The work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.
What is the work done?In Physics, we define the work done as the product of the of force and distance. Hence, we generally define the work done as that which occurs when the force applied moves a distance in the direction of the force. This implies that the work done is not a vector but a scalar quantity.
In relation to the man and the bag, the work done is the product of the force that the man applies and the displacement of the bag. As such, the reason for the tiredness of the man is that the internal energy that he possesses is transferred to moving the bag.
Thus, the work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.
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A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
a)What is the velocity of the space capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut.
b)If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other?
Express your answer to three significant figures and include the appropriate units.
c)What is the kinetic energy of the astronaut after the push in the reference frame?
Express your answer to three significant figures and include the appropriate units.
d)What is the kinetic energy of the capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units.
The change in the speed of the space capsule will be -0.189 m/s.
The average force exerted by each on the other will be 567 N.
The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
Given:Mass of the astronaut, [tex]m_a[/tex] = 126 kg
Speed he acquires, [tex]v_{a}[/tex] = 2.70 m/s
Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg
The initial momentum of the astronaut-capsule system is zero due to rest.
[tex]P_f = m_av_a + m_cv_c[/tex]
[tex]P_I[/tex] = 0
[tex]m_av_a + m_cv_c = 0[/tex]
[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]
[tex]= \frac{126* 2.70}{1800}[/tex]
[tex]= - 0.189[/tex] m/s
Therefore,
According, to the impulse-momentum theorem;
FΔt = ΔP
ΔP = m Δv
ΔP = 126×2.70
= 340.2 kgm/sec
t is time interval = 0.600s
F = ΔP/Δt
F = 340.2/0.600
= 567 N
Therefore, the average force exerted by each on the other will be 567 N.
The Kinetic Energy of the astronaut;
K.E = [tex]\frac{1}{2} m v^2[/tex]
[tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]
= 459.27 J
The Kinetic Energy of the capsule;
K.E = [tex]\frac{1}{2} m v^2[/tex]
= [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]
= 32.14 J
Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
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ASAP NEED HELP !! :(
Why are temperatures more moderate around the fall and spring equinoxes?
C The angle at which Earth's axis tilts changes.
Neither end of Earth's axis is tilted toward the Sun.
The north end of Earth's axis is tilted toward the Sun.
C
The Earth briefly wobbles on its axis.
Answer:C is the answer
Explanation:
It is the most reasonable and the answer that makes most sense
Answer:
The angle at which Earth's axis tilts changes.
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.
1. The vertical axis is the maximum displacement of the air, and the horizontal axis is along the length of the pipe. What is the harmonic number for the mode of oscillation illustrated?
2. The length of the pipe is 0.440 m. What is the pitch (frequency) of the sound? Use 340 m/s for the speed of sound in air.
3. The pipe is now held vertically, and a small amount of water is poured inside the pipe. The first harmonic frequency is then measured to be 251.1 Hz. Measuring from the bottom of the pipe, what is the level of the water?
(1) The harmonic number for the mode of oscillation is 3.
(2) The pitch (frequency) of the sound is 579.55 Hz
(3) The level of the water inside the vertical pipe is 0.1 m.
The harmonic number
The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.
Frequency of the waveThe pitch (frequency) of the sound is calculated from third harmonic formula;
f = 3v/4L
where;
v is speed of soundL is length of the pipef = (3 x 340) / (4 x 0.44)
f = 579.55 Hz
level of the waterwave equation for first harmonic of a closed pipe is given as
f = v/(4L)
251.1 = 340/(4L)
4L = 340/251.1
4L = 1.35
L = 1.35/4
L = 0.34 m
level of water = 0.44 m - 0.34 m = 0.1 m
Thus, the level of the water inside the vertical pipe is 0.1 m.
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A mechanic changing the spark plugs in a car notes that the instruction manual calls for a torque with a magnitude of
39 N · m.
If the mechanic grasps the wrench as shown in the figure below, determine the magnitude (in N) of the force she must exert on the wrench.
