(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
Learn more:
https://brainly.com/question/13033911
you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you or the friends you left behind back on earth
Your friend would have a better experience of this event, than you .
What is an eclipse?An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,
there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.
For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.
The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.
To learn more about the eclipse from here, refer to the link;
brainly.com/question/4279342
#SPJ2
g A thin uniform film of oil that can be varied in thickness covers a sheet of glass of refractive index 1.52. The refractive index of the oil is 1.64. Light of wavelength 555 nm is shone from air at normal incidence on the film. Starting with no oil on the glass, you gradually increase the thickness of the oil film until the first interference maximum in the reflected light occurs. What is the thickness of the oil film at that instant
Answer:
The right solution is "84.09 nm".
Explanation:
The given values are:
Refractive index of glass,
= 1.52
Refractive index of oil (n),
= 1.64
Wavelength (λ),
= 555 nm
Now,
The thickness of the film (t) will be:
= [tex]\frac{\lambda}{4n}[/tex]
= [tex]\frac{555}{4\times 1.65}[/tex]
= [tex]\frac{555}{6.6}[/tex]
= [tex]84.09 \ nm[/tex]
A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213
Answer:
4√6 rad/s
Explanation:
Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.
So, W = F
mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft
making k the spring constant subject of the formula, we have
k = mg/x
substituting the values of the variables into the equation, we have
k = mg/x
k = 4 lb × 32 ft/s² ÷ 1/3 ft
k = 32 × 4 × 3
k = 384 lbft²/s²
Now, assuming there is no friction and no external force, we have an undamped system.
So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb
So, substituting the values of the variables into the equation, we have
ω = √(k/m)
ω = √(384 lbft²/s² ÷ 4 lb)
ω = √96
ω = √(16 × 6)
ω = √16 × √6
ω = 4√6 rad/s
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.
Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.
Explanation:
Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:
[tex]\omega = 2\pi\cdot f[/tex] (1)
Where [tex]f[/tex] is the frequency, in hertz.
If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:
[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]
[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]
The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:
[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)
Where:
[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.
[tex]g[/tex] - Free-fall acceleration, in meters per square second.
[tex]A[/tex] - Amplitude, in meters.
If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:
[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]
[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]A \approx 0.146\,m[/tex]
The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.
1. A sequence of potential differences v is applied accross a wire (diameter =0.32 mm length = 11 cm and the resulting current I are measured as follows: V 0.1 0.2 0.3 0.4 0.5 I (MA) 72 144 216 288 360 2) a) plot a graph of v against I.
b) determine the wire's resistence , R.
c) State ohm's law and try to relate it . your results.
Answer:
a. Find the graph in the attachment
b. 720 kΩ
c. The ratio V/I gives us our resistance which is 720 kΩ
Explanation:
a) plot a graph of V against I.
To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.
b) Determine the wire's resistance , R.
The resistance of the wire is determined as the gradient of the graph.
R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)
Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA
R = (144 V - 72 V)/(0.2 - 0.1) mA
R = 72 V/0.1 mA
R = 72 V/(0.1 × 10⁻³ A)
R = 720 × 10³ V/A
R = 720 kΩ
c) State ohm's law and try to relate it your results.
Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.
Mathematically, V ∝ I
V = kI
V/I = k = R
Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.
Since V/I is constant, we thus verify Ohm's law.
A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in the capacitor is
Answer:
The energy stored is 1.4 x 10^-9 J.
Explanation:
Side of square, L = 10 cm = 0.1 m
Distance, d = 2 mm = 0.002 m
Electric field, E = 4000 V/m
The energy stored in the capacitor is
[tex]U = 0.5 C V^2[/tex]
The capacitance is given by
[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]
calculate the value of 200°C in Kelvin
Answer:
473.15
Explanation:
basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains
Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
What star is known as the "cold planet"?
Explanation:
OGLE-2005-BLG-390Lb.
PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...
Neptune. Surface Temperature: 72 Kelvin. ...
Uranus. Surface Temperature: 76 Kelvin. ...
Saturn. Surface Temperature: 134 Kelvin. ...
Jupiter. Image Courtesy: NASA. ...
OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
Answer:
K = 2 10⁻⁸ J
Explanation:
Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor
C = Q / ΔV
C = ε₀ A / d
ε₀ A / d = Q / ΔV
Q = ε₀ A ΔV / d (1)
indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,
as the power supply is disconnected and the capacitor is ideal the charge remains constant
in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1
ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]
we substitute the equation for Q
ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]
ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]
in the third part we use the concepts of energy
starting point. Test charge near positive plate
Em₀ = U = q ΔV₂
final point. Test charge near negative plate
Em_f = K
energy is conserved
Em₀ = Em_f
q ΔV₂ = K
K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]
we calculate
K = 1 10⁻⁹ 12 0.005/0.003
K = 2 10⁻⁸ J
1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3
Solution :
1. We know that : Buoyant force = weight of the liquid displace
= volume displaced x density of the fluid
Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]
Mass = weight / g
[tex]$=\frac{940}{9.8}$[/tex]
= 95.92 kg
And density = 1000 [tex]kg/m^3[/tex]
Therefore,
[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]
[tex]$=\frac{95.92}{1000}$[/tex]
= 0.0959 [tex]m^3[/tex]
We know density of air = 1.225 [tex]kg/m^3[/tex]
∴ Mass of air displaced = 0.0959 x 1.225
= 0.1175 kg
Weight of the air displaced = 1.1515 N
Therefore, the buoyant force = 1.1515 N
2). As the balloon is not accelerated, the net force acting on it is zero.
Thus the weight that acts downwards = buoyant force upwards
So, the weight of the air displaced = weight of the balloon
= 17000 N
Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air
[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]
[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]
= 1416.0766 [tex]m^3[/tex]
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.
