Answer:
(a) 1.3 x 10^6 Hz
(b) 76.73 cm
Explanation:
(a)
the formula for the frequency is given by
f = B q / 2 π m
where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.
B = 46.7 micro tesla = 46.7 x 10^-6 T
q = 1.6 x 10^-19 C
m = 9.1 x 10^-31 kg
f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz
(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J
K = 1/2 mv^2
182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 6.3 x 10^6 m/s
r = m v / B q
Where, r be the radius of circular path
r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)
r = 0.7673 m = 76.73 cm
A Galilean telescope with two lenses spaced 30 cm apart has an objective of 50 cm focal length. (i) What is the focal length of the eyepiece? (ii) What is the magnification of the telescope? Assume the object to be very far away. (iii) What must be the separation between the two lenses when the subject being viewed is 30 m away? Assume the viewing is done with a relaxed eye.
Answer:
i think 7
Explanation:
The maximum potential energy of a spring system (mass 15 kg, spring constant 850 N/m) is 6.5 J. a) What is the amplitude of the oscillation? b) What is the maximum speed? c) Setting φ = 0, write the equation for the potential energy as a function of time.
Answer:
a) 0.124 m
b) 0.93 ms⁻¹
c) 0.5 k A² cos ² ( ωt )
Explanation:
1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.
0.5 ( 850) (A² ) = 6.5
⇒ A = 0.124 m = Amplitude.
b) From energy conservation, 0.5 m v² = 6.5
⇒ speed = v = 0.93 ms⁻¹
c) If x = A cos ωt ,
Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )
If a 2 inch diameter rod is subjected to a centric tensile axial load of 15.71 kip, what is the average normal stress (ksi) to three significant figures?
Answer:
The average normal stress is 5 ksi.
Explanation:
Given that,
Diameter = 2 inch
Load = 15.71 kip
We need to calculate the average normal stress
Using formula of stress
Average normal stress [tex]\sigma =\dfrac{F}{A}[/tex]
Where, F = load
A = area
Put the value into the formula
[tex]\sigma=\dfrac{15.71}{\pi\times(\dfrac{2}{2})^2}[/tex]
[tex]\sigma = 5\ kip/inc^2[/tex]
[tex]\sigma=5\ ksi[/tex]
Hence, The average normal stress is 5 ksi.
A 5601 turn, 9.1 cm long solenoid carries a current of 18.2 Amperes. What is the magnetic field inside this solenoid?
Answer:
Magnetic field = 1.41 T
Explanation:
Magnetic field of solenoid, B = μnI
μ = 4π x 10⁻⁷N/A²
Number of turns per meter, [tex]n=\frac{N}{L}=\frac{5601}{9.1\times 10^{-2}}=6.15\times 10^4turns/m[/tex]
Current, i = 18.2 A
B = μnI = 4π x 10⁻⁷ x 6.15 x 10⁴ x 18.2 = 1.41 T
Magnetic field = 1.41 T
Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F> 32° C>32 K 32°C 32° F 32 K 32° F 32 K 32° c 32° F 32° c 32 K 32° C 32° F 32 K All are the same temperature
Answer:
32 C > 32 F > 32 K
Explanation:
32 F, 32 C, 32 K
Let T1 = 32 F
T2 = 32 C
T3 = 32 K
Convert all the temperatures in degree C
The relation between F and C is given by
(F - 32) / 9 = C / 100
so, (32 - 32) / 9 = C / 100
C = 0
So, T1 = 32 F = 0 C
The relation between c and K is given by
C = K - 273 = 32 - 273 = - 241
So, T3 = 32 K = - 241 C
So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C
Thus, T2 > T1 > T3
32C > 32 F > 32 K
Relative to the ground, a car has a velocity of 15.3 m/s, directed due north. Relative to this car, a truck has a velocity of 22.5 m/s, directed 52.0° north of east. What is the magnitude of the truck's velocity relative to the ground
Answer:
The magnitude of the truck's velocity relative to the ground is 35.82 m/s.