______N
The magnitude (in N) of the force she must exert on the wrench is 150.1 N.
Force exerted by the wrench
The force exerted by the wrench is calculated using torque formula as follows;
torque, τ = F x r x sinθ
where;
F is the applied forcer is the perpendicular distance if force appliedF = τ /(r sinθ)
F = (39) / (0.3 sin 60)
F = 150.1 N
Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.
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Circle the anomalous result in the table below. An anomaly is a piece of data that doesn't
fit in with the rest of the data.
1. Concentration of
sucrose (M)
0.25
0.5
1
1.5
2
2. Change in mass of piece of potato (g)
Group B's
Group A's
measurements
0.02
0.05
0.12
0.17
0.21
3. measurements
0.03
0.05
2.3
0.14
0.2
Average?
Pls help I need to get this done really soon otherwise I will get a detention for two hours after school with my least fave teacher plss it’s for a test!!!!!!
Answer:
Explanation:
yes
A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as indicated below. Its center of mass is a distance d=0.300 m from the pivot axis and its moment of inertia about an axis through the CM and parallel to the pivot axis is ICM=9.00×10-3 kg m2.
The plate is released from an angle θ=15°. Calculate the period of the oscillations.
The period of the oscillations.T = 1.2042s
Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.
Typically, the mathematical formula for the moment of inertia is
T = 2 π √(I / mgd)
Therefore, a moment of inertia
I = 9.00×10-3 + md^2 ;
I=9.00*10^{-3}+ 0.5 * 0.3^2
I=0.054
T=2[tex]\pi \sqrt{0.5*9.8*0.3}[/tex]
T=1.2042s
The period of the oscillations.T = 1.2042s
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Consider the ballistic pendulum collision. The projectile, of mass m, is fired into a large block of mass M. (Figure 1)
a) Derive a formula for the fraction of the magnitude of kinetic energy lost
Express your answer in terms of the variables m and M .
b)Evaluate the fraction for m = 18.0 g and M = 380 g .
Express your answer using three significant figures.
The fraction of the magnitude of the kinetic energy lost is [tex]\frac{(change) KE }{KE} = 1 - \frac{m}{m + M }[/tex]. = 0.955.
using the law of conservation of momentum,
[tex]mv=(m+M)V[/tex]
[tex]V= m ( \frac{1}{M+m} ) v\\[/tex]
kinetic energy lost,
Δ[tex]KE=KE_{i} -KE_{f}[/tex]
(see image )
now, for the other part
the fraction of the kinetic energy lost,
ΔKE/ KE = [tex]1- m (\frac{1}{m+M} )[/tex]
ΔKE/ KE = [tex]1 - 18 ( \frac{1}{18 + 380 } )[/tex]
ΔKE/ KE = 0.955.
what is kinetic energy ?
The energy that an object has as a result of motion is known as kinetic energy. It is stated as the effort needed to transfer a bulk body from rest to the given velocity. The body holds onto the kinetic energy it gained during its acceleration until its speed changes.
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The y-position of a damped oscillator as a function of time is shown in the figure.
This function can be described by the y(t) = [tex]A_{0}[/tex][tex]e^{-btx}[/tex]cos(ωt) formula, where [tex]A_{0}[/tex] is the initial amplitude, b is the damping coefficient and ω is the angular frequency.
1. What is the period of the oscillator? Please, notice that the function goes through a grid intersection point.
2. Determine the damping coefficient.
(1) The period of the oscillator is 1 second.
(2) The damping coefficient is 0.93.
What is period of oscillation?
The period of oscillation is the time taken to make one complete cycle.
From the graph, the time taken to make one complete oscillation is 1 second.
