Answer:
The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.
[tex]s\frac{d}{t}[/tex]
Explanation:
hope this helps you my friend
K donates, or transfers, one electron to bromine, which has 7 electrons. Both K and Br are now stable with 8 electrons. K becomes a positive ion and Br becomes a negative ion. The positive K ion and the negative Br ion attract each other to form an ionic bond.
K is cation by losing of electron whereas Br is anion due to accepting of electrons.
Is charge appears when an atom lose or accept electron?Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.
So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.
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A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glass when it warms up, in such a way that the volume increases by 0.6% (that is, ΔV/V0 = 6 ✕ 10-3) relative to the space available. Calculate the force exerted by the moonshine per square centimeter if the bulk modulus is 2.1 ✕ 109 N/m2, assuming the jar does not break.
The force exerted per square centimeter is 126 N/cm².
What is pressure?Pressure is the force acting per unit area.
Pressure = force/areaBased on the data given:
volume increase, ΔV/V0 = 6 * 10⁻³
Bulk Modulus, B = 2.1 * 10⁹ N/m²
Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:
B = ΔP/ [(ΔV/V)]Solving for ΔP;
ΔP = B * [(ΔV/V)]
ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)
ΔP = 1.26 * 10⁶ N/m²
Converting to per square centimeter
ΔP = (1.26 * 10⁶ N/m²)/10⁴
ΔP = 126 N/cm²
In conclusion, the force exerted per square centimeter is a measure of the pressure.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Tension in the cable
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, and thus the magnetic field it produces. The second coil is connected only to an ammeter. The ammeter will indicate that a current is flowing in the second coil:.
The ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.
Induced emf
An emf is induced in a coil placed in a magnetic field when a current carrying conductor moves in the field.
emf = NdФ/dt
where;
dФ is change in flux of the fieldN is number of turnsdt is change in timeThus, the ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.
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An Airbus A380-800 passanger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? The average density of the air around the airplane is ρ^air = 1.21 kg/m^3.
The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.
To find the answer, we need to know about the Bernoulli's principle.
How to find the mass of the airplane?The Bernoulli's principle states that, the sum of pressure energy, kinetic energy and potential energy of an incompressible, non-viscous, fluid in streamlined flow is a constant.It can be expressed as,P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.
Instead of ρ we take d as density.
We have given that,[tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]
We equate the principle for lower and upper surfaces of the wing like,[tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]
Thus, the mass of the airplane from the above equation will be,[tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]
Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.
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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.
We substitute d for to represent density.
We've done that,[tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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Drag each label to the correct location on the image. Jessica is visiting a park with her mother. Jessica sits on a swing. Her mother pulls the swing to a height of 3 meters above the ground and lets it go. The image shows Jessica at three positions on the swing. Jessica‘s mass is 44 kilograms and the maximum velocity of the swing is 5 meters/second. What’s the energy she has at each position shown? Ignore friction and air resistance. Use g = 9.8 m/s2, PE = m × g × h, and KE = 0 joules KE = 550 joules PE = 862.4 joules
The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JWhat does velocity implies?Velocity is known to be a term that connote the direction of any kind of a moving body or an object.
Note that the Speed is known to be a a scalar quantity and as such, Velocity is said to be a vector quantity.
Note also that from the question given, Jessica's height of the swing 3 meters above the ground, therefore:
Jessica Position at maximum height :
PE = mgh
PE = 44kg x 9.8m/s² x 3
PE = 1,294 J
Jessica Position at minimum height:
PE = 0 J
Jessica Position at maximum velocity:
KE = 1/2 x mv²
KE = 1/2 x 44kg x (5m/s²)²
KE = 550 J
Jessica Position at minimum velocity:
KE = 0 J
Therefore, The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JLearn more about velocity from
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Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car at an angle of 5.23 ∘ relative to the vertical.
A) Find the tension in the string holding the tassel.
B) At what angle to the vertical will the tension in the string be twice the weight of the tassel?