Explanation:
Given that,
Velocity of car relative to ground = 15.3 m/s
Velocity of truck relative to car = 22.5 m/s
We need to calculate the magnitude of the truck's velocity relative to the ground
We need to calculate the x component of the velocity
[tex]v_{x}=22.5\cos\theta[/tex]
[tex]v_{x}=22.5\cos52^{\circ}[/tex]
[tex]v_{x}=13.852\ m/s[/tex]
We need to calculate the y component of the velocity
[tex]v_{y}=15.3+22.5\sin\theta[/tex]
[tex]v_{y}=15.3+22.5\sin52^{\circ}[/tex]
[tex]v_{y}=33.030\ m/s[/tex]
Using Pythagorean theorem
[tex]|v|=\sqrt{v_{x}^2+v_{y}^2}[/tex]
[tex]|v|=\sqrt{(13.852)^2+(33.030)^2}[/tex]
[tex]|v|=35.82\ m/s[/tex]
Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.
The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and the walls thickness is 1.3 cm. Assume steady state conditions exist. Determine the temperature gradient in units of K/m and degree C/m. If the temperature gradient were larger what happens to the heat flux and why? Comment on the direction of heat flux given a negative temperature gradient and again for the case of a positive temperature gradient.
Answer:
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
Explanation:
As we know that heat flux is given by the formula
[tex]q^n = K\frac{dT}{dx}[/tex]
here we know that
K = thermal conductivity
[tex]\frac{dT}{dx}[/tex] = temperature gradient
now we know that
[tex]q^n = 11 W/m^2[/tex]
also we know that
K = 1.7 W/mK
now we have
[tex]11 = 1.7 \frac{dT}{dx}[/tex]
so temperature gradient is given as
[tex]\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m [/tex]
also in other unit it will be same
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
A spring has a natural length of 28 cm. If a 27-N force is required to keep it stretched to a length of 32 cm, how much work W is required to stretch it from 28 cm to 30 cm? (Round your answer to two decimal places.) W = 13.5 Incorrect: Your answer is incorrect. J
Answer:
0.14 J
Explanation:
Use the force to calculate the spring constant.
F = k Δx
27 N = k (0.32 m − 0.28 m)
k = 675 N/m
Work is the change in energy:
W = PE
W = ½ k (Δx)²
W = ½ (675 N/m) (0.30 m − 0.28 m)²
W = 0.135 Nm
W = 0.135 J
Rounding to two decimal places, W = 0.14 J.
Your answer was correct, but it was in units of Ncm, and you needed to answer in units of J.
The work done to stretch the spring from 28 cm to 30 cm is 39.00 Joules, computed using the principles of Hooke's Law and the concept of work done.
Explanation:In this problem, we are dealing with the concept of work done on a spring, which falls under Physics principles. Hooke's Law states that the force to compress or extend a spring by a distance x from its natural length is proportional to x. It can be written as F = kx, where F is the force, k is the spring constant, and x is the distance.
In this case, the force (F) is 27 N, and the length of stretch (x) is 32 cm - 28 cm = 4 cm. We can find the spring constant (k) using the formula k = F / x = 27 N / 4 cm = 6.75 N/cm.
The work done (W) to stretch the spring from 28 cm to 30 cm is the area under the force/displacement graph from '28 cm' to '30 cm'. Since the force is linear with displacement for a spring, this area can be found using the formula for the area of a trapezoid: W = ½ (F1 + F2) x d. F1 is the initial force (k*28 cm), F2 is the final force (k*30 cm), and d is the displacement (30 cm - 28 cm). Substituting the values, W = ½ [(6.75 N/cm*28 cm)+(6.75 N/cm*30 cm)]*(2 cm) = 39.00 J.
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1/32nd of a radioactive element's atoms remain in a lunar rock sample. How many half lives old is the rock?
A. 1
B. 2
C. 3
D. 4
E. 5
Answer:
E. 5
Explanation:
N₀ = initial total number of radioactive elements number
N = Number of atoms of radioactive element after "n" half lives = N₀ /32
n = number of half lives
Number of atoms of radioactive element after "n" half lives is given as
[tex]N = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]
inserting the values
[tex]\frac{N_{0}}{32} = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]
[tex]\frac{1}{32} = \left ( \frac{1}{2} \right )^{n}[/tex]
n = 5
The electrostatic force between two charges is 1.8 × 10–6 N when q1 = q2; that is, when the charges are equal in magnitude. If the magnitude of q1 only is now doubled, what happens to the force between them if the distance between them is not changed?
Answer:
Doubled
Explanation:
F = 1.8 x 10^-6 N, q1 = q2
The force between the two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them.
