Answer:
A. 16 J
B. 16 J
C. 8 N
Explanation:
A. Determination of the kinetic energy.
Mass (m) = 0.5 Kg
Velocity (v) =. 8 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 8²
KE = ½ × 0.5 × 64
KE = 0.5 × 32
KE = 16 J
B. Determination of the Workdone by the force.
Kinetic energy (KE) = 16 J
Workdone =.?
Workdone and kinetic energy has the same unit of measurement. Thus,
Workdone = kinetic energy
Workdone = 16 J
C. Determination of the force.
Workdone (Wd) = 16 J
Displacement (s) = 2 m
Force (F) =?
Wd = F × s
16 = F × 2
Divide both side by 2
F = 16 / 2
F = 8 N
Nhiệt dung riêng của một chất là ?
Answer:
enchantment table language
Explanation:
Can an electron be diffracted? Can it exhibit interference?
Answer:
Yeah, it can be diffracted. Though it depends on a diffracting medium.
It must have some magnetic fields .
Forexample; X-ray diffraction where electrons are diffracted to the target filament.
The distance between the two object is fixed at 5.0 m. The uncertainty distance measurement is? The percentage error in the distance is?
Two distinct systems have the same amount of stored internal energy. 500 J are added by heat to the first system and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?
Answer:
The change in the internal energy of the first system is 300 J
The second system will do zero work in order to have the same internal energy.
Explanation:
Given;
heat added to the first system, Q₁ = 500 J
heat added to the second system, Q₂ = 300 J
work done by the first system, W₁ = 200 J
The change in the internal energy of the system is given by the first law of thermodynamics;
ΔU = Q - W
where;
ΔU is the change in internal energy of the system
The change in the internal energy of the first system is calculated as;
ΔU₁ = Q₁ - W₁
ΔU₁ = 500 J - 200 J
ΔU₁ = = 300 J
The work done by the second system to have the same internal energy with the first.
ΔU₁ = Q₂ - W₂
W₂ = Q₂ - ΔU₁
W₂ = 300 J - 300 J
W₂ = 0
The second system will do zero work in order to have the same internal energy.
• Explain how sound travels
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
A long, current-carrying solenoid with an air core has 1550 turns per meter of length and a radius of 0.0240 m. A coil of 200 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
[tex]M=7.05*10^{-4}[/tex]
Explanation:
From the question we are told that:
Coil one turns N_1=1550 Turns/m
Radius [tex]r=0.0240m[/tex]
Turns 2 [tex]N_2=200N[/tex]
Generally the equation for area is mathematically given by
[tex]A=\pi*r^2[/tex]
[tex]A=\pi*0.024^2[/tex]
[tex]A=\1.81*10^{-3} m^2[/tex]
Therefore
The mutual inductance of this system is
[tex]M=\mu*N_1*N_2*A[/tex]
[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]
[tex]M=7.05*10^{-4}[/tex]
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way? answer in seconds.
Answer:
Time to move out of the way = 1.74 s
Explanation:
Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.
Time to move out of the way = 1.74 s
The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off
Answer:
In order to lift off the ground, the air in the balloon must be heated to 710.26 K
Explanation:
Given the data in the question;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
let F represent the force acting upward.
Now in a condition where the hot air balloon is just about to take off;
F - Mg - m[tex]_g[/tex]g = 0
where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.
the force acting upward F = Vρg
so
Vρg - Mg - m[tex]_g[/tex]g = 0
solve for m[tex]_g[/tex]
m[tex]_g[/tex] = ( Vρg - Mg ) / g
m[tex]_g[/tex] = Vρg/g - Mg/g
m[tex]_g[/tex] = ρV - M ------- let this be equation 1
Now, from the ideal gas law, PV = nRT
we know that number of moles n = m[tex]_g[/tex] / μ
where μ is the molecular mass of air
so
PV = (m[tex]_g[/tex]/μ)RT
solve for T
μPV = m[tex]_g[/tex]RT
T = μPV / m[tex]_g[/tex]R -------- let this be equation 2
from equation 1 and 2
T = μPV / (ρV - M)R
so we substitute in our values;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]
T = 1405920 / 1979.442
T = 710.26 K
Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K
The temperature required for the air to be heated is 710.26 K.
Given data:
The mass of a hot air-balloon is, m = 381 kg.
The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].
The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].
The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].
