Answer:
1.656 × 10^(-21) J
Explanation:
Formula for the average translational kinetic energy per atom is;
K.E = (3/2)kΔt
Where;
k is boltzmann constant = 1.38 × 10^(-23) J/K
ΔT is change in temperature = 373 - 293 = 80 K
Thus;
K.E = (3/2) × 1.38 × 10^(-23) × 80
K.E = 1.656 × 10^(-21) J
what is projectile motion
[tex]\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}[/tex]
[tex]\sf{\qquad{\qquad{\underline{\underline{ Projectile~motion }}}}}[/tex]
If an object is given an initial velocity in any direction and then allowed to travel freely under gravity only, it is called a projectile motion.
It is basically 3 types
horizontally projectile motion oblique projectile motion included plane projectile motionThe path followed by a projectile is called its trajectory.
Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path.
The path that the object follows is called its trajectory.
Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity
The search for black holes involves searching for Group of answer choices single stars that emit large amounts of X-rays. a stellar core greater than 3 solar masses large spherical regions from which no light is detected. pulsars with periods less than one millisecond. pulsars that are orbited by planets.
Answer:
A stellar core greater than 3 solar masses
Explanation:
The process of the evolution of a star in which the star changes form with time, includes phases of evolution which are determined by the mass of the core of the star, which involves a process of conversion of elements and the contraction of the star
Where the mass of the core of the star is more than the mass of three Suns, (3 solar masses) the contraction of the star continues, due to the gravitational force being larger than the pressure exerted by neutron, such that the star collapses to a black hole
Therefore, the search for black holes involves searching for a stellar core greater than 3 solar masses.
A thunderclap sends a sound wave through the air and the ocean below. The
thunderclap sound wave has a constant frequency of 100 Hz. What is the
wavelength of the sound wave in water? (The equation for the speed of a
wave is v= f x 1.)
Water
Diamond
Glass
Air
1,493
12,000
5,640
346
Speed of
sound
(m/s)
A. 11.00 m
B. 12.00 m
C. 14.93 m
D. 3.46 m
Answer:
C. 14.93 m
Explanation:
The given frequency of the wave, f = 100 Hz
The given equation for the wave speed, v, is presented as follows;
v = f × λ
The speed of sound in water, v = 1,493 m/s
Therefore, we get;
The wavelength, λ = v/f
∴ λ = 1,493 m/s/(100 Hz) = 14.93 m
The wavelength, λ = 14.93 m.
What is the final step in the fourth stage of technological design, after a product has been improved abs approved
Answer:
Evaluate the solution.
Explanation:
A technological design is designed as the design and study of a solution that can be provided from the solution by identifying the root cause or problem and trying to solve by various means.
A good technological design requires the minimum effort and resources while meeting the requirement of the problem.
The steps involved in the technological design are :
1. search and identify the problem or need.
2. design a solution
3. Implement a solution.
4. Evaluate the solution.
Therefore, the final step or the fourth step in the process of a technological design is " evaluating or communicating the final design solution".
an object 5 cm in size is placed at 30 cm in front of a concave mirror of focal length 45 CM at what distance from the mirror should a screen be placed in order to obtain a shop image find the magnification size and nature of the image
Answer: Given: h₀=5cm
p=30cm
hi
f=45cm
require: q=?
hi=?
formulas:
1/f=1/p+1/q
hi/h₀=q/p
calculations:1/q=1/f-1/p
1/q=1/45-1/30
1/q=0.022=0.033
1/q= = -0.011
q=1/-0.011
q= -90.91cm
hi=p × h₀/q
hi=30ₓ5/-90.91
hi=150/-90.91
hi= -1.65cm
image is virtual , erect and diminished
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A force of 825N is required to push
a car across a lot.Tow students push the car 35m.How much work is done?
Answer:
the work done by the two students is 28,875 J
Explanation:
Given;
applied force, f = 825 N
distance through which the car is pushed, d = 35 m
The work done by the two students who pushed the 825 N across 35 m lot is calculated as follows;
Work done = force x distance
Work done = 825 N x 35 m
Work done = 28,875 Nm = 28,875 J
Therefore, the work done by the two students is 28,875 J
A parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor is connected in series to a 4.89 μF capacitor. This three‑capacitor combination is connected to a 15.5 V battery. Determine the charge on each capacitor.
Answer:
a. i. 35.96 μC b. i. 11.98 μC ii. 24.04 μC
Explanation:
We need to find the total capacitance of the system C.
