It is used internationally.
The SI units of the fundamental quantities are given below
Mass - Kilogram
Length - metre
Time - second
temperature - Kelvin
Amount of substance - mole
Electric current - Ampere
Luminous Intensity - Candela
So, The SI unit of work is Joule
SI unit of acceleration is m/s^2
SI unit of power is watt
SI unit of momentum is kg m /s
SI unit of force is newton
Thus, the last option is incorrect.
Find the volume of a cylinder of height 10 cm if its base is a square of side 4 cm
Answer:
[tex]Volume=160 cm^3[/tex]
Explanation:
The volume of the cylinder is given by the area of the base multiplied by the height.
In this case:
Height:
[tex]h = 10 cm\\[/tex]
Base area = area of the square ([tex]area=side*side=side^2[/tex])
the side of the square is:
[tex]side=4cm[/tex]
thus, the area of the base:
[tex]area=(4cm)^2 = 16 cm^2[/tex]
Now we multiply this quantities, to find the volume:
[tex]Volume= 16 cm^2*10cm=160 cm^3[/tex]
An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1 microsecond is 1x10 65). (Express your answers in vector form.) Part 1 (a) What is the average velocity of the electron? Vavg = < > m/s Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER Part 2 (b) If the electron continues to travel at this average velocity, where will it be in another 9 us? 7 = < > m
Final answer:
The electron's average velocity is found to be (0 m/s, 316,000 m/s, -146,000 m/s), and after another 9 microseconds, it will be at the position (0.02 m, 4.464 m, -2.104 m).
Explanation:
To calculate the average velocity of an electron, we use the formula:
Vavg = (rf - ri) / Δt, where rf is the final position, ri is the initial position, and Δt is the time interval between the positions.
Given the initial position (0.02 m, 0.04 m, -0.06 m) and the final position (0.02 m, 1.62 m, -0.79 m), with a time difference of 5 microseconds (μs), which is 5 x 10-6 seconds:
The position change in vector form is
Δr = (0.02 m - 0.02 m, 1.62 m - 0.04 m, -0.79 m - (-0.06 m))
= (0 m, 1.58 m, -0.73 m).
Thus, the average velocity is
Vavg = Δr / Δt
= (0 m, 1.58 m, -0.73 m) / (5 x 10-6 s)
= (0 m/s, 316,000 m/s, -146,000 m/s)
The electron's new position after another 9 μs, moving with the same average velocity, is calculated by:
rnew = rf + Vavg × Δtnew
Here, Δtnew is 9 μs, which is 9 x 10-6 seconds, so:
rnew = (0.02 m, 1.62 m, -0.79 m) + (0 m/s, 316,000 m/s, -146,000 m/s) × (9 x 10-6 s)
= (0.02 m, 1.62 m + (2.844 m), -0.79 m - (1.314 m))
= (0.02 m, 4.464 m, -2.104 m).
The gauge pressure in your car tires is 2.40 x 10^5 N/m^2 at a temperature of 35.0°C when you drive it onto a ferry boat to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to −42.0°C? (Assume that their volume has not changed.)
Answer:
The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
Explanation:
Given that,
Gauge pressure of car tires [tex]P_{1}=2.40\times10^{5}\ N/m^2[/tex]
Temperature [tex]T_{1}=35.0^{\circ}C = 35.0+273=308 K[/tex]
Dropped temperature [tex]T_{2}= -42.0^{\circ}C=273-42=231 K[/tex]
We need to calculate the gauge pressure P₂
Using relation pressure and temperature
[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{2.40\times10^{5}}{308}=\dfrac{P_{2}}{231}[/tex]
[tex]P_{2}=\dfrac{2.40\times10^{5}\times231}{308}[/tex]
[tex]P_{2}=180000 = 1.8\times10^{5}\ N/m^2[/tex]
Hence, The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
A 3.00 kg steel ball strikes a massive wall at 10.0m/s at
anangle of 60.0 degree with the plane of the wall. It bouncesoff
the wall with the same speed and angle. If the ball is incontact
with the wall for 0.200s, what is the average force exertedby the
wall on the ball?