The condition where the hot air balloon is just about to take off is as follows:
[tex]F-mg - m'g =0[/tex]
Here,
m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,
[tex]F = V \times \rho \times g[/tex]
Solving as,
[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]
Now, apply the ideal gas law as,
PV = nRT
here, R is the universal gas constant and n is the number of moles and its value is,
[tex]n=\dfrac{m'}{M}[/tex]
M is the molecular mass of gas. Solving as,
[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]
Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,
[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]
Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.
Learn more about the ideal gas equation here:
https://brainly.com/question/18518493
Question 4 of 5
How can the Fitness Logs help you in this class?
O A. They can't; the Fitness Logs are only useful to your teacher.
B. They show your parents how much you're learning.
C. They let you keep track of your thoughts, feelings, and progress.
D. They help you evaluate yourself for your final grade.
SUBMIT
Answer:
C is the right answer
Explanation:
fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success
hope it was useful for you
A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.30 m/s. If, instead, a 0.40 kg mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]
Therefore, the maximum speed is 0.20 m/s
Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way
Answer:
Move towards the wall.
Explanation:
When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.
As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.
If you tethered a space station to the earth by a long cable, you could get to space in an elevator that rides up the cable much simpler and cheaper than riding to space on a rocket. There's one big problem, however: There is no way to create a cable that is long enough. The cable would need to reach 36,000 km upward, to the height where a satellite orbits at the same speed as the earth rotates; a cable this long made of ordinary materials couldn't even support its own weight. Consider a steel cable suspended from a point high above the earth. The stress in the cable is highest at the top; it must support the weight of cable below it.
What is the greatest length the cable could have without failing?
Answer:
[tex]l=12916.5m[/tex]
Explanation:
Distance [tex]d=3600km[/tex]
Since
Density of steel [tex]\rho=7900kg/m^3[/tex]
Stress of steel [tex]\mu= 1*10^9[/tex]
Generally the equation for Stress on Cable is mathematically given by
[tex]S=\frac{F}{A}[/tex]
[tex]S=\frac{\rho Alg}{A}[/tex]
Therefore
[tex]l=\frac{s}{\rhog}[/tex]
[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]
[tex]l=12916.5m[/tex]
An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
Answer:
[tex]t=0.50cm[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lamda=3c[/tex]m
Refraction Index [tex]n=1.50[/tex]
Generally the equation for Destructive interference for Normal incidence is mathematically given by
[tex]2nt=m(\frac{1}{2})\lambda[/tex]
Since Minimum Thickness occurs at
At [tex]m=0[/tex]
Therefore
[tex]t=\frac{\lambda}{2}[/tex]
[tex]t=\frac{3}{4(1.50)}[/tex]
[tex]t=0.50cm[/tex]
True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
Answer:
false?
Explanation:
The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.
Answer:
I think the answer is False.
Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A
Answer:
The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.
3. If you change the resistance of the resistor:
a. How does the current through the circuit change? (answer, explain, evidence)
b. How does the voltage of the battery change? (answer, explain, evidence)
Answer:
Explanation:
Changing the resistance of a resistor means the resistance is either increased or decreased.
a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.
Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;
V = IR
6 = I x 30
I = 0.2 A
If the resistance is changed to 50 Ohm's, then:
I = 0.12 A
(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.
In the previous example, if the resistance is changed to 5 Ohm's, then:
V = IR
6 = I x 5
I = 1.2 A
(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.
The thermal efficiency (in %) of a system that undergoes a power cycle while receiving 1000 kJ of energy by heat transfer from a hot reservoir at 1000 K and discharging 500 kJ of energy by heat transfer to a cold reservoir at 400 K is:
Answer:
η = 0.5 = 50%
Explanation:
The efficiency of the power cycle is given by the following formula:
[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]
where,
where,
η = efficiency = ?
Q₁ = heat received from hot reservoir = 1000 KJ
Q₂ = heat discharged to cold reservoir = 500 KJ
Therefore,
[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]
η = 0.5 = 50%
The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time
Answer:
0.28 J
Explanation:
Let the mass of the object is 0.5 kg
The maximum velocity of the object is 1.06 m/s.
We need to find the kinetic energy at that time. It is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]
So, the required kinetic energy is equal to 0.28 J.
Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.
Answer:
1) Law of Universal Gravitation Gravity is an attractive force
5) General relativity Gravity is due to the curvature of spacetime
Explanation:
In this exercise you are asked to relate the correct theory and its explanation
Theory Explanation
1) Law of Universal Gravitation Gravity is an attractive force
2) Law of universal gravitation Mass and distance affect force
3) Classical mechanics time and space are absolute
4) Special relativity Time and space are relative
5) General relativity Gravity is due to the curvature of
spacetime
6) General relativity Mass affects the curvature of space - time
Answer:
Explanation:
edge2022
The human eye can readily detect wavelengths from about 400 nm to 700 nm. Part A If white light illuminates a diffraction grating having 910 lines/mm , over what range of angles does the visible m
Answer:
The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".