The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 NФ = 34.4 °What is the tension in the string holding the tassel. ?Generally, the equation for Tension is mathematically given as
[tex]TCos\theta = mg[/tex]
Therefore
[tex]TCos6.58^{o} = 19.8*10^{-3}*9.8[/tex]
T = 0.1953 N
b).
Where
[tex]T* sin \theta = ma[/tex]
[tex]0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a[/tex]
a = 1.13 m/s^2
In conclusion
T* sinФ = ma
2msinФ = ma
2sinФ = a
[tex]sin\theta = \frac{a}{2}[/tex]
[tex]\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}[/tex]
Ф = 34.4 °
In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 N
Ф = 34.4 °
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A 1400-kg sports car (including the driver) crosses the rounded top of a hill (radius = 87.0 m ) at 18.0 m/s .
a) Determine the normal force exerted by the road on the car.
b)Determine the normal force exerted by the car on the 65.0- kg driver.
c)Determine the car speed at which the normal force on the driver equals zero.
(a) The normal force exerted by the road on the car is 8,506.2 N.
(b) The normal force exerted by the car on the driver is 394.9 N.
(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.
Normal force exerted by the road on the carThe normal force exerted by the road on the car is calculated as follows;
Normal force = weight of the car - centripetal force of car
Weight of the car = (1400 x 9.8) = 13,720 N
Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N
Normal force = 13,720 N - 5,213.8 N
Normal force = 8,506.2 N
Normal force exerted on the driverNormal force = weight of driver - centripetal force of driver
Weight of driver = (65 x 9.8) = 637 N
Centripetal force of driver = (65² x 18²)/(87) = 242.07 N
Normal force = 637 N - 242.07 N = 394.9 N
Speed at which normal force on driver is zeroN = mg - mv²/r
0 = mg - mv²/r
mv²/r = mg
v²/r = g
v² = rg
v = √rg
v = √(87 x 9.8)
v = 29.2 m/s
Thus, the normal force exerted by the road on the car is 8,506.2 N.
The normal force exerted by the car on the driver is 394.9 N.
The speed of the car at which the normal force on driver is zero is 29.2 m/s.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
26.10 N is the vertical component of the force.
Rx represents the Horizontal component of force
Ry represents The Vertical component of force
According to the given diagram
Rx - Tcosθ = 0
Rx = Tcosθ
And,
Ry + Tsinθ = mg
Ry = mg - Tsinθ
The horizontal component of force =The Vertical component of force
Rx = Ry
Tcosθ = mg - Tsinθ
T(cosθ + sinθ) = 29 × 9.8 = 284.2 N
T√2 cosθ = 284.2 N
T × √2 ×0.544 = 284.2 N
T × 0.769 = 284.2 N
T = 370 N (app)
So,
Ry = 284.2 - 370 (sin 57°)
= 284.2 - 310.3 = -26.10 N
Hence, 26.10 N is the vertical component of the force exerted.
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A 1300-kg car moving on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.3 m .
What is the spring stiffness constant of the spring?
Express your answer to two significant figures and include the appropriate units.
The stiffness constant of the spring is 68,290.3 N/m
Stiffness constant of the spring
Apply the principle of conservation of energy;
U = K.E
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
where;
m is massv is speed = 60 km/h = 16.67 m/sx is the distancek = (1300 x 16.67²)/(2.3²)
k = 68,290.3 N/m
Thus, the stiffness constant of the spring is 68,290.3 N/m.
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Why should you use an iron hammer instead of a wooden
hammer?
Answer:
more durability and easier to pound a hammer with the heavy weight.
Explanation:
One should use an iron hammer instead of the wooden hammer because the force developed with the help of the iron hammer is far greater than the wooden hammer
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
The mathematical expression for Newton's second law is as follows
F = m*a
where F represents the force applied
m is the mass of the object
a is the acceleration of the object
As we know that force is a product of mass and acceleration, let us assume that both iron and wooden hammers have the same dimension , considering the acceleration applied is the same while hammering ,
As for the same dimension the mass of iron hammer would be greater as result force applied through the iron hammer would be grater
Since the iron hammer's force is so much stronger than the wooden hammer's force, one should use an iron hammer instead of a wooden hammer.