Now one of the charge is doubled but the distance remains same so the force between the two charges becomes doubled.
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a) What is the half-life of radon-222? (Round your answer to two decimal places.) days (b) How long will it take the sample to decay to 15% of its original amount? (Round your answer to two decimal places.) days
Final answer:
The half-life of radon-222 is 3.82 days.
It take approximately 9.34 days for the sample to decay to 15% of its original amount.
Explanation:
To answer the student's question regarding the half-life of radon-222, we turn to the information provided which states that radon-222 (Rn-222) has a half-life of 3.823 days. Considering this data:
(a) The half-life of radon-222 is 3.82 days.
(b) To determine how long it will take for the sample to decay to 15% of its original amount, we can utilize the radioactive decay formula:
N(t)=N_0(1/2)^(t/T)
Where:
N(t) = remaining quantity after time t
N_0 = initial quantity
T = half-life of the substance
t = time elapsed
For radon-222, substituting N(t)/N_0 = 0.15 and T = 3.823 days into the formula and solving for t, we can generate an accurate answer. The calculation would reveal that it takes approximately 9.34 days for the sample to decay to 15% of its original amount.
What is the work of the force F (6.0N)(4.0N)j(-2.0N)k when the object moves from an initial point with coordinates (1.5 m, 3.0m, -4.5 m) to a final point with coordinates (4.0m, -2.5 m, -3.0m)? (Answer: C) (d) 35.J (a) 35J (b) 10J (c) - 10J (e) Can not tell since the path along which the object moves is not identified.
Answer:
option (c) - 10 j
Explanation:
F = (6 i + 4 j - 2 k) N
r1 = (1.5, 3, -4.5) m = (1.5 i + 3j - 4.5 k) m
r2 = (4, -2.5, - 3) m = (4 i - 2.5 j - 3 k) m
displacement, r = r2 - r1 = ( 2.5 i - 5.5 j + 1.5 k) m
Work done is defined as the dot product of force vector and teh displacement vector.
[tex]W = \overrightarrow{F}.\overrightarrow{r}[/tex]
W = (6 i + 4 j - 2 k) . ( 2.5 i - 5.5 j + 1.5 k)
W = 15 - 22 - 3 = - 10 J
A series circuit consists of a 50-Hz ac source, a 40-Ω resistor, a 0.30-H inductor, and a 60-μF capacitor. The rms current in the circuit is measured to be 1.6 A. What is the power factor of the circuit?
Answer:
0.7
Explanation:
f = 50 hz, R = 40 ohm, L = 0.3 h, C = 60 uC = 60 x 10^-6 c
XL = 2 x 3.14 x f x L = 2 x 3.14 x 50 x 0.3 = 94.2 ohm
Xc = 1 / ( 2 x 3.14 x 50 x 60 x 10^-6) = 53.078 ohm
Impedance is Z.
Z^2 = R^2 + ( XL - Xc)^2
Z^2 = 40^2 + (94.2 - 53.078)^2
Z^2 = 1600 + 1691.019
Z = 57.37 ohm
The power factor is given by
CosФ = r / Z = 40 / 57.37 = 0.697 = 0.7
The power factor of a circuit is the ratio of real power to apparent power and is calculated by the cosine of the phase angle between voltage and current. With the information given, we cannot calculate the power factor of the RLC series circuit because the rms voltage of the source is required but not provided.
Explanation:The power factor of a circuit represents the cosine of the phase angle between the voltage and the current in an AC circuit. In an RLC circuit, like the one described, the power factor can be calculated by dividing the real power (measured in watts) by the apparent power (volt-amps). To find the power factor, we need to first calculate the impedance (Z) of the circuit using the formula Z = √(R² + (XL - XC)²), where XL is the inductive reactance and XC is the capacitive reactance.
Inductive reactance (XL) is given by XL = 2πfL, and capacitive reactance (XC) by XC = 1/(2πfC). Since we know that f = 50 Hz, L = 0.30 H, and C = 60 μF, we can calculate XL and XC. Substituting R for the resistance in the circuit, we can find the impedance. After finding the impedance, we can calculate the real power (P) using P = I²R, where I is the rms current.
From the real power and the apparent power, which is IZ, we can find the power factor by calculating P/(IZ). However, we need the actual values of all elements, including the voltage, to complete these calculations. With the information provided in the question, we can't calculate the power factor because the rms voltage of the source isn't given.
Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Answer:
[tex]strain = 1.4 \times 10^{-3} [/tex]
Explanation:
As we know by the formula of elasticity that
[tex]E = \frac{stress}{strain}[/tex]
now we have
[tex]E = 110 GPA[/tex]
[tex]E = 110 \times 10^9 Pa[/tex]
Area = 15.2 mm x 19.1 mm
[tex]A = 290.3 \times 10^{-6}[/tex]
now we also know that force is given as
[tex]F = 44500 N[/tex]
here we have
stress = Force / Area
[tex]stress = \frac{44500}{290.3 \times 10^{-6}}[/tex]
[tex]stress = 1.53 \times 10^8 N/m^2[/tex]
now from above formula we have
[tex]strain = \frac{stress}{E}[/tex]
[tex]strain = \frac{1.53 \times 10^8}{110 \times 10^9}[/tex]
[tex]strain = 1.4 \times 10^{-3} [/tex]
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 31.5 m/s31.5 m/s (about 7.0×101 mph7.0×101 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
Answer:
Centripetal acceleration of the car is 17.4 m/s²
Explanation:
It is given that,
Radius of the track, r = 57 m
Speed of car, v = 31.5 m/s
We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
[tex]a=\dfrac{(31.5\ m/s)^2}{57\ m}[/tex]
[tex]a=17.4\ m/s^2[/tex]
So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.
n ice skater has a moment of inertia of 5.0 kg-m2 when her arms are outstretched. At this time she is spinning at 3.0 revolutions per second (rps). If she pulls in her arms and decreases her moment of inertia to 2.0 kg-m2, how much work will she have to do to pull her arms in?
Answer:
2440.24 J
Explanation:
Moment of inertia, I1 = 5 kg m^2
frequency, f1 = 3 rps
ω1 = 2 x π x f1 = 2 x π x 3 = 6 π rad/s
Moment of inertia, I2 = 2 kg m^2
Let the new frequency is f2.
ω2 = 2 x π x f2
here no external torque is applied, so the angular momentum remains constant.
I1 x ω1 = I2 x ω2
5 x 6 π = 2 x 2 x π x f2
f2 = 7.5 rps
ω2 = 2 x π x 7.5 = 15 π
Initial kinetic energy, K1 = 1/2 x I1 x ω1^2 = 0.5 x 5 x (6 π)² = 887.36 J
Final kinetic energy, K2 = 1/2 x I2 x ω2^2 = 0.5 x 3 x (15 π)² = 3327.6 J
Work done, W = Change in kinetic energy = 3327.6 - 887.36 = 2440.24 J
(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/ s?
Answer:
1.4719 m per sec
Explanation:
Hello
Kinetic energy is the energy associated with the movement of objects. Although there are many forms of kinetic energy
the formula to use is
[tex]E=\frac{mv^{2} }{2}[/tex]
where m is the mass of the object and v the velocity
lets see the kinetic energy of the sprinter running
[tex]E=\frac{65 Kg*10(\frac{m}{s} ^)){2} }{2} \\\\E=\frac{65 *100 }{2} \\E=3250 Joules\\\\[/tex]
Now, the elephant must have the same kinetic energy
[tex]E=\frac{m*v_{2} ^{2} }{2} \\\\E*2=m*v_{2} ^{2}\\ \frac{2E}{m} =v_{2} ^{2} \\\sqrt{\frac{2E}{m} } =v_{2} \\\\\\v_{2} =\sqrt{\frac{2*3250}{3000} }\\ \\v_{2} =1.4719 \frac{m}{s} \\\\[/tex]
it works only the positive root, so the elephant must to walk to 1.4719 m/s to have the same kinetic energy.
Have a great day
A 3000-kg elephant needs to move at a speed of approximately 1.183 m/s to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s.
Explanation:The subject of the question falls into the category of physics, specifically dealing with the concept of kinetic energy and motion. Kinetic energy is given by the equation KE = 1/2 m v², where m is the mass and v is the velocity of the object. In this case, we want the elephant and the sprinter to have the same kinetic energy.
To find the velocity at which the elephant must move, we set the kinetic energy of the elephant (1/2 * 3000 kg * v²) equal to the kinetic energy of the sprinter (1/2 * 65.0 kg * (10.0 m/s)²) and solve for v (velocity of the elephant).