Explanation:
According to the question,
Slit width,
[tex]d=\frac{1}{910 \ lines/mm}[/tex]
[tex]=\frac{1}{910\times 10^3}[/tex]
[tex]=1.099\times 10^{-6} \ m[/tex]
The condition far first order maxima will be:
⇒ [tex]d Sin \theta = 1 \lambda[/tex]
Now,
⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=21.344^{\circ}[/tex]
⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=39.56^{\circ}[/tex]
HEELLPPPPPpppppppppppppppp
Explanation:
Given:
[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]
[tex]v_1[/tex] = 40 cm/s
[tex]v_2[/tex] = 90 cm/s
[tex]A_2[/tex] = ?
a) The continuity equation is given by
[tex]A_1v_1 = A_2v_2[/tex]
Solving for [tex]A_2[/tex],
[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]
[tex]= 2\:\text{cm}^2[/tex]
b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of
[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]
The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?
a. WA = WB
b. WA = 2WB.
c. WA=4WB
d. WB= 2WA
Answer:
Option (c).
Explanation:
Let the mass of each cart is m and the force is F.
Time for cart A is 2t and for cart B is t.
Work done is given by the
W= force x displacement
As the distance is given by
S= u t +0.5 at^2
So, when the time is doubled the distance is four times.
So, WA = F x 4 S
WB = F x S
WA= 4 WB
Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.
Required:
Rank from smallest to largest.
Answer:
Momentum of object A = Momentum of object C < momentum of B.
Explanation:
The momentum of an object is equal to the product of mass and velocity.
Object A has a mass m and a speed v. Its momentum is :
p = mv
Object B has a mass m/2 and a speed 4v. Its momentum is :
p = (m/2)×4v = 2mv
Object C has a mass 3m and a speed v/3. Its momentum is :
p = (3m)×(v/3) = mv
So,
Momentum of object A = Momentum of object C < momentum of B.
A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is
[tex]\implies F_1 < F_2[/tex]
[tex] \implies F_{net} = F_2 - F1[/tex]
[tex]\implies F_{net} = 42 -15[/tex]
[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]
The net force on the 4.0 kg box is 27 N towards EAST.
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN
Answer:
1.621 kN
Explanation:
Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).
The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).
So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N
So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N
The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)
= √(R² + 0²) (since R' = 0)
= √R²
= R
= 1620.82 N
= 1.62082 kN
≅ 1.621 kN
So, the sum of these two forces on the barge is 1.621 kN
A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper
Answer:
The spring constant of the spring is 10.3 N/m.
Explanation:
Given that,
Mass of a bungee jumper, m = 59 kg
The period of oscillation, T = 0.25 min = 15 sec
We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Where
k is the spring constant
[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]
So, the spring constant of the spring is 10.3 N/m.
A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s. If the object is replaced by one with mass 0.400 kg, what is the period for small amplitude of swing? (a) 1.5 s (b) 3.05 (c) 6.0 s (d) 12.0 s (e) none of the above answers
Answer:
The correct option is (e) "none of the above".
Explanation:
Given that,
A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.
If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.
We know that the time period can be calculated as :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Where
l is the length
g is acceleration due to gravity
It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.
A baseball pitcher throws a fastball by spinning his arm at 27.7m/s. The ball has a mass of 0.700kg and experiences a net centripetal force of 625N. How long is the pitchers arm (the radius of the curve)?
In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.
In a physics lab, light with a wavelength of 560 nm travels in air from a laser to a photocell in a time of 17.2 ns . When a slab of glass with a thickness of 0.810 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 20.8 ns to travel from the laser to the photocell.
Required:
What is the wavelength of the light in the glass?
Answer:
Distance traveled = 3 * 10E8 * 17.2 * 10E-9 = 5.16 m
.81 / 3 * 10E8 = 2.7 * 10E-9 normal time thru glass
(20.8 - 17.2) * E10-9 = 3.6 * 10E-9 additional time due to glass
c tg = c n ta where tg and ta are the times spent in glass and air
(Note you can also write Va = n Vg or D / ta = n D / tg)
n = tg / ta = 3.6 / 2.7 = 1.33 the index of refraction of the glass
Wavelength (air) = Wavelength (glass) * n
Wavelenght = 560 nm / 1.33 = 421 nm