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Calculate the magnitude of the electric field at one corner of a square 1.10 m on a side if the other three corners are occupied by 4.05×10−6 C charges. What is the direction of the electric field?
The direction of magnetic field is south-east and the magnitude is
[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].
Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.
magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.
To find the magnitude of the electric field use the formula
[tex]E = kq/ r^{2}[/tex]
inserting the values,
[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]
[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]
[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C
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If the velocity of an object is 9 m/s and its momentum is 72 kgm/s, what is its mass
An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.
What is momentum?In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.
It is a vector quantity, possessing a magnitude and a direction.
The mathematical expression for momentum is:
p = m . v
where,
p is the linear momentum of the object.m is the mass of the object.v is the velocity of the object.An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.
p = m . v
m = p / v
m = (72 kg.m/s) / (9 m/s) = 8 kg
An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.
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A 40.0-kg child running at 8.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?
The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
What is principle of conservation of angular momentum?
The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.
Li = Lf
Li = Ii ωi + Ic ωc
Li = Iiωi + MR²(V/R)
Li = Iiωi + MRV ----- (1)
Angular speed of the merry go round after the child jumps on itLf = If ωf
ωf = Lf/If
If = Im + MR²
ωf = Lf / ( Im + MR²)
Recall, Lf = Li
ωf = (Iiωi + MRV) / ( Im + MR²)
where;
Iiωi is the initial angular momentum of the merry - go round = 0M is mass of the childR radius of rotationV is tangential speed of the childIm is the moment of inertia of the merry go roundωf = (Iiωi + MRV) / ( Im + MR²)
ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)
ωf = (480) / (690)
ωf = 0.696 rad/s
Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
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A 69-kg base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
J
(b) How far does he slide?
m
The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.
we know:-
Mass = 69 kg
Speed = 3.2 m/s
Coefficient of friction ( ratio of friction force and normal force ) = 0.70
Acceleration due to gravity, g = 9.8 m/s^2
(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:
[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]
[tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]
[tex]= 353.28[/tex] Joules
Mechanical energy = 353.28 Joules
(b) To determine how far (distance) the runner slide:
acceleration = ug
[tex]= 3.2[/tex] × [tex]9.8[/tex]
[tex]= 31.36[/tex] [tex]m/s^{2}[/tex]
distance ,
[tex]V^{2} = U^{2} + 2aS[/tex]
[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]
[tex]10.24 = 62.72 S[/tex]
[tex]S = {\frac{10.24}{62.72} }[/tex]
Distance, S = 0.16 m
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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.
The horizontal distance traveled by the venom before it hits the ground is 0.6 m
What is Projectile ?A stone or ball or anything projected is known as a projectile.
Given that a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon.
Assuming air resistance is neglected, the parameters to be considered are;
Height h = 0.42 mVelocity v = 2.5 m/sAngle Ф = 35°The ball we reach the ground at the same time when it is dropped vertically or horizontally.
In a vertical direction,
h = 1/2g[tex]t^{2}[/tex]
Substitute all the parameters into the formula
0.42 = 1/2 x 9.8 x [tex]t^{2}[/tex]
0.42 = 4.9[tex]t^{2}[/tex]
[tex]t^{2}[/tex] = 0.42/4.9
[tex]t^{2}[/tex] = 0.0857
t = [tex]\sqrt{0.0857}[/tex]
t = 0.29 s
The horizontal distance traveled by the venom before it hits the ground will be
Distance = ucosФ x t
Distance = 2.5cos35 x 0.29
Distance = 0.599 m
Distance = 0.6 m
Therefore, the horizontal distance (in m) traveled by the venom before it hits the ground is 0.6 m
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10. When the force applied on an object is doubled, how does the pressure exert on the object change?
Answer:
It doubles as well.
Explanation:
Assuming that the area on which the force acts remains constant, and knowing that by definition:
P = F/A
We see that if we have a double force:
p = 2 F/A
Then the pressure is doubled.