Answer:259.80 N
Explanation:
Given
mass of ball=3 kg
ball velocity =10 m/s
angle made by ball with plane of wall [tex]\theta [/tex]=60
Momentum change in Y direction remains same and there is change only in x direction
therefore
initial momentum[tex]=mvsin\theta [/tex]
=30sin60
Final momentum=-30sin60
Change in momentum is =30sin60+30sin60
=60sin60
and Impulse = change in momentum
Fdt=dP
where F=force applied
dP=change in momentum
[tex]F\times 0.2=60sin60[/tex]
[tex]F\times 0.2=51.96[/tex]
F=259.80 N
The average force exerted by the wall on the 3.00 kg steel ball after it bounces off is 150 N. We calculated this using the principle of conservation of momentum and Newton's second law.
Explanation:The subject of this question is Physics, specifically, the topic of motion and force. To answer this question, we should apply the law of conservation of momentum and Newton's second law (Force = change in Momentum / change in Time).
Firstly, the ball is projected against the wall at an angle of 60 degrees, but we are concerned with the component of velocity perpendicular to the wall. This means we will consider the velocity component of the ball towards the wall, which is 10 cos(60), giving us 5.0 m/s. The incoming momentum of the ball can then be calculated as the mass times the velocity (3.0 kg * 5.0 m/s = 15 kg*m/s).
Since the ball bounces off with the same speed at the same angle, its outgoing momentum is -15 kg*m/s. The change in momentum is therefore Outgoing momentum - Incoming momentum = -15 kg*m/s - 15 kg*m/s = -30 kg*m/s. The force exerted by the wall on the ball equals the Change in momentum divided by the time it takes for the change to occur (-30 kg*m/s / 0.200 s = -150 N). Given that force is a vector and we are asked for the magnitude of the force, the answer is 150 N.
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Multiple Concept Example 9 deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) 40.0 cm, due west; (2) 26.0 cm, 32.0 ° south of west; (3) 19.0 cm, 50.0 ° south of east; and (4) 18.0 cm, 60.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.
Answer:
Explanation:
We shall convert the movement of grasshopper in vector form. Suppose the grass hopper is initially sitting at the origin or (00) position .
It went 40 cm due west so
D₁ = -40 i
It then moves 26 cm 32 ° south of west so
D₂ = - 26 Cos32i - 26 Sin32 j = - 22 i -13.77 j
Then it moves 19 cm 50° south of east
D₃ = 19 Cos 50 i - 19 Sin 50 j = 12.2 i - 14.55 j
Then it moves 18 cm 60° north of east
D₄ = 18 Cos 60 i + 18 Sin 60 j = 9 i + 15.58 j
Total displacement = D₁ +D₂+D₃+D₄
= - 40i -22 i - 13.77 j + 12.2 i - 14.55 j + 9 i + 15.58 j
= - 40.8 i - 12.74 j
Magnitude of displacement D
D² = ( 40.8 )² + ( 12.74)²
D = 42 .74 cm
If ∅ be the required angle
Tan∅ = 12.74 / 40.80 = .31
∅ = 17 ° positive angle with respect to due west.
A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g salt/100 g of water) of the salt? b. How much water would it take to dissolve 25 g of this salt? c. If 10.0 g of this salt is mixed with 15.0 g of water, what percentage of the salt dissolves?
Answer:
The solubility of the salt is 35.16 (g/100 g of water).It would take 71.09 grams of water to dissolve 25 grams of salt.The percentage of salt that dissolves is 52.7 %Explanation:
a.We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.