Damping coefficientequation of the wave is given as;
y(t) = Ae^(-btx) cos(ωt)
at time, t = 0, y = 3.53.5 = Ae^(-0) cos(0)
3.5 = A x 1
A = 3.5 cm
at time, t = 1 cm, y = - 3cm-3 = 3.5e^(-bx) cos(ω)
-3/3.5 = e^(-bx) cos(ω)
-0.857 = e^(-bx) cos(ω)
-0.857 / cos(ω) = e^(-bx)
ln[-0.857 / cos(ω)] = -bx
ln[-0.857 / cos(ω)] / b = - x ---- (1)
at time, t = 2 cm, y = - 2cm-2 = 3.5e^(-2bx) cos(2ω)
-0.57 = e^(-2bx) cos(2ω)
ln[-0.57 / cos(2ω)] = -2bx
ln[-0.57 / cos(2ω)] /2b = - x ------(2)
solve (1) and (2)
ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b
-0.57 / cos(ω) = 2(-0.857 / cos(ω))
2(-0.857/cosω) = -0.57/cos2ω
-(2 x 0.857) / (-0.57) = cosω/cos 2ω
3 = cosω/cos 2ω
3(cos 2ω) = cosω
3(2cos²ω - 1) = cos ω
6cos²ω - 6 = cosω
6cos²ω - cosω - 6 = 0
let cosω = y
6y² - y - 6 = 0
solve the quadratic equation;
y = 1.1 or -0.92
cosω = -0.92
ω = arc cos(-0.92)
ω = 2.74 rad/s
From equation (1)
ln[-0.857 / cos(ω)] / x = -b ---- (1)
let x = 1
ln(-0.857/cos(2.74) = -b
-0.93 = -b
b = 0.93
Thus, the damping coefficient is 0.93.
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The distance between a carbon atom ( m = 12 u ) and an oxygen atom ( m = 16 u ) in the CO molecule is 1.13 × 10-10 m .
How far from the carbon atom is the center of mass of the molecule?
Express your answer to two significant figures and include the appropriate units.
6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.
The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.
The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms, [tex]x_{c}[/tex] and [tex]x_{o}[/tex] their distances relative to C atom.
Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )
[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]
[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m
Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.
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Write a properly formatted hypothesis statement to answer this question: How does the amount of salt added to ice affect the rate at which the ice will melt?
Specify how you plan to change the independent variable by using terms such as increase or decrease. Also, specify how the dependent variable will change in response by using terms such as increase, decrease, or stays the same.
Criteria pts
Correct placement of IV 5
Correct placement of DV 5
If, then format 5
IV indicates either "increases" or "decreases" 5
DV indicates either "increases", "decreases", or "stays the same" 5
The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
The independent variable which is can be changed by increasing the rate of salt added to the equation.
The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.
What is the effect of salt on the melting temperature of ice?Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.
Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.
If the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.
Hence, the hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
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Compute your average velocity in the following two cases: (a) You
walk 73.2 m at a speed of 1.22 m/s and then run 13.2 m at a speed
of 3.02 m/s along a straight track. (b) You walk for 1.00 min at a
speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a
straight track. (c) Graph x versus t for both cases and indicate how
the average velocity is found on the graph.