The total capacitance of the parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor is C' = 1.47 μF + 2.95 μF = 4.42 μF.
C' = 4.42 μF is in series with the 4.89 μF capacitor and for a series combination of capacitors, we have the total capacitance, C from
1/C = 1/4.42 μF + 1/4.89 μF
1/C = (4.42 μF + 4.89 μF)/(4.42 μF × 4.89 μF)
1/C = 9.31 μF/21.6138 μF²
C = 21.6138/9.31 μF
C = 2.32 μF
So, the total charge in the circuit Q = CV where C = total capacitance = 2.32 μF and v = voltage = 15.5 V
So, Q = CV
Q = 2.32 μF × 15.5 V
Q = 35.96 μC
i. The charge on the 4.89 μF capacitor
Since the 4.89 μF is in series with C', the total charge flowing i the circuit is the total charge in the 4.89 μF capacitor. So, its charge Q = 35.96 μC
b. The charge in the 1.47 μF and 2.95 μF capacitors.
To find the charge in the 4.89 μF and 2.95 μF capacitors, we need to find the voltage across the combined parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor. The voltage, V' across the 4.89 μF capacitor, since Q = CV', V' = Q/C = 35.96 μC/4.89 μF = 7.35 V
So, the voltage V" across the combined parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor, C' is V" = 15.5 V - V' (since V' + V" = 15.5 V).
So, V" = 15.5 V - V'
V" = 15.5 V - 7.35 V
V" = 8.15 V
i. The charge on the 1.47 μF capacitor
Using Q' = CV" where Q' = charge across capacitor, C = 1.47 μF and V" = 8.15 V.
So, Q' = CV"
Q' = 1.47 μF × 8.15 V
Q' = 11.98 μC
ii. The charge on the 2.95 μF capacitor
Using Q" = CV" where Q' = charge across capacitor, C = 2.95 μF and V" = 8.15 V.
So, Q" = CV"
Q" = 2.95 μF × 8.15 V
Q" = 24.04 μC
if the pelican in item 3 was traveling at the same speed but was only 2.7m above the water, how far would the fish travel horizontally before hitting the water?
Answer:
5.66 m
Explanation:
From online sources, the speed in item 3 being referred to was discovered to be 7.62 m/s
Now, let's get the time of flight from one of Newton's equation of motion;
S = ut + ½gt²
Considering the vertical component, we have u = 0 m/s.
Thus;.
S = ½gt²
Plugging in the relevant values;
2.7 = ½ × 9.8 × t²
t² = 2.7/4.9
t = √(2.7/4.9)
t = 0.7423 s
Now, when we consider the horizontal component of the motion, we have;
S = vt
Where;
S is the distance the fish will travel horizontally before hitting the water.
v = 7.62 m/s
t = 0.7423
Thus
s = 7.62 × 0.7423
s ≈ 5.66 m
How can we measure electromotive force of a battery illustrated answer with the help of circuit diagram
Answer:
the emf of the cell can be determined by measuring the voltage across the cell using a voltmeter and the current in the circuit using an ammeter for various resistances.
absolute potential difference ,due of point charge of 1C at a distance of 1 m is given by
Answer:
[tex] \implies U = \dfrac{kq}{r} [/tex]
[tex]\implies U = \dfrac{9 \times {10}^{9} \times 1}{1} [/tex]
[tex]\implies U = 9 \times {10}^{9} \: J[/tex]
We have that absolute potential difference ,due of point charge of 1C at a distance of 1 m is given by
[tex]\rho=9x10 ^{10}J[/tex]
From the question we are told that
point charge of 1C at a distance of 1 m
Generally the equation for the Electrostatic potential energy is mathematically given as
[tex]\rho=\frac{kq_1q_2}{r}[/tex]
Where k is a constant
[tex]k=9*10^9Nm^2/c^2[/tex]
Therefore
[tex]\rho=\frac{(9*10^9)1*10^(-6)*1*10^{-6}}{1}[/tex]
[tex]\rho=9x10 ^{10}J[/tex]
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For saving lives, what is the most important safety feature on a car? A. Air bag B. Safety Belt C. Anti-lock brakes
For saving lives the most important safety feature on a car is B. Safety Belt
What are safety features of a car ?Safety features of a car is a feature of a product designed to ensure or increase safety.
Air bag and Anti-lock brakes are the supplemental protection and designed to work best with combination with seat bells.