[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}[/tex]
Working it a little this gives us :
[tex] x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]
[tex] x = 35.16 \ g \ salt [/tex]
So, the solubility of the salt is 35.16 (g/100 g of water).
b.Using the rule of three, we got:
[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}[/tex]
Working it a little this gives us :
[tex] x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}} [/tex]
[tex] x = 71.09 g \ water [/tex]
So, it would take 71.09 grams of water to dissolve 25 grams of salt.
C.Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:
[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}[/tex]
Working it a little this gives us :
[tex] x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]
[tex] x = 5.27\ g \ salt [/tex]
This is the salt dissolved
The percentage of salt dissolved is:
[tex]percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}[/tex]
[tex]percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}[/tex]
[tex]percentage \ salt \ dissolved = 52.7 \% [/tex]
a. The solubility (in g salt/100 g of water) of the salt is 35.16 g/100 g water. b. Amount of water it would take to dissolve 25 g of this salt is 71.10 g. c. If 10.0 g of the salt is mixed with 15.0 g of water, 52.7% of the salt will dissolve.
To solve the problem, we need to determine the solubility of the salt at 25°C using the given data:
Part a: Solubility
The solubility of the salt is calculated as follows:A 3.20 g sample dissolves in 9.10 g of water to form a saturated solution.Solubility (g/100 g water) = (3.20 g salt / 9.10 g water) * 100 = 35.16 g/100 g water.Part b: Amount of Water Needed to Dissolve 25 g of Salt
First, use the solubility obtained in part a:Solubility = 35.16 g salt / 100 g water.To find how much water is needed to dissolve 25 g of salt: (100 g water / 35.16 g salt) * 25 g salt = 71.10 g water.Part c: Percentage of Salt Dissolved
Given 10.0 g of salt mixed with 15.0 g of water:Because we know the solubility is 35.16 g/100 g water, we find the amount that will dissolve in 15.0 g water: (35.16 g salt / 100 g water) * 15.0 g water = 5.27 g salt.Percentage dissolved = (5.27 g dissolved / 10.0 g total) * 100% = 52.7%.Therefore, 52.7% of the 10.0 g of salt will dissolve in 15.0 g of water.
If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of one arcminute, equal to 1.67×10^−2 degrees . If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume that the wavelength of the light is 540 nm . Express your answer in millimeters to three significant figures.
Answer:
2.26 mm.
Explanation:
According to Rayleigh criterion , angular rosolution of eye is given by the expression
Angular resolution ( in radian ) = 1.22 λ / D
λ wave length of light, D is diameter of the eye
Given
angular resolution in degree = 1.67 x 10⁻²
= 1.67 x 10⁻² x π / 180 radian ( 180 degree = π radian )
= 29.13 x 10⁻⁵ radian
λ = 540 x 10⁻⁹ m
Put these values in the expression
29.13 x 10⁻⁵ = 1.22 x 540 x 10⁻⁹ / D
D = [tex]\frac{1.22\times540\times10^{-9}}{29.13\times10^{-5}}[/tex]
D = 2.26 mm.
The effective diameter of an eye's optical system that corresponds to a resolving power of one arcminute, calculated using Rayleigh's criterion and the given values of resolving power and wavelength of light, is approximately 2.27 mm.
Explanation:The effective diameter of an eye's optical system that corresponds to a resolving power of one arcminute can be determined using Rayleigh's criterion for diffraction limit and the given values of the resolving power and wavelength of light.
Rayleigh's criterion states that the minimum resolvable angle for a diffraction-limited system is θmin = 1.22*λ/D where λ is the wavelength of light and D is the diameter of the optical system. Given that the resolving power of the eye is 1.67×10^−2 degrees and the wavelength of light is 540 nm, we can rearrange Rayleigh's formula to solve for D.
Converting 1.67×10^−2 degrees to radians gives us 0.00029 rad. Plugging in the values into Rayleigh's formula and solving for D gives us D = 1.22*λ/θmin. Substituting λ=540*10^−9 m and θmin=0.00029 into the equation, we get D = 2.27 mm to three significant figures.