This gives us v = sqrt((1/2 * 65 kg * (10 m/s)²) / (1/2 * 3000 kg)) = 1.183 m/s. Therefore, a 3000-kg elephant must move at a speed of approximately 1.183 m/s to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s.
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BRAIN BURNER! You observe a hockey puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec. The interaction of the puck and the ice results in a frictional force on the puck, f = 0.15 N. Calculate: the stopping distance for this puck. Type in the numeric part of your answer to the nearest 0.1 m of stopping distance. E.g., if your answer works out to be 2.337 m, then type 2.3 in the answer box. Note: this brain burner calculation puts together
The distance at which the puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec will stop is 131.2 meters.
Given to us
Mass of the puck, m = 0.13kg
The velocity of the ice, u = 17.4 m/sec
Friction force, f = 0.15 N
What is the final velocity of the puck?We know we want to stop the puck, therefore, the final velocity of the puck will be 0.
v = 0
What is the deceleration of the puck?We know that according to the first law of motion,
Force = mass x acceleration
F = m x a
Substitute the value,
[tex]0.15 = 0.13 \times a[/tex]
[tex]a = 1.1538\rm\ m/s^2[/tex]
As we know that the final velocity of the puck will be 0, therefore, there will be a deceleration in the puck.
a = -1.1538 m/s².
Thus, the acceleration of the ice puck is -1.1538 m/s².
What is the stopping distance for this puck?
We know that according to the third equation of the motion,
[tex]v^2-u^2 = 2as[/tex]
substitute the values,
[tex]0^2-(17.4)^2 = 2(-1.1538)s[/tex]
s = 131.2012 = 131.2 meters
Hence, the distance at which the puck of mass 0.13 kg, traveling across the ice at speed 17.4 m/sec will stop is 131.2 meters.
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Final answer:
The stopping distance of a hockey puck is 131.1 m after rounding to the nearest tenth of a meter.
Explanation:
To calculate the stopping distance of a hockey puck, we first need to determine the deceleration caused by the frictional force.
The formula for deceleration due to friction is a = f/m, where a is the acceleration (deceleration, in this case, as it's negative), f is the frictional force, and m is the mass of the puck. Given that the frictional force f is 0.15 N and the mass m of the puck is 0.13 kg, we can calculate the deceleration as follows:
a = f/m = 0.15 N / 0.13 kg ≈ 1.15 m/s2
Now, to find the stopping distance we can use the formula d = v2 / (2×a), where d is the stopping distance, v is the initial speed, and a is the deceleration. Plugging in the initial speed v = 17.4 m/s and the deceleration a = 1.15 m/s2, we get:
d = (17.4 m/s)2 / (2 × 1.15 m/s2) ≈ 131.06 m
To round to the nearest 0.1 m, the stopping distance of the puck is 131.1 m.
Why do we use the two-body problem to solve interplanetary trajectories, instead of including all of the appropriate gravitational forces that actually apply?
Answer:
ur mom
hsheu7shrbrjxbfbbrndnifidfjf
A bullet of mass 0.093 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 2.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block?
Explanation:
It is given that,
Mass of the bullet, m₁ = 0.093 kg
Initial speed of bullet, u₁ = 100 m/s
Mass of block, m₂ = 2.5 kg
Initial speed of block, u₂ = 0
We need to find the speed of the block after the bullet embeds itself in the block. Let it is given by V. On applying the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]V=\dfrac{0.093\ kg\times 100\ m/s+0}{(0.093\ kg+2.5\ kg)}[/tex]
V = 3.58 m/s
So, the speed of the bullet is 3.58 m/s. Hence, this is the required solution.
A car battery has a rating of 270 ampere-hours. This rating is one indication of the total charge that the battery can provide to a circuit before failing. (a) What is the total charge (in coulombs) that this battery can provide? (b) Determine the maximum current that the battery can provide for 42 minutes.
Answer:
Part a)
charge through the battery is
[tex]Q = 9.72 \times 10^5 C[/tex]
Part b)
Maximum current is
i = 386 A
Explanation:
Part a)
As we know that the rating of battery is given as
[tex]R = 270 A h[/tex]
here we also know that the charge given by the battery is same as the capacity of the battery
so we will have
[tex]Q = i t[/tex]
[tex]Q = (270 A)(3600 s)[/tex]
[tex]Q = 9.72 \times 10^5 C[/tex]
Part b)
Now we know that current in the wire is given by
[tex]i = \frac{Q}{t}[/tex]
now plug in all values in it
[tex]i = \frac{9.72 \times 10^5}{42\times 60}[/tex]
[tex]i = 386 A[/tex]
A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of chloroform is found to contain 30.0 g of carbon, what is the total mass of chloroform in the second sample?