1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J
We know
[tex]\boxed{\sf E=hv}[/tex]
[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]
[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]
[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]
[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]
Answer:
D!!!!!
Explanation:
Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?
Answer:
[tex]_{18}^{39} } Ar[/tex]
Explanation:
The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.
[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]
Argon is a stable gas and is found in the group 8 on the periodic table of elements.
Answer:
Answer is below
Explanation:
39 18 Ar
If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?
a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.
Answer: They are equal in magnitude.
- The signs of the two changes in potential energy are opposite
Explanation:
When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.
Also, it should be noted that the signs of the two changes in potential energy are opposite.
1. To avoid getting hurt, do not play with sharp simple machines. The highlighted statements best reminds us about ____.
A. Safety measures in handling machines
B. Gentle reminders when playing
C. Benefits of using machines
D. Disadvantages of simple machines
2. Road signs are example of advertisements that play important role in the promotion of road safety. If you are going to make an advertisement, which do you think is appropriate to make part of a road is temporarily closed for repair or renovation?
A. End of 4 lanes
B. No Parking
C. One lane, follow the traffic signal
D. Stop, look and listen
3. Wherever you go, you see mountains, plains and volcanoes. Which among the given ideas is being supported by the highlighted statement?
A. Earth's landmasses before and at present are the same.
B. Earth's land features are not charging
C. Internal forces such as plate tectonic movements do not result to various surface features
D. Internal and external forces contribute to the formation of various surface features of earth
4. Landslides, avalanche, tsunami, seiche and fire are some effects of strong earthquakes. If the huge inland wave produced during an earthquake is known as SEICHE (SAYSH) then what refers to the huge underwater wave produced when an earthquake occurs under the sea?
A. Avalanche
B. Landslide
C. Seiche
D. Tsunami
Answer:
1. A. Safety measures in handling machines
2. C. One lane, follow the traffic signal
3. D. Internal and external forces contribute to the formation of various surface features of the earth.
4. D. Tsunami
Explanation:
1. Safety measures are precautions that must be taken to avoid injuries when handling some potentially dangerous equipment. They are meant to protect us from harm.
2. If a part of the road is to be sealed up temporarily, road users should be alerted that there is just one lane ahead. Therefore, they should follow the traffic signal.
3. The pressure exerted on the earth through phenomena like plate tectonics, earthquakes, etc., causes the eruption and formation of different features on the earth.
4. A tsunami is a giant ocean wave caused by earthquakes and volcanic eruptions that occur under the sea. They occur with a lot of speed and cause an overflow of water onto land.
The exponent of the exponential function contains RC for the given circuit, which is called the time constant. Use the units of R and C to find units of RC. Write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify until you get something simple. Show your work below.
Answer:
The unit of the time constant RC is the second
Explanation:
The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.
Since R = V/I
Unit of R = unit of V/unit of I
Unit of R = V/A
Ω = V/A
Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V
Since C = Q/V
Unit of C = unit of Q/unit of V
Unit of C = C/V
F = C/V
Now the time constant equals RC.
So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A
Also. we know that the 1 Ampere = 1 Coulomb per second
1 A = 1 C/s
So, substituting 1 A in the denominator, we have
unit of RC = C/A = C ÷ C/s = s
So, the unit of RC = s = second
So, the unit of the time constant RC is the second
What type of wave is a microwave?
O heat
O longitudinal
sound
transverse
Answer:
Microwave is a types of a electromagnetic radiation
Answer:
Transvers
Explanation:
Because microwave is electromagnetic waves and all electromagnetic waves are transvers.
Which circuit element is of special importance in AC circuits?
A. Resistor
B. Ammeter
C. Battery
D. Capacitor
Answer:
Explanation:capacitor
Answer:
Ammeter
Explanation:
pls mark me as a brainlist
what is environment in world
Answer:
World Environment Day has been celebrated every year on 5 June, engaging governments, businesses and citizens in an effort to address pressing environmental issues
Explanation:
Answer:
The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial. The term is most often applied to the Earth or some parts of Earth.
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
Find out other examples of bodies showing more than one type of motion Tabulate your findings.
Answer:
down below
Explanation:
Image 1- wheels of train showing both translatory motion as well as rotatory motion.
Image 2- rotation of ball shows both rotatory motion as well as translatory motion.
Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)
Image 4- while cutting wood, the
carpenter's saw has both
translatory motion and oscillatory
motion, as it moves down while
oscillating.
The water evaporated. How did that help ?
Answer:
hellooooo. hi hellooooo and programming laptops
Answer:
The molecules move and vibrate so quickly that they escape into the atmosphere as molecules of water vapor. Evaporation is a very important part of the water cycle. The Heat from the sun, or solar energy, powers the evaporation process. ... Once the water evaporates, it also helps form clouds.
Explanation:
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.
Answer:
The correct answer is "[tex](0,300\times 10^{-3} \ N/C)[/tex]".
Explanation:
The given problem seems to be incomplete. Please find the attachment of the complete query.
According to the question,
At point A, we have
⇒ [tex]E_x = \frac{k q_1}{d_1^2} Cos \theta_1 - \frac{k q_2}{d_2^2} Cos \theta_2[/tex]
or,
⇒ [tex]E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ][/tex]
[tex]=0[/tex]
and,
⇒ [tex]E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2[/tex]
or,
⇒ [tex]E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ][/tex]
[tex]=0.3 \ N/C[/tex]
An object moving with initial velocity 10 m/s is subjected to a uniform acceleration of 8 m/s ^² . The displacement in the next 2 s is: (a) 0m (b) 36 m (c) 16 m (d) 4 m
Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