Answer:
pressure doubles if area is constant
Explanation:
we know,
p=F/A
when force is double by keeping area constant
F=2F
Then,
Change in pressure (p') will be
P'=2F/A
or,p'=2×F/A
or,p'=2p(since, P=F/A)
therefore, when force is double by keeping area constant the pressure will increase by 2 times
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Avogadro‘s number was calculated by determining The number of atoms in
Answer:
12 grams of the isotope carbon-12.
Explanation:
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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b).
By what angle has planet Y rotated through during this time?
Planet Y has rotated by 135.5° through during this time.
To find the answer, we need to know about the relation between angle and radius of orbit.
What's the expression of angle in terms of radius?Angle= arc/radiusAs arc = orbital velocity × time,angle= (orbital velocity × time)/radius
Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sunSo, angle = (√(GM)× time)/radius^3/2What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet XAs time = 5 years ( a constant)Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂= (4/3)^(3/2) × 88°
= 135.5°
Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.
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A first year student projected a farm business brochure to a farmer at 30 degrees to horizontal. calculate the maximum height attained by the projectile if it was launched at 400m/s
Answer:
Maximum height = 2040 m
Explanation:
We can solve the problem using kinematics.
Consider the vertical motion of the object and use the equation:
[tex]\boxed{v^2 = u^2 + 2as}[/tex]
where:
• v = final velocity (0 m/s, because when the object is at max. height, it has no vertical velocity)
• u = initial velocity (400sin30° m/s ⇒ vertical component of 400 m/s at 30° to horizontal)
• a = acceleration (-9.81 m/s²; considering upward acceleration to be negative)
• s = displacement (? m; this represents the max. height of the object),
Substitute the values into the equation and solve for s :
[tex]0^2 = (400 sin (30 \textdegree))^2 + 2(-9.81)(s)[/tex]
⇒ [tex]2(9.81)(s) = (400 sin (30 \textdegree))^2[/tex]
⇒ [tex]s = 2040 \space\ m[/tex] (3 s.f.)
Four canisters contain helium gas.
If all the canisters had the same amount of particles, which canister would have the fastest moving particles?
W
X
Y
Z
D. The canister with the fastest moving particle is Z due to concentration of particle.
Canister with the fastest moving particle
The speed of the particles depend on the mean distance of the particles.
The canister with the largest concentration per particle will contain particles with the greatest speed.
If the particle concentration is increasing from W to Z, then Z will have fastest moving particle.
Thus, the canister with the fastest moving particle is Z due to concentration of particle.
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Answer:
d. Z
Explanation:
If all the canisters had the same amount of particles, which canister would have the fastest moving particles?
Z
The Venn diagram compares protons with electrons. Which shared property belongs in the region marked "B”?
The shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.
What charge is present on proton and electron?We know that proton is positively charge particle while on the other hand, electron is negatively charge particle. Due to opposite charges, these particles attract each other.
So we can conclude that the shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.
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Answer:
electrically charged
Explanation:
got it right
A golf ball is driven from a point A with a speed of 40 m/s at an angle of elevation of 30°. On its downward flight, the ball hits an advertising hoarding at a height 15.1 m above the level of A, as shown in the diagram below. Find A. The time taken by the ball to reach its greatest height above A, B. The time taken by the ball to travel from A to B, C. The speed with which the ball hits the hoarding. 20
(a) The time taken by the ball to reach its greatest height is 2 seconds.
(b) The time taken by the ball to travel from A to B, C is 4.08 m/s.
(c) The speed with which the ball hits the hoarding is 69.3 m/s.
Maximum height of the projectile
H = u²sin²θ/2g
H = (40² x (sin 30)²) / (2 x 9.8)
H = 20.4 m
Time for the ball to travel 20.4 mh = ut - ¹/₂gt²
20.4 = (40 x sin 30)t - (0.5)(9.8)t²
20.4 = 20t - 4.9t²
4.9t² - 20t + 20.4 = 0
solve the quadratic equation using formula method;
t = 2 seconds
Time taken for the ball to travel A, B , CThis is the time of motion of the ball;
T = 2usinθ/g
T = (2 x 40 sin 30)/9.8
T = 4.08 s
Speed of the ball when it hits the hoardingv = u + gt
vy = 40 sin30 + (9.8 x 4.08)
vy = 59.98 ms
vx = 40 x cos30
vx = 34.64 m/s
vf = √(vy² + vx²)
vf = √(59.98² + 34.64²)
vf = 69.3 m/s
Thus, the time taken by the ball to reach its greatest height is 2 seconds.