(a) Walking 73.2 m at 1.22 m/s would take
[tex]\dfrac{73.2\,\rm m}{1.22 \frac{\rm m}{\rm s}} = 60 \,\rm s[/tex]
and running 13.2 m at 3.02 m/s would take
[tex]\dfrac{13.2\,\rm m}{3.02\frac{\rm m}{\rm s}} \approx 4.37\,\rm s[/tex]
You've undergone a total displacement of 73.2 + 13.2 = 86.4 m in a matter of approximatly 64.37 s, so your average velocity is
[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{86.4\,\mathrm m}{64.37\,\rm s} \approx \boxed{1.34\dfrac{\rm m}{\rm s}}[/tex]
(b) In the first 1.00 min = 60 s, you undergo a displacement of
[tex](60\,\mathrm s) \left(1.22 \dfrac{\rm m}{\rm s}\right) = 73.2 \,\rm m[/tex]
and in the second minute, you undergo a displacement of
[tex](60\,\mathrm s) \left(3.05\dfrac{\rm m}{\rm s}\right) = 183 \,\rm m[/tex]
Your total displacement is then 73.2 + 183 = 256.2 m in a matter of 2.00 min = 120 s, so your average velocity is
[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{256.2\,\mathrm m}{120\,\rm s} \approx \boxed{2.14\dfrac{\rm m}{\rm s}}[/tex]
(c) For part (a), your displacement [tex]x(t)[/tex] (in m) at time [tex]t[/tex] (in s) is given by
[tex]x(t) = \begin{cases}1.22t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.02 (t-60) & \text{for } t > 60\end{cases}[/tex]
and for part (b), your displacement is given by the very similar
[tex]x(t) = \begin{cases}1.22 t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.05(t-60) & \text{for } t > 60 \end{cases}[/tex]
See the attached plots. The average velocity for the given situation is the slope of the dotted line.
(refer to photos attached. Example of previous question with wrong/correct answers example, and current question needing to be solved)
Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C
If a charge of −3.94 µC is placed at this point, what are the magnitude and direction of the force on it?
Magnitude _______N
Direction?
- toward the left
- upward
-downward
- toward the right
(a) The electric field strength at a point 1.00 cm to the left of the middle is 2.0 x 10⁷ N/C.
(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.
Electric field strengthThe electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first chargeE1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second chargeE2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third chargeE3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
Net electric fieldE = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)
E = +2.0 x 10⁷ N/C
Force on the charge −4.72 µCF = Eq
F = 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = -94.4 N
Thus, the direction of the force will be towards the left.
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formula for conductivity
Answer:
Thermal conductivity formula
k is the thermal conductivity (Wm -1 K -1)Q is the amount of heat transferred through the material (Js -1)A is the area of the body (m 2)ΔT is the temperature difference (K)Explanation:
The mass of a coin is measured to be 12.5±0.1 g. The diameter is 2.8±0.1 cm and the thickness 2.1 ±0.1 mm. Calculate the average density of the material from which the coin is made with its uncertainty. Give your answer in kg m-³.
with proper explanation
answer only if u know or else u know the consequences
The average density of the material from which the coin is made is 9.67 g/cm³.
Volume of the coinThe volume of the coin at the given diameter is calculated as follows;
V = Ah
where;
A is area of the coinh is the thickness of the coinV = πd²/4 x h
V = π(2.8)²/4 x (0.21 cm)
V = 1.293 cm³
average density of the coinThe average density of the material from which the coin is made is calculated as follows;
density = mass/volume
density = 12.5 g / (1.293 cm³)
density = 9.67 g/cm³
Thus, the average density of the material from which the coin is made is 9.67 g/cm³.
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If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
Answer:
The binding energy of sodium Na=5.407791×10⁹JExplanation:
Greetings !Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.
Formula : Eb=(Δm)c²where:Eb= binding energy .Δm= mass defect(kg) c= speed of light 3.00×10⁸ms¯¹Given valuesm= 18.02597c=3.00×10⁸ms¯¹required valueEb=?Solution:Eb=(Δm)c²Eb=(18.02597)*(3.00*10⁸ms¯¹Eb=5.407791*10⁹J*photo attached* The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.67 m and 1.01 m, respectively. The respective rotational speeds are 444 rev/min and 4,130 rev/min. Calculate the speeds of the tips of both rotors.
main rotor ______m/s
tail rotor _______m/s
Compare these speeds with the speed of sound, 343 m/s.
vmain rotor = _______ vsound
vtail rotor = _______ vsound
(a) The speeds of the tips of both rotors; main rotor 178.3 m/s and tail rotor 218.4 m/s.