To find the total mass of chloroform in the second sample, the mass of carbon is scaled proportionally to find the corresponding masses of hydrogen and chlorine. By summing these, the total mass of chloroform is calculated to be 298.525 grams.
Explanation:To determine the total mass of chloroform in the second sample, we must first understand that chloroform has a known molecular formula of CHCl3. Given that the first sample contains 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen, we can deduce the mass ratios of the elements within chloroform. Using the molecular mass of chloroform, which is 119.37 g/mol, we can calculate the masses of hydrogen and chlorine in the second sample based on the given mass of carbon.
For the second sample: If 12.0 g of carbon is accompanied by 1.01 g of hydrogen and 106.4 g of chlorine, then 30.0 g of carbon should be accompanied by:
Hydrogen: (1.01 g H / 12.0 g C) × 30.0 g C = 2.525 g HChlorine: (106.4 g Cl / 12.0 g C) × 30.0 g C = 266.0 g ClThus, the total mass of chloroform in the second sample would be the sum of the masses of carbon, hydrogen, and chlorine: 30.0 g C + 2.525 g H + 266.0 g Cl = 298.525 g of chloroform.
Starting from rest, a 1.8 x 10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.42 N on it. This force does +2.7 x 10^-4 J of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.
Answer:
a)
1.73 m/s
b)
6.43 x 10⁻⁴ m
Explanation:
m = mass of the flea = 1.8 x 10⁻⁴ kg
v₀ = initial speed of the flea = 0 m/s
v = final speed of the flea
W = work done by the force on the flea = 2.7 x 10⁻⁴ J
Using work-change in kinetic energy, Work done is given as
W = (0.5) m (v² - v₀²)
Inserting the values
2.7 x 10⁻⁴ = (0.5) (1.8 x 10⁻⁴) (v² - 0²)
v = 1.73 m/s
b)
d = distance moved by the flea while pushing off
F = Upward force applied on the flea by ground = 0.42 N
Work done is also given as
W = F d
2.7 x 10⁻⁴ = (0.42) d
d = 6.43 x 10⁻⁴ m
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.60 kg and 16.0 g whose centers are separated by about 4.00 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Answer:
[tex]1.0672\times 10^{-9}N[/tex]
Explanation:
[tex]G[/tex] = Gravitational constant = 6.67 x 10⁻¹¹
[tex]F[/tex] = Gravitational force between these spheres
[tex]m_{1}[/tex] = mass of first sphere = 1.60 kg
[tex]m_{2}[/tex] = mass of second sphere = 16 g = 0.016 kg
[tex]r[/tex] = distance between the centers of the sphere = 4 cm = 0.04 m
Gravitational force between these spheres is given as
[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]
[tex]F = \frac{(6.67\times 10^{-11})(1.60)(0.016)}{0.04^{2}}[/tex]
[tex]F = 1.0672\times 10^{-9}N[/tex]
Determine the COP. for this thermodynamic refrigerator
Answer:
Explanation:
The efficiency of a refrigerator is defined in the terms of coefficient of performance (COP).
The ratio of amount of heat in cold reservoir to the work done is termed as the COP.
COP = QL / W
COP = T2 / (T1 - T2)
Where, T1 be the temperature of hot reservoir, T2 be the temperature of cold reservoir.
How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature? A. 334,584 J B. 167,292 J C. 100,375 J D. 450,759 J
Answer:
Option C is the correct answer.
Explanation:
Heat required to melt solid in to liquid is calculated using the formula
H = mL, where m is the mass and L is the latent heat of fusion.
Latent heat of fusion for water = 333.55 J/g
Mass of ice = 0.3 kg = 300 g
Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature
H = mL = 300 x 333.55 = 100,375 J
Option C is the correct answer.
A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 5.21 m/s at an angle of 29.7° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.
Answer:
Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.
Explanation:
Let line of action be horizontal axis , mass of ball be m and unknown velocity be v.
Here momentum is conserved.