Air bag reduce the chance that upper body or head will strike the vehicle's interior during a crash alongside with belt that will also hold your upper body
so, the primary safety feature is seat belt and Air bag and Anti-lock brakes comes in secondary safety feature as they increases the safety and risk of getting an injury during any accident
correct answer is B. Safety Belt
learn more about Safety features
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An object 20mm high is placed 10cm from a sperical mirror and forms a virtual image which is 40mm high. what is the radius of curvature off the mirrors
Answer:
R = 40 cm
Explanation:
From the formulae of magnification:
[tex]\frac{q}{p}=\frac{image\ height}{object\ height}\\\\\frac{q}{10\ cm} = \frac{4\ cm}{2\ cm}\\\\q = (10\ cm)(2)\\\\q = 20\ cm[/tex]
where,
q = image distance from mirror
p = object distance from mirror
Using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f}=\frac{1}{10\ cm}+\frac{1}{-20\ cm}\\\\\frac{1}{f} = 0.05\\\\f = 20\ cm[/tex]
q is negative for the virtual image.
Now, the radius of the spherical mirror is double the focal length (f):
R = 2f
R = 2(20 cm)
R = 40 cm
Help
The Euler buckling load of a 160-cm-long column will be _____ times the Euler buckling load of an equivalent 120-cm-long column.
1.78
0.56
0.75
2.37
1.33
0.42
Answer:
Length in times = 0.75 times
Explanation:
Given:
Length of Euler buckling load = 160 cm
Equivalent load length = 120 cm
Find:
Length in times
Computation:
Length in times = Equivalent load length / Length of Euler buckling load
Length in times = 120 / 160
Length in times = 12 / 16
Length in times = 3 / 4
Length in times = 0.75 times
In certain ranges of a piano keyboard, more than one string istuned to the same note to provide extra loudness. For example, thenote at 1.10 x 102 Hz has two strings at this frequency.If one string slips from its normal tension of 6.00 x102 N to 5.40 x 102 N, what beatfrequency is heard when the hammer strikes the two stringssimultaneously?
Answer:
The beat frequency is 5.5 Hz.
Explanation:
f = 110 Hz, T = 600 N , T' = 540 N
let the frequency is f'.
[tex]\frac{f'}{f}=\sqrt\frac{T'}{T}\\\\\frac{f'}{f}=\sqrt\frac{540}{600}\\\\\frac{f'}{f}=0.95\\\\f'= 104.5 Hz[/tex]
So, the beat frequency is f - f' = 110 - 104.5 = 5.5 Hz
8. A copper container of 84g mass contains 84g of water at 20°C. 46g of water at 200°C is mixed with water in the copper ontainer. What is the final temperature of the water? Specific heat capacity of water = 4200 J kg-1 °C-1, Specific heat capacity of copper = 400 J kg-1 °C-1
Answer:
80 °C
Explanation:
The heat transfer parameters for the water and copper container are;
Mass of the copper container, m₁ = 84 g
Mass of the water in the container, m₂ = 84 g
Initial temperature of the water in the container, T₂ = 20°C
Mass of the hot water added, m₃ = 46 g
Initial temperature of the hot water, T₃ = 200°C
Specific heat capacity of water, c₂ = 4,200 J·kg⁻¹·°C⁻¹
Specific heat capacity of copper, c₁ = 400 J·kg⁻¹·C⁻¹
The formula for the specific heat, ΔQ = m·c·ΔT
The heat lost by the hot water = The heat gained by the container the and the cold water
The formula for the specific heat of the mixture is presented as follows;
m₃ × c₃ × (T₃ - T) = m₁ × c₁ × (T - T₁) + m₂ × c₂ × (T - T₂)
Where T represents the final temperature of the water
Therefore, by plugging in the values, we get;
46 × 4200 × (200 - T) = 84 × 400 × (T - 20) + 84 × 4200 × (T - 20)
38640000 - 193200·T = 386400·T - 7728000
38640000 + 7728000 = 46368000 = 386400·T + 193200·T = 579,600·T
∴ T = 46368000/579,600 = 80
The final temperature of the water, T = 80°C
Use the information below the answer the following 3 questions.
A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40o from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N.