The time taken by the ball to travel from A to B, C is 4.08 m/s.
The speed with which the ball hits the hoarding is 69.3 m/s.
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The small amount of missing mass that occurs when _____ atoms fuse to form helium
atoms is converted into energy in the form of heat and light.
nitrogen
hydrogen
barium
Ochromium
hydrogen
Explanation:Atomic fusion occurs when 2 or more atoms combine to create a new element.
Helium
Before figuring out what elements create helium, we need to understand its properties. Helium is the second element on the periodic table. It has 2 protons and usually 2 electrons in the base state.
There are technically ions but base helium will have 2 electrons, especially because it is a noble gas.
Fusion
In most cases, including this one, when 2 atoms fuse together you can add their protons together to find the atomic number of the new element. Since helium only has 2 protons, the elements fused to form helium must both have 1 proton.
The element with 1 proton is hydrogen. So, 2 hydrogen atoms fuse to form helium.
Real World Applications
Fusion can be seen across the universe. One of the most common examples of hydrogen fusing to become helium is in stars. Due to the heat within stars, 2 hydrogen atoms can fuse together to create helium. This process also creates energy in the form of heat and light that the star is made of.
an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the electron and photon.
Answer:
1.85c
Explanation:
a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c
The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.
The formula for adding velocities in special relativity is given by:
V(relative) = (V₁ - V₂) ÷ (1 - (V₁ × V₂ ÷ c²))
where:
V₁ = velocity of the electron
V₂ = velocity of the photon
c = speed of light in a vacuum
The relative velocity is:
V(relative) = (0.85c - c) ÷ (1 - (0.85c × c ÷ c²))
V(relative) = (0.85c - c) ÷ (1 - 0.85)
V(relative) = (-0.15c) ÷ (0.15)
V(relative) = -c
Therefore, The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.
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A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
Tension in rope A:
7.42×102 N
Tension in rope B:
7.52×102 N
What is the tension in rope C?
Tension in the rope C is 1.24× 10² N.
To find the answer, we need to know about the horizontal component of tension in the rope B.
What's the angle made by the rope B by horizontal?From the figure of the answer, in the triangle PQR, tan(θ)= PQ/QR = 6/1= 6 θ= tan inverse of 6 = 80.5°What's the horizontal component of the tension in rope B?Horizontal component= tension in rope B × cos80.5°
= 7.52×10² N × cos80.5°
= 1.24×10² N
What's the tension in the rope C?From the figure, we have found that the tension in rope C = horizontal component of the tension in rope BSo, tension in rope C= 1.24×10² NThus, we can conclude that the tension in the rope C is 1.24× 10² N.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
The minimum speed required for a satellite in order to break free permanently from the planet and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.
To find the answer, we need to know about the escape velocity and time period of revolving satellite.
What's the expression of escape velocity of a satellite?Mathematically, escape velocity= √(2GM/R)G = gravitational constant, M = mass of planet, R= radius of the planetHere, M = 4.74×10²⁴kg, R = 5870 kmEscape velocity= √(6.67×10^(-11)×4.74×10²⁴/5.870×10⁶)= 7.3 Km/s
What's the expression of time period of a circularly orbiting satellite?T= {2π×r^(3/2)}/√(GM)r= (T/2π)⅔× (GM)^(1/3)r is the radius of the orbitWhat's the radius of the circular orbit, if the time period of the satellite is 16.6 hours?T = 16.6 hours = 16.6×3600 second = 59760sr = (59760/2π)^⅔× (6.67×10^(-11)×4.74×10²⁴)^(1/3)= 3.1 × 10⁴ km
Thus, we can conclude that the escape velocity and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.
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What affects fuel consumption in automobiles?
A. Drag
B. Nothing
C. Air resistance
D. Time of day
Answer:
A and C
Explanation:
drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).