Initial momentum =Final momentum
Initial momentum = m x 6i + m x 0i = 6m i
Final momentum = m x (5.21cos 29.7 i + 5.21sin 29.7 j) + m x v = 4.26 m i + 2.58 m j + m v
4.26 m i + 2.58 m j + m v = 6m i
v = 1.74 i - 2.58 j
Magnitude of velocity [tex]=\sqrt{1.74^2+(-2.58)^2}=3.11m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-2.58}{1.74}\right )=--56^0[/tex]
Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.
Part C Suppose 1 kg each of water (4.19 J/(g ⋅ ∘C)), brick (0.90 J/(g ⋅ ∘C)), iron (0.46 J/(g ⋅ ∘C)), and olive oil (1.79 J/(g ⋅ ∘C)) were held at the same initial temperature and heated for an equivalent amount of time. Indicate their relative final temperatures from lowest (left) to highest (right). Assume no heat is lost to the surroundings. You can use the Intro tab of the PhET to help visualize the temperature changes. (Assume that the mas of the water, iron, and olive oil is 1 kg each and that the mass of the brick is 0.5 kg.) Rank from lowest resulting temperature to highest resulting temperature. To rank items as equivalent, overlap them. View Available Hint(s) ResetHelp IronWaterOlive oilBrick
Answer:
Water < Olive oil < Brick < IronExplanation:
The change in temperature may be calculated from the formula:
Q = m × C × ΔT ⇒ ΔT = Q / (m × C)Where:
Q = amount of heat energy supplied (or released in case of cooling)m = massC = specific heat (a different constant, property, for every matter)ΔT = increase of temperatute (decrease when it is cooling)Then, you can make these assumptions or inferences from the conditions stated in the problem:
Q is the same for all the matters because it is supplied by a external source, taking into account that all the substances are held the same time.Initial temperature is the same for all the substances (given)The mass is the same (1 kg each sample)Then, you can state that, for those samples, ΔT = k / C, i.e. the increase in temperature is inversely related to the specific heat.
That means that the higher the specific heat the lower ΔT, and the lower the specific heat the higher ΔT.
The ranking in decrasing order of specific heat is:
Water (4.19) > Olive oil (1.79) > Brick (0.9) > Iron (0.46)Ranking in increasing order of ΔT:
Water < Olive oil < Brick < IronAnd since all of them started at the same temperature, that is the ranking in resulting temperature from lowest to highest:
Water < Olive oil < Brick < Iron ← answerThat means that the sample of water, the matter with the highest specific heat capacity (4.19 J/g°C), will reach the lowest temperature, and the sample of iron, the matter with the lowest heat capacity (0.46 J/g°C) will reach the highest temperature.
A 20 m high filled water tank develops a 0.50 cm hole in the vertical wall near the base. With what speed does the water shoot out of the hole? a) 30 m/s
b)15 m/s
c) 25m/s
d) 20 m/s
Answer:
The speed of the water shoot out of the hole is 20 m/s.
(d) is correct option.
Explanation:
Given that,
Height = 20 m
We need to calculate the velocity
Using formula Bernoulli equation
[tex]\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Where,
v₁= initial velocity
v₂=final velocity
h₁=total height
h₂=height of the hole from the base
Put the value into the formula
[tex]v_{1}^2=2g(h_{2}-h_{1})[/tex]
[tex]v_{1}=\sqrt{2g(h_{2}-h_{1})}[/tex]
[tex]v_{1}=\sqrt{2\times9.8\times(20-0.005)}[/tex]
[tex]v_{1}=19.7\ m/s= approximate\ 20\ m/s[/tex]
Hence, The speed of the water shoot out of the hole is 20 m/s.
wo charged spheres are 1.5 m apart and are exerting an electrostatic force (Fo) on each other. If the charge on each sphere decreases by a factor of 9, determine (in terms of Fo) how much electrostatic force each sphere will exert on the other.
Answer:
F0 / 81
Explanation:
Let the two charges by Q and q which are separated by d.
By use of coulomb's law
F0 = k Q q / d^2 ......(1)
Now the charges are decreased by factor of 9.
Q' = Q / 9
q' = q / 9 ......(2)
Now the Force is
F' = k Q' q' / d^2
F' = k (Q /9) (q / 9) / d^2
F' = k Q q / 81d^2
F' = F0 / 81