(b) The speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
Linear speed of main motor and tail rotorv = ωr
where;
ω is the angular speed (rad/s)r is radius (m)v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)
v(main rotor) = 178.3 m/s
v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)
v(tail rotor) = 218.4 m/s
Speed of the rotors with respect to speed of sound% speed (main motor) = 178.3/343 = 0.52 = 52 %
% speed (tail motor) = 218.4/343 = 0.64 = 64 %
Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
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Summarize the step you might use to Carry out an investigation using scientific method
The summary of the step you might use to carry out an investigation using scientific method include:
ObservationHypothesisExperimentConclusionWhat is Scientific method?This is defined as a systematic approach which helps to predict how several elements in the universe works.
This usually starts with observing a phenomenon and making a hypothesis. This is usually followed by conducting experiment to ascertain its authenticity before a conclusion is drawn.
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4. Four objects are situated along the y axis as follows: a 2.00kg object is at +3.00m, a 3.00kg object is at +2.50m, a 2.50kg object is at the origin, and a 4.00kg object is at -0.500m. Where is the center of mass of these objects?
Answer: 1.348 meters
Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point c . Adding the four moments together, we get:
(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0
6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0
11.5c=15.50
c= 1.348 metres
The center of mass is on the y axis at y = 1.348 metres.
Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.52 cm. If the potential difference across the plates was 24.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
________V/m
The electric filed in V/m is 1.58 * 10^6 V/m
What is the electric field?We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.
Thus;
E = V/d
E = electric field
V = voltage
d = distance of separation
E = 24 * 10^3 V/1.52 * 10^-2 m
E = 1.58 * 10^6 V/m
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the earth orbits is oval in shape.explain how the mangnitude of gravitational between the earth and sun changes as the eart moves from position a to b
As the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.
What is gravitational force?
Gravitational force is a force that attracts any two objects with mass.
According to Newton's law of universal gravitation, the force of attraction between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects.
F = Gm₁m₂/R²
where;
m₁ is mass of Earthm₂ is mass of sunR is the distance between the Earth and SunThus, as the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.
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what the answer???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
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a man carries a hand bag by hanging on his hand moves horizantaly wher the bag does not up or down what is the work done on the bag
Since the displacement is completely perpendicular to the direction of the applied force, the work done on the bag is zero.
When is the Work done on an object ?The work is done on an object when the force applied is multiply by the distance moved by the object in the direction of the force applied.
Given that a man carries a hand bag by hanging on his hand moves horizontally where the bag does not up or down.
What is work if the displacement is not in the direction of force ?
The work done can only be zero if the displacement is perpendicular to the direction of force. otherwise, it will not be equal to zero.
Also, the work done will be zero, if the displacement is zero.
In the question above, the displacement is completely perpendicular to the direction of the applied force.
Therefore, the work done on the bag is zero.
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What is an example of total internal reflection at work?
A.
A ray of light has the same intensity both entering and exiting a fiber optic cable.
B.
A ray of light entering a glass cube gets refracted.
C.
A ray of light in air hits a shiny surface and bounces off.
D.
A ray of light entering a ruby gets refracted.
Answer:
A i think...
Explanation:
Sorry if its wrong
A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .
If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Express your answer to two significant figures and include the appropriate units.
The body moves at a velocity of 1.62m/s after the bullet emerges.
Given:Mass of bullet, [tex]m_1[/tex] = 22g
= 0.022 kg
Mass of the block, [tex]m_2[/tex] = 1.9 kg
Velocity of bullet , [tex]v_1[/tex] = 265 m/s
[tex]v_2 = 0[/tex]
According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.
After penetration;
[tex]v^{'}_1 =125 m/s[/tex]
[tex]v^{'}_2=?[/tex]
The formula for calculating the collision of a body is expressed as:
p = mv
m is the mass of the body
v is the velocity of the body
∴ Momentum before = Momentum after
Substitute the given parameters into the formula as shown:
[tex]m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s[/tex]
Therefore, It moves with a velocity of 1.62 m/s.
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*According to Bohr's Theory, what is the maximum number of electron orbital layers?*
Four are known: s, p, d, and f.