1. Is the image projected on a movie screen real or virtual? What about the image of yourself seen in a bathroom mirror?
2. Hold a shiny spoon in front of you. What differences do you notice about the image of your face seen in the convex and concave sides?
3. Where are the images formed by each side of the spoon? In front or behind the spoon? (Try the parallax method. Look at the image of an overhead light. Hold the tip of a pencil where you think the image is. Move your head from side to side. If the image and pencil tip appear to move relative to each other, adjust the position of the pencil back and forth until they appear as one)
Answer: 1. The movie one is virtual and the bathroom mirror is real
2. The image is distorted in a way
3. Behind the spoon
Explanation:
The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.
What is virtual screen?In virtual desktops, the desktop environment is segregated from the physical device being used to access it. They are preloaded images of operating systems and apps. Over a network, users can remotely view their virtual desktops.
A virtual desktop can be accessed from any endpoint device, including a laptop, smartphone, or tablet. The user interacts with the client software that was installed on the endpoint device by the virtual desktop provider.
A virtual desktop mimics the appearance and feel of a real workstation. Because robust resources, like storage and back-end databases, are easily accessible, the user experience is frequently even better than that of a physical workstation.
Therefore, The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.
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A crate with a mass of M = 62.5 kg is suspended by a rope from the endpoint of a uniform boom. The boom has a mass of m = 116 kg and a length of l = 7.65 m. The midpoint of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 57.7° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?
Answer:
Tension in the vertical rope: approximately [tex]613 \; {\rm N}[/tex].
Tension in the horizontal rope: approximately [tex]3.74 \times 10^{3}\; {\rm N}[/tex].
Assumption: [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
Explanation:
Since the system is not moving, the tension in the vertical rope would be equal to the weight of the crate:
[tex]\begin{aligned}\text{weight of crate} &= m\, g \\ &= 62.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 613.25 \; {\rm N}\end{aligned}[/tex].
Note that [tex]\theta = 57.7^{\circ}[/tex] is the angle between the beam (the lever) and the vertical rope. The torque that this vertical rope exert on the beam would be:
[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &=(7.65\; {\rm m}) \, (613.25\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.965 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].
This torque is in the clockwise direction.
The weight of the beam ([tex]m = 116\; {\rm kg}[/tex]) would be:
[tex]\begin{aligned}\text{weight of beam} &= m\, g \\ &= 116 \; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 1.138 \times 10^{3}\; {\rm N}\end{aligned}[/tex].
Note that [tex]\theta = 57.7^{\circ}[/tex] is also the angle between the beam and the direction of the (downward) gravitational pull on this. Since this beam is uniform, it would appear as if the weight of this beam is applied at the center of this beam (with a distance of [tex](7.65\; {\rm m}) / 2[/tex] from the pivot.) Thus, the torque gravitational pull exerts on this beam would be:
[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &= \genfrac{(}{)}{}{}{7.65\; {\rm m}}{2} \, (1.138 \times 10^{3}\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.679 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].
This torque is also in the clockwise direction.
The tension in the horizontal rope would need to supply a torque in the counterclockwise direction. The magnitude of that torque would be approximately:
[tex]\begin{aligned} & 3.965\times 10^{3}\; {\rm N \cdot m} + 3.679\times 10^{3}\; {\rm N \cdot m} \\ \approx \; & 7.645 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].
Note the angle between the direction of this tension and the beam is [tex](90^{\circ} - \theta) = 32.3^{\circ}[/tex]. This force is applied [tex](7.65\; {\rm m}) / 2[/tex] from the pivot. Hence, achieving that torque of [tex]7.645 \times 10^{3}\; {\rm N \cdot m}[/tex] would require:
[tex]\begin{aligned} F &= \frac{\tau}{r\, \sin(90^{\circ} - \theta)} \\ &\approx \frac{7.645\times 10^{3}\; {\rm N \cdot m}}{((7.65\; {\rm m}) / 2) \times \sin(32.3^{\circ})} \\ &\approx 3.74 \times 10^{3}\; {\rm N} \end{aligned}[/tex].