Answer:
Different surfaces have different impact force during collision which depends on the time it takes the person to come to rest after collision.
Explanation:
Given;
speed on concrete = 12 m/s (27 mi/h)
speed on soil = 15 m/s (34 mi/h)
speed on water = 34 m/s (76 mi/h)
The impact force on this person during collision is rate of change of momentum;
[tex]F = \frac{\delta P}{\delta t}[/tex]
During collision, the force exerted on this person depends on how long the collision lasts; that is, how long it takes for this person to come to rest after collision with each of the surfaces.
The longer the time of collision, the smaller the force exerted by each.
It takes shorter time for the person to come to rest on concrete surface than on soil surface, also it takes shorter time for the person to come to rest on soil surface than on water surface.
As a result of the reason above, the force exerted on the person during collision by the concrete surface is greater than that of soil surface which is greater than that of water surface.
You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = [tex]\frac{W}{F}[/tex]
d = [tex]\frac{3}{9}[/tex]
d = 0.33 m
The distance d the block ice moved is 33 cm.
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
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A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
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What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Answer:
The highest rms voltage will be 8.485 V
Explanation:
For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load
If the peak or maximum voltage should not exceed 12 V, then from the relationship
[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]
where [tex]V_{rms}[/tex] is the rms voltage
[tex]V_{p}[/tex] is the peak or maximum voltage
substituting values into the equation, we'll have
[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration
Answer:
The average velocity for this trip is 0 km/hr
Explanation:
We know that average velocity = total displacement/total time.
Now, its displacement is d = final position - initial position.
Since the car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.
So, its displacement is d = 0 km - 0 km = 0 km.
Since the total time for the round trip is 2 hours, the average velocity is
total displacement/ total time = 0 km/2 hr = 0 km/hr.
So the average velocity for this trip is 0 km/hr
A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?
a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.
Answer:
the greatest intensity is obtained from c
Explanation:
An electromagnetic wave stagnant by the expression
E = E₀ sin (kx -wt)
when two waves meet their electric fields add up
E_total = E₁ + E₂
the intensity is
I = E_total . E_total
I = E₁² + E₂² + 2E₁ E₂ cos θ
where θ is the phase angle between the two rays
Let's examine the two waves
in this case E₁ = E₂ = E₀
I = Eo2 + Eo2 + 2 E₀ E₀ coasts
I = E₀² (2 + 2 cos θ )
I = 2 I₀ (1 + cos θ )
let's apply this expression to different cases
a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀
b) cos π = -1 this implies that I_total = 0
c) the cosine is 1,
I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ
I = E₀² (5 +4 cos θ)
I = E₀² 9
I = 9 Io
d) in this case the cos pi = -1
I = E₀² (5 -4)
I = I₀
e) we rewrite the equation
I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ
I = Eo2 (10 + 6 cos θ)
cos π = -1
I = E₀² (10-6)
I = 4 I₀
the greatest intensity is obtained from c
The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.
What is an amplitude?An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.
In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.
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A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.
Required:
a. How much work is done by the tension?
b. How much work is done by gravity?
The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
The given parameters;
mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 NThe work done by the tension is calculated as follows;
W = Fd
W = 15 x 0.15
W = 2.25 J
The work done by gravity is calculated as;
W = (25 x 9.8) x 0.15
W = 36.75 J
Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
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Three point charges (some positive and some negative) are fixed to the corners of the same square in various ways, as the drawings show. Each charge, no matter what its algebraic sign, has the same magnitude. In which arrangement (if any) does the net electric field at the center of the square have the greatest magnitude?
Answer:
The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.
An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
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A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
please brainliest!!!
Explanation:
V1/√T1 =V2/√T2
V1 = 331m/s
T1 = 0°C = 273k
V2 = ?
T2 = 35°c = 308k
331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/s21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)
Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?
Answer:
Explanation:
Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction , so their relative velocity
= u + v = 2.3 m /s ( given )
We shall apply conservation of momentum law for the movement of astronaut and metal plate
mu = M v where m is mass of astronaut , M is mass of metal plate
71 u = 230 x v
71 ( 2.3 - v ) = 230 v
163.3 = 301 v
v = .54 m / s
u = 1.76 m / s
honeycomb will be at rest because honeycomb surface is frictionless . Plate will slip over it . Over plate astronaut is walking .
a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s
b ) velocity of astronaut relative to honeycomb will be + .54 m /s
Here + ve direction is assumed to be the direction of astronaut .
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.
Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface
Answer:
Explanation:
jjjjjjjjjjjjjjjj
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?
Answer:
The glider’s speed after the skydiver lets go is 26 m/s
Explanation:
To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum
Mathematically;
mv = mv + mv
so 680 * 26 = (680-60)v + 60 * 26
17680 = 620v + 1560
17680-1560 = 620v
16120 = 620v
v = 16120/620
v = 26 m/s