Thank you,
Eddie
A centrifuge in a medical laboratory rotates at an angular speed of 3,750 rev/min. When switched off, it rotates through 48.0 revolutions before coming to rest. Find the constant angular acceleration (in rad/s²) of the centrifuge.
______rad/s²
The constant angular acceleration (in rad/s²) of the centrifuge is 255.66 rad/s².
Constant angular acceleration of the centrifuge
The constant angular acceleration of the centrifuge is calculated as follows;
ωf² = ωi² - 2αθ
where;
ωf is the final angular velocity ωi is initial angular velocityθ is angular displacementα is angular accelerationWhen the centrifuge is switched off, the final angular velocity = 0
Initial angular velocity: ωi = 3,750 rev/min x 2π rad/rev x 1 min/60 s = 392.7 rad/s
angular displacement: θ = 48 rev = 48 rev x 2π rad/rev = 301.6 rad
0 = ωi² - 2αθ
2αθ = ωi²
α = ωi²/2θ
α = (392.7²) / (2 x 301.6)
α = 255.66 rad/s²
Thus, the constant angular acceleration (in rad/s²) of the centrifuge is 255.66 rad/s².
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On space missions with international teams, every country involved wants the flight
director to be chosen from their country because their national space organization
receives a significant amount of funding if they lead the mission.
True
False
False, their national space organization receives a significant amount of funding if they lead the mission.
The team typically sends about 30,000 remote control commands from the ground to the space station to manage the platform. Highly interactive, hands-on, tactile, and highly rewarding.
A mission control center (MCC, sometimes called a flight control center or operations center) is a facility that typically manages spaceflight from the launch point to landing or the end of the mission. This is part of the ground segment of spacecraft operations.
The station serves as a microgravity and space laboratory where scientific research in fields such as astrobiology, astronomy, meteorology, and physics is conducted.
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Three bees are oriented as shown in the figure. B1 has +17 µC of charge, B2 has −5 µC of charge, and B3 has +26 µC of charge.
B1 to b2 5 cm
B2 to b3 10 com
Bee number 3 is stationed at the observation location for this problem, and we want to find the net electric field at B3. We'll do this in a few steps.
(a) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B1? Remember that the components can be positive or negative depending on their directions along the x or y axis.
E1x =
E1y =
(b) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B2? Again, remember that the components can be positive or negative depending on their directions along the x or y axis.
E2x = N/C
E2y =
(c) What are the magnitude and direction of the net electric field at the observation location where B3 is resting?
magnitude
N/C
direction
° counterclockwise from the +x-axis
(d) What is the magnitude of the force on B3 due to this net electric field?
Given the following data:
Charge of B₁ = +17 µC to C = 17 × 10⁻⁶ C.Charge of B₂ = -5 µC to C = -5 × 10⁻⁶ C.Charge of B₃ = +26 µC to C = 26 × 10⁻⁶ C.Radius of B₁ to B₂ = 5 cm to m = 0.05 meter.Radius of B₂ to B₃ = 10 cm to m = 0.1 meter.How to determine the x and y components of the electric field?First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.
The electric field due to B₁ to B₃ is given by:
E₁ = kq₁/r₁²
E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)
E₁ = 153 × 10³/0.0125
Electric field, E₁ = 12.24 × 10⁶ N/C.
The electric field due to B₂ to B₃ is given by:
E₂ = kq₂/r₂²
E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)
E₂ = 45 × 10³/0.01
Electric field, E₂ = 4.5 × 10⁶ N/C.
Also, the magnitude of the angle formed is given by tan trigonometry:
Tanθ = 5/10
Tanθ = 0.5
θ = tan⁻¹(0.5)
θ = 26.56°.
Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁x = E₁cosθ - E₂
E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶
E₁x = 10.95 × 10⁶ - 4.5 × 10⁶
E₁x = 6.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁y = -E₁sinθ
E₁y = -12.24 × 10⁶ × sin(26.56)
E₁y = -12.24 × 10⁶ × 0.4471
E₁y = -5.5 × 10⁶ N/C.
Also, the magnitude of the electric field is given by:
Exy = √(E₁x² + E₁y²)
Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)
Exy = √4.225 × 10¹³ + 3.025 × 10¹³)
Exy = √(7.25 × 10¹³)
Exy = 8.5 × 10⁶ N/C.
Part B.We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:
E₂x = -E₂sinθ
E₂x = -4.5 × 10⁶ × sin(90)
E₂x = -4.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₂y = E₂cosθ
E₂y = -4.5 × 10⁶ × cos(90)
E₂y = 0 N/C.
How to calculate the net electric field at the observation location?The magnitude of the net electric field at B₃ is given by:
Eₙ = √(E₁x + E₂x)² + E₁y²)
Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)
Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)
Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)
Eₙ = √(3.65 × 10¹³)
Eₙ = 6.04 × 10⁶ N/C.
For the direction, we have:
Tanθ = x/y
Tanθ = 6.5/-5.5
Tanθ = -1.1818
θ = tan⁻¹(-1.1818)
θ = 49.76°.
Part C.The magnitude of the force on B₃ due to this net electric field is given by:
F = B₃ × Eₙ
F = 26 × 10⁻⁶ × 6.04 × 10⁶
F = 157.04 Newton.
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Which one of the following statements concerning weight and energy balance is most accurate?
A. People generally need the same amount of physical activity to maintain weight stability.
B. Regular physical activity doesn’t impact the percentage of body fat in children and adolescents.
C. It’s possible to achieve weight stability by doing the equivalent of 60–120 minutes a week of moderate-intensity walking.
D. The optimal amount of physical activity needed to maintain weight is unclear.
Answer: D. I took the test and got it right
The correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.
What is energy balance?Energy balance refers to the way in balance is achieved when intake of energy is equal to energy expended.
Energy refers to the impetus behind all motion and all activity. If is also the capacity to do work. Energy is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent.
So therefore, the correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.
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High-power laser in factories are used to cut through cloth and metal. One such laser has a beam diameter of 1.00mm and generates an electric field having an amplitude 0.800MV/m at the target. Find(a) the amplitude of the magnetic field produced,(b) the intensity of the laser and (c) the power delivered by the laser.
(a) The amplitude of the magnetic field produced is 2.667 mV/m.
(b) The intensity of the laser is 2.832 W/m².
(c) The power delivered by the laser is 2.22 x 10⁻⁶ W.
Amplitude of the magnetic field producedIn electromagnetic waves, the amplitude of magnetic field is the maximum field strength of the magnetic fields.
B₀ = E₀/c
where;
E₀ is the amplitude electric fieldB₀ is the amplitude magnetic fieldc is speed of lightB₀ = (0.8 MV/m) / (3 x 10⁸)
B₀ = (0.8 x 10⁶ V/m) / (3 x 10⁸)
B₀ = 2.667 x 10⁻³ V/m
B₀ = 2.667 mV/m
Intensity of laserThe intensity of the laser is calculated as follows;
I = ¹/₂ε₀E₀²
I = (0.5)(8.85 x 10⁻¹²)(0.8 x 10⁶)²
I = 2.832 W/m²
power delivered by the laserP = IA
where;
A is the area of the beamA = πd²/4
where;
d is diameterA = π(1 x 10⁻³)²/4
A = 7.854 x 10⁻⁷ m²
Power = (2.832 W/m²) x (7.854 x 10⁻⁷ m²)
Power = 2.22 x 10⁻⁶ W
Thus, the amplitude of the magnetic field produced is 2.667 mV/m.
The intensity of the laser is 2.832 W/m².
The power delivered by the laser is 2.22 x 10⁻⁶ W.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres is 46622.72N
Calculation and Parameters( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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