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Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about 150 N/C at the Earth's surface and points inward toward the Earth's center. What is the magnitude of the electric charge on the Earth? Express your answer using two significant figures.
Answer:
Magnitude of electric field on Earth = Q = 6.8 × 10⁵ C
Explanation:
Electric field = E = 150 N/C
Distance from the center of the earth to the surface = Radius of the earth
Radius of the earth = R = 6.38× 10⁶ m
E = k Q / R² is the basic formula for the electric field. k = 9 × 10⁹ N m²/C²
150 = (9 × 10⁹)(Q) / (6.38× 10⁶ )²
⇒ Charge = Q = (150)(6.38× 10⁶ )²/(9 × 10⁹)
= 6.8 × 10⁵ C(2 significant figures).
Answer:
The charge of earth is[tex]-6.8\times 10^{5}Columbs[/tex]
Explanation:
Assuming earth as a spherical body we have
For a sphere of radius 'r' and charge 'q' the electric field generated at a distance 'r' form the center of sphere is given by the equation
[tex]E=\frac{1}{4\pi \epsilon _o }\cdot \frac{Q}{r^{2}}[/tex]
where
'Q' is the total charge on sphere
Now at a distance 'r' equal to radius of earth(6371 km) we have the electric field strength is 150N/C
Using the given values we obtain
[tex]150=\frac{1}{4\pi \epsilon _o}\frac{Q}{(6371\times 10^{3})^2}\\\\\therefore Q=150\times (6371\times 10^{3})^{2}\times 4\pi \epsilon _o\\\\\therefore Q=6.8\times 10^{5}Columbs[/tex]
Now since the electric field is inwards thus we conclude that this charge is negative in magnitude.
An aeroplane flies in a loop (a circular path in a vertical plane) of radius 200 m. The pilot's head always points toward the centre of the loop. The speed of the aeroplane is not constant; the aeroplane goes slowest at the top of the loop and fastest at the bottom. At the top of the loop, the pilot feels weightless. What is the speed of the aeroplane at this point?
At the bottom of the loop, the speed of the aeroplane is 280 km/h . What is the apparent weight of the pilot at this point? His true weight is 710 N .
Answer:
2899.24 N
Explanation:
W = Weight of pilot
r = Radius
v = Velocity
g = Acceleration due to gravity = 9.81 m/s²
[tex]mg=m\frac{v^2}{r}\\\Rightarrow v=\sqrt{gr}\\\Rightarrow v=\sqrt{9.81\times 200}\\\Rightarrow v=44.3\ m/s[/tex]
Speed of the aeroplane at the top of the loop is 44.3 m/s
Now, v = 280 km/h = 280/3.6 = 77.78 m/s
Apparent weight
[tex]A=W+\frac{W}{g}\frac{v^2}{r}\\\Rightarrow A=710+\frac{710}{9.81}\times \frac{77.78^2}{200}\\\Rightarrow A=2899.24\ N[/tex]
Apparent weight at the bottom of the loop is 2899.24 N
f the electric field is zero at a particular point, must the electric potential be zero at the same point? Explain
Answer:
No
Explanation:
As we know that the electric field nullity does not define that the electric potential will be zero at that point.
For example consider the two positive charge at the mid point of these charge electric field is zero but potential is finite.The electric potential has two contribution means it is positive if charges are positive and it is negative if charges are negative.Question #1: Consider Eratosthenes's experiment to measure the size of the Earth. Suppose the Earth were a smaller planet -- but the sun were still directly overhead in Syene at noon on the Summer Solstice, and it was still 500 miles from Syene to Alexandria. Would the shadow of the stick in Alexandria at noon on the Summer Solstice have been longer, shorter, or the same as it was on our Earth? Briefly explain your reasoning.
Answer:
It would have been longer.