Question 2 (2 points)
Question 2 options:
The work done on the crate by the applied force is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 3 (2 points)
Question 3 options:
The work done on the crate by the frictional force is -___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 4 (2 points)
Question 4 options:
The net work done on the crate is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Hint: Do not use rounded answers in subsequent calculations
Answer:
2. 8.62×10² Nm
3. 2.30×10² Nm
4. 6.32×10² Nm
Explanation:
2. Determination of the work done by the applied force.
Force (F) = 225 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fd × Cos θ
Wd = 225 × 5 × Cos 40
Wd = 8.62×10² Nm
3. Determination of the work done by the frictional force.
Frictional Force (Fբ) = 60 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fբd × Cos θ
Wd = 60 × 5 × Cos 40
Wd = 2.30×10² Nm
4. Determination of the net work done.
We'll begin by calculating the net force acting on the crate
Force applied (F) = 225 N
Frictional Force (Fբ) = 60 N
Net force (Fₙ) =?
Fₙ = F – Fբ
Fₙ = 225 – 60
Fₙ = 165 N
Finally, we shall determine the net Workdone. This can be obtained as follow:
Net force (Fₙ) = 165 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fₙd × Cos θ
Wd = 165 × 5 × Cos 40
Wd = 6.32×10² Nm
Which statement describes why energy is released in a nuclear fission reaction based on mass-energy equivalence?
A. For large nuclei, the mass of the original nucleus is greater than the mass of the products.
B. For large nuclei, the mass of the original nucleus is less than the mass of the products.
C. For small nuclei, the binding energy of the lighter nuclei is greater than the binding energy of the heavier nucleus.
D. For small nuclei, the binding energy of the lighter nuclei is less than the binding energy of the heavier nucleus.
Answer:
A is the answer!
Explanation:
Edge 2021
Answer:
A
Explanation:
Edge
Draw a closed circuit diagram of the battery of 2 cells arranged in series, connecting wire, switch and bulb; mark the direction of the current.
pls ill give brainly
Which of the following is evidence for continental drift?
Pls help asap!!
A bucket contains hot water at 95°c. A man wants to bath with water at 40°c. What is the ratio of the mass of hot water to the mass of cold water that he needs.
Answer:
55
Explanation:95-40=55
i hope i did the math right if i didnt please tell me
Photosynthesis is a process in which plants prepare food using carbon dioxide, chlorophyll, and water in the presence of sunlight.
Which activity is best described as a scientific endeavor?
A. Designing devices astronauts can use to help them stay healthy
while in space
B. Developing a procedure for testing how extended stays in space
affect the human body
C. Performing tests on devices used by astronauts to ensure they
can be used safely
D. Creating a process for improving the design of equipment used to
monitor astronauts' health
Answer:
B
Explanation:
BECAUSE TO DO THE TESTS YOU NEED TO DO THE SCIENTIFIC METHOD.
FOR EXAMPLE: OBSERVATIONS AND EXPERIMENTS TO OBTAIN RESULTS.
ANYWAY I LEAVE YOU THE LINK:
https://gscourses.thinkific.com
it will be easier to lift a load in wheel barrow if the load is moved towards the wheel. Give Reason.
Answer:
Here's your answer: The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.
Explanation:
Answer:
Explanation:
The wheel barrows wheel and axle helps the wheelbarrow to move without friction this making it easier to push or pull.thats why it will be easier to lift a load in a wheelbarrow if it's transferred towards the wheel..
Hope it helps
Two astronomy students travel to South Dakota. One stands on Earth’s surface and enjoys some sunshine. At the same time, the other descends into a gold mine where neutrinos are detected, arriving in time to detect the creation of a new radioactive argon nucleus. Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences.
Answer:
The photon takes millions of years to reach the Surface of the sun while the Neutrinos travelling at the speed of light reaches the surface of the sun in approximately 2 seconds
The Photon is million year old while the neutrino is just some minutes old as observed by the student .
Explanation:
Although The Photon ( sunshine from the sun's surface ) heating up the student standing on the Earth's surface and the neutrinos discovered by the other student inside the gold mine are both formed in the Sun's core.