Explanation:
Lets assume the Sun angle = θ
Distance between Syene and Alexandria = D
Circumference of Earth = C
As per Eratosthenes' calculations,
[tex]\frac{\theta}{360} =\frac{D}{C}[/tex]
From the above equation it is evident that if the circumference decreases value of θ will increase which implies that the shadow length would be longer as compared to that on the Earth.
A closed system consisting of 2 lb of a gas undergoes a process during which the relation between pressure and volume is pVn = constant. The process begins with p1 = 15 lbf/in.2, ν1 = 1.25 ft3/lb and ends with p2 = 60 lbf/in.2, ν2 = 0.5 ft3/lb. Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value of n.
Answer:
V1=2.5ft3
V2=1ft3
n=1.51
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=2lb*1.25ft3/lb=2.5ft3
state 2
V2=m.v2
V2=2lb*0.5ft3/lb= 1ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/60)/ ln (1/2.5)
n=1.51
In your first lab, you will measure the diameter and height of a cylinder. The diameter will be measured with a micrometer and the height will be measured with a vernier caliper. Say you measure a diameter of 5.1±0.0005 cm and a height of 37.6±0.005 cm. What will be the uncertainty in your volume?
Answer:
ΔV = ±0.175 cm
Explanation:
The equation for volume is
V = π/4 * d^2 * h
All the measurements are multiplied. To propagate uncertainties in multiplication we add the relative uncertainties together.
The relative uncertainty of the diameter is:
εd = Δd/d
εd = 0.0005/5.1 = 0.000098
The relative uncertainty of the height is:
εh = Δh/h
εh = 0.005/37.6 = 0.00013
Then, the relative uncertainty of the volume is:
εV = 2 * εd + εh
εV = 2 * 0.000098 + 0.00013 = 0.000228
Then we get the absolute uncertainty of the volume, for that we need the volume:
V = π/4 * 5.1^2 * 37.6 = 768.1 cm^3
So:
ΔV = ±εV * V
ΔV = ±0.000228 * 768.1 = ±0.175 cm
A centrifuge in a medical labarotary rotates at an
angularspeed of 3600 rev/min. When switch off, it rotates
through50.0 revolutions before coming to rest. Find the
constantangular acceleration of the centrifuge.
Answer:
angular acceleration = - 217.5 rad/s²
Explanation:
given data
angular speed = 3600 rev/min
rotate = 50 revolution
to find out
angular acceleration
solution
we know here no of rotation n = 3600 rev/min i.e 60 rev/s
so initial angular velocity will be ω(i) = 2π× n
ω(i) = 2π× 60 = 376.9 rad/s
and
final angular velocity will be ω(f) = 0
so
angular displacement will be = 2π × 52 = 326.56 rad
and angular acceleration calculated as
angular acceleration = [tex]\frac{\omega(f)^2-\omega(i)^2}{2*angular displacement}[/tex]
put here value
angular acceleration = [tex]\frac{-376.9^2}{326.56}[/tex]
angular acceleration = - 217.5 rad/s²
In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver’s motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. How long is the diver in the air, from the moment he leaves the board until
Answer:
The diver is in the air for [tex]1.65s[/tex].
Explanation:
Hi
Known data [tex]v_{i}=6.3m/s, y_{i}=3.0m[/tex] and [tex]g=9.8m/s^{2}[/tex].We need to find the time when the diver reaches the highest point, so we use the following equation [tex]v_{f} =v_{i}-gt[/tex] with [tex]v_{f}=0ms[/tex] so [tex]t_{1}=\frac{v_{i}-v_{f} }{g}=\frac{6.3m/s}{9.8m/s^{2} }=0.64 s[/tex]. Then we need to find the highest point, so we use [tex]y=v_{i}t-\frac{1}{2} gt^{2}=(6.3m/s)(0.64s)-\frac{1}{2} (9.8m/s^{2})(0.64s)^{2}=2.03m[/tex], this is above the springboard so the highest point is [tex]y_{max}=5.03m[/tex].Finally, we need to find the time in free fall, so we use [tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2}gt^{2}[/tex], at this stage [tex]v_{i}=0m/s, y_{i}=5.03m[/tex] and [tex]y_{f}=0m[/tex], therefore we have [tex]0m=5.03-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex], and solving for [tex]t_{2}=\sqrt{\frac{5.03m}{4.9m/s^{2}}} =\sqrt{1.02s^{2}}=1.01s[/tex].
Last steep is to sum [tex]t_{1}[/tex] and [tex]t_{2}[/tex], so [tex]t_{T}=t_{1}+t_{2}=0.64s+1.01s=1.65s[/tex].
The total time spent by the diver in the air from the moment he leaves the board until he gets to the water is 1.65 s
We'll begin by calculating the time taken to get to the maximum height from the board.
Initial velocity (u) = 6.3 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Time to reach maximum height (t₁) =?
v = u – gt (since the diver is going against gravity)
0 = 6.3 – 9.8t₁
Collect like terms
0 – 6.3 = –9.8t₁
–6.3 = –9.8t₁
Divide both side by –9.8
t₁ = –6.3 / –9.8
t₁ = 0.64 sNext, we shall determine the maximum height reached by the diver from the board
Initial velocity (u) = 6.3 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum Height from the board (h₁) =?
v² = u² – 2gh (since the diver is going against gravity)
0² = 6.3² – (2 × 9.8 × h₁)
0 = 39.69 – 19.6h₁
Collect like terms
0 – 39.69 = –19.6h₁
–39.69 = –19.6h₁
Divide both side by –19.6
h₁ = –39.69 / –19.6
h₁ = 2.03 mNext, we shall determine the height from the maximum height reached by the diver to the water.
Maximum height from the board (h₁) = 2.03 m
Height of board from water (h₂) = 3 m
Height of diver from maximum height to water (H) =?
H = h₁ + h₂
H = 2.03 + 3
H = 5.03 mNext, we shall determine the time taken by the diver to fall from the maximum height to the water.
Height (H) = 5.03 m
Acceleration due to gravity (g) = 9.8 m/s²
Time to fall from maximum height to water (t₂) =?
H = ½gt²
5.03 = ½ × 9.8 × t₂²
5.03 = 4.9 × t₂²
Divide both side by 4.9
t₂² = 5.03 / 4.9
Take the square root of both side
t₂ = √(5.03 / 4.9)
t₂ = 1.01 sFinally, we shall determine the total time spent by the diver in the air.
Time to reach maximum height (t₁) = 0.64 s
Time to fall from maximum height to water (t₂) = 1.01 s
Total time in air (T) =?T = t₁ + t₂
T = 0.64 + 1.01
T = 1.65 sTherefore, the total time spent by the diver in the air is 1.65 s
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Two tiny conducting spheres are identical and carry charges of -23.0C and +67.2C. They are separated by a distance of 3.18 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.18 cm. Determine the magnitude of the force that each sphere now experiences.
Explanation:
Given that,
Charge 1, [tex]q_1=-23\ C[/tex]
Charge 2, [tex]q_1=+67\ C[/tex]
Distance between charges, r = 3.18 cm = 0.0318 m
(a) Let F is the magnitude of force that each sphere experiences. The force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{-23\times 67}{(0.0318)^2}[/tex]
[tex]F=-1.37\times 10^{16}\ N[/tex]
(b) The spheres are brought into contact and then separated to a distance of 3.18 cm. When they are in contact, both possess equal charges. Net charge is :
[tex]q=\dfrac{q_1+q_2}{2}[/tex]
[tex]q=\dfrac{-23+67}{2}=22\ C[/tex]
Electric force is given by :
[tex]F=9\times 10^9\times \dfrac{22^2}{(0.0318)^2}[/tex]
[tex]F=4.307\times 10^{15}\ N[/tex]
Hence, this is the required solution.