The difference between both are
The photon takes millions of years to reach the Surface of the sun while the Neutrinos travelling at the speed of light reaches the surface of the sun in approximately 2 seconds
The Photon is million year old while the neutrino is just some minutes old as observed by the student .
how far should an object be from a converging mirror of radius 36cm to form a real image one - ninth of its size
Answer:
focal length :
[tex]{ \tt{f = \frac{r}{2} }} \\ focal \: length = \frac{36}{2} = 18 \: cm[/tex]
From linear magnification:
[tex]{ \tt{ \frac{1}{m} + 1 = \frac{u}{f} }} \\ \frac{1}{9} + 1 = \frac{u}{18} \\ \\ u = \frac{18 \times 10}{9} \\ u = 20 \: cm[/tex]
The object must be at 20 cm
A toy airplane is flying at a speed of 3 m/s with an acceleration of 1.1 m/s squared how fast is it flying after five seconds
Answer:
[tex]8.5\frac{m}{s}[/tex]
Explanation:
Use Kinematics:
[tex]v = v_0 + at[/tex]
[tex]v = 3 + 1.1 * 5[/tex]
[tex]v = 8.5 \frac{m}{s}[/tex]
The net charge difference across the membrane, just like the charge difference across the plates of a capacitor, is what leads to the voltage across the membrane. How much excess charge (in picocoulombs, where 1 pc = 1x10-12 C) must lie on either side of the membrane of an axon of length 2 cm to provide this potential difference (0.07 V) across the membrane? You may consider that a net positive charge with this value lies just outside the axon cell wall, and a negative charge with this value lies just inside cell wall, like the equal and opposite charges on capacitor plates. Again, make sure to include LaTeX: \kappa=7κ = 7 for the lipid bilayer.
Answer:
a) Q = 1.24 10⁻² pC, b) Q = 8.68 10⁻² pC
Explanation:
a) the capacitance is defined
C = [tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]
Q = ε₀ [tex]\frac{A}{d} \ \Delta V[/tex]
let's calculate
Q = 8.85 10⁻¹² 0.07 [tex]\frac{A}{d}[/tex]
Q = 0.6195 10⁻¹² [tex]\frac{A}{d}[/tex]
where a is the area of the membrane
A = d L
Q = 0.6195 10⁻¹² Ll
Q = 0.6195 10⁻¹² 0.02
Q = 1.24 10⁻¹⁰ C
Q = 1.24 10⁻² pC
B) the membrane is full of fat with k = 7
C = [tex]\frac{Q}{\Delta V} = k \epsilon_o \ \frac{A}{d}[/tex]
Q = k ε₀ [tex]\frac{A}{d} \ \Delta V[/tex]
Q = k Q₀
Q = 7 1.24 10⁻²
Q = 8.68 10⁻² pC
A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.40 A, is oriented with the plane of its loop perpendicular to a uniform 1.50-T magnetic field pointing into the plane of the loop. Let the loop be in x-y Cartesian plane so that the long and short sides of the loop are parallel to x- and y-axis, respectively. The loop center is at the origin of x-y Cartesian plane. Note that the magnetic field is in the direction of the negative z-axis.a. Calculate: (i) the net force that the magnetic field exerts on the coil; (ii) the torque about the z-axis that the magnetic field exerts on the coil.b. The plane of the coil is now rotated through +30º from its initial orientation (the x-y plane of the Cartesian coordinate system that remains the same). Calculate: (i) the net force that the magnetic field exerts on the coil; (ii) the torque about the rotation axis that the magnetic field exerts on the coil.
Answer:
a) [tex]F_{net}=0[/tex]
b) [tex]T=0[/tex]
Explanation:
From the question we are told that:
Dimensions:
[tex]L*B=22.0*35.0cm[/tex]
Current [tex]I=1.40A[/tex]
Magnetic field [tex]B=1.40[/tex]
Therefore
[tex]Area=L*B[/tex]
[tex]A=22.0*35.0cm[/tex]
[tex]A=770cm=>770*0^{-4}[/tex]
a)
Generally Force on Looping gives
[tex]F_1-F_2[/tex]
[tex]F_3=F_4[/tex]
Therefore
[tex]F_{net}=0[/tex]
b)
Generally the equation for Torque is mathematically given by
[tex]T=i*Asin \theta[/tex]
Since A and B are on opposite direction
[tex]\theta=180[/tex]
Therefore
[tex]T=1.40*770*10^{-4}sin 180[/tex]
[tex]T=0[/tex]
What is the relationship between electric field lines and equipotential lines that you observed in doing the lab
Answer:
Explained below
Explanation:
Generally speaking, we know in physics that Electric field lines are lines which usually start at positive charges and deflect away from them to terminate at the negative charges. Meanwhile Equipotential lines are lines that are used to connect points located on the same electric potential.
Finally, in conclusion, electric field lines are usually lines that go through in a perpendicular manner across every equipotential lines.
HELP ASAP
A. 1.09 A
B. 1.20 A
C.0.910 A
D. 0.830 A
Answer:
The answer should be: 1.20 A
Explanation:
