The bomb strike the ground relative to the point at 1km . B
How to determine the distanceUsing the equation
h = 1/2 gt^2
500 = 1/2 * 10* t^2
500 = 5t^2
t = √500/5
t = √100
t = 10seconds
To find the distance,
Distance = velocity * time
Distance = 100 ÷ 10
Distance = 1000m = 1km
Therefore, the bomb strike the ground relative to the point at 1km . B
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A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 41,200 ly (light-years) from the galactic center and has a speed of 275 km/s. Estimate the total mass of the galaxy based on the motion of the star.
Gravitational constant is 6.674×10−11 m3/(kg·s2) and mass of the Sun Ms=1.99 × 1030 kg.
*Answer in billion solar mass
The total mass of the galaxy is 443.4 Solar mass
Orbital velocity ([tex]v[/tex]) = [tex]\sqrt{\frac{MG}{R} }[/tex]
where M= weight of galaxy
G= gravitational constatnt = [tex]6.674*10^-^1^1[/tex] (given)
R = distance from centre = [tex]41200[/tex] Light years (given)= [tex]4.12*9.5*10^1^6[/tex] km (1 ly= [tex]9.5*10^3[/tex] billion km)
v= orbital velocity = [tex]275[/tex] [tex]km/s[/tex] (given)
∴ According to the formula
[tex](2.75*10^2)^2[/tex] = [tex]\frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}[/tex]
⇒ [tex]7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1[/tex] (cross multiplying and expanding)
⇒ [tex]29.59*10^2^1=M*6.674*10^-^1^1[/tex]
⇒ [tex]\frac{29.59*10^2^1*10^1^1}{6.674}=M[/tex]
⇒ [tex]4.434*10^3^2=M[/tex]
1 solar mass = [tex]1.989*10^3^0 kg[/tex]
⇒ Mass in solar mass =443.4 Solar mass
⇒ M = 443.4 Solar mass
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A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed when it strikes the ground below? (Use conservation of energy.)
The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
Time of motion of the projectileThe time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
h is height of the cliffv is velocityt is time of motion265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
Final velocity of the projectilevyf = vyi + gt
where;
vyf is the final vertical velocityvyi is initial vertical velocityvyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
vxf is the final horizontal velocityvxi is the initial horizontal velocityvxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
vf is the speed of the projectile when it strikes the ground belowvf = √(149.322² + 130.8²)
vf = 198.51 m/s
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A 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle as seen in the figure.
If the ball is in contact with the wall for 0.207 s, what is the average force exerted on the ball by the wall?
The magnitude of the average force exerted on the ball by the wall is 225.469 N.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s
Substitute the values into the expression, we get
Impulse = 2 x mvcosθ
Impulse= 2 x 3.42 x 14.3 x cos 61.5°
Impulse = 46.672 kg.m/s
The impact force can be written as
F.t = I
Put the given values, we have
F = 46.672 / 0.207
F = 225.469 N
Thus, the magnitude of force exerted by the wall is 225.469 N
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Water flows with a volume flow rate of 1.44 m³/s in a pipe. Find the water speed where the pipe radius is 0.430 m.
Answer:
0.6192
Explanation:
1.44 x 0.430=0.6192
Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in Figure 1, the pressure is and the pipe diameter is 6.0 cm. At another point higher, the pressure is and the pipe diameter is 3.0 cm. Find the speed of flow (a) in the lower section and (b) in the upper section. (c) Find the volume flow rate through the pipe
Answer:
full explanation and answer is on the picture
A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift the bucket at the other end?
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[tex]\large \bold {ANSWER}[/tex]
Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.
[tex]\large \bold {EXPLANATION}[/tex]
Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.
As the statement only tells us a pulley, we must consider that it is fixed, because generally when it is mobile, this information is highlighted in the question.
In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.
Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.
Help me ASAP!!!!
If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density?
A. Ball A
B. Ball B
C. Their Densities are equal.
Answer: A) Ball A
Explanation:
Density = m/v where m is mass and v is volume
This means the greater the mass, the greater the density.
Because of Ball A's greater mass it would have the greater density.
A small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.
The picture is to scale. It shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V. Click here to view the motion of the mass m.
(options are: less than, greater than, equal to)
• The speed change between T and S is (blank) between S and R.
• The speed of m at X is (blank) that at Q.
• The size of the total force on m at P is (blank) at U.
• The mechanical energy of m at P is (blank) that at V.
• The size of the total force on m at S is (blank) at P.
• The velocity of m at X is (blank) that at V.
• The speed change between T and S is greater between S and R.
• The speed of m at X is greater that at Q.
• The size of the total force on m at P is less at U.
• The mechanical energy of m at P is equal to that at V.
• The size of the total force on m at S is greater at P.
• The velocity of m at X is equal to that at V.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.
M.E = KE +PE
Given is a small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.
The picture shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V.
Between T to S and S to R, the mass is under constant acceleration. Time taken to move from T to S is greater than S to R. Thus, the speed change between T and S is greater than between S and R.
At Q, there is only a horizontal velocity component, but at X. the speed will be greater and has both vertical and horizontal component. Thus, the speed of m at X is greater than that at Q.
Force is given as the rate of change of momentum with time. At U, change in momentum is large compared to P. Thus, the size of the total force on m at P is less at U.
There is no friction acting on the system. So the energy remains conserved. Mechanical energy at P = V.
The force on mass m at S is only the gravity force. The remaining forces are cancelled by the normal force. Thus, size of the total force on m at S is greater at P.
The energy is conserved at each point of motion of mass. If X and V are at same height, they have same potential energy and so their kinetic energy. Thus, velocity of m at X is equal to that at V.
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Explain why aircraft are carefully designed so that parts do not resonate.
Answer:
See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.
A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.
The stress on the wire is determined as 6.37 x 10⁶ N/m².
Stress on the wireThe stress on the wire is calculated as follows;
σ = F/A
where;
F is the force applied on the wireA is area of the wireLet the diameter of the wire = 2 mm
Area = πr²
Area = π(1 x 10⁻³)²
Area = 3.142 x 10⁻⁶ m²
Stress = 20/(3.142 x 10⁻⁶)
Stress = 6.37 x 10⁶ N/m²
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A skateboarder starts from rest and maintains a constant acceleration of 0.50 m/s² for 8.4 s. What is the rider's displacement during this time
meters
Answer:
4.2m/s
Explanation:
0.50x8.4=4.2m/s
A bumblebee
is flying towards a flower in a
straight line at 4.09 m/s when it begins to
accelerate at 1.01 m/s².
How long does it take the bee to reach the
flower if it is 23.4 m away?
Answer:
given -
initial velocity = 4.09 m/s
acceleration = 1.01 m/s²
distance = 23.4 m
time = ?
using second formula of motion,
s = ut + 1/2 at².
where, s = distance
u = initial velocity
t = time
a = acceleration
23.4 = 4.09(t) + 1/2(1.01)(t) ²
23.4 = 4.09t + 2.02t²
2.02t² + 4.09t - 23.4 = 0
solving the equation by using quadratic formula
Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :
a = 2.02
b = 4.09
c = -23.4
we get t=2.539 or t= -4.563
time cannot be negative so
t= 2.539 sec = 2.6 Sec is the answer
An object weighs 200N, what is its mass?
Answer:
kg
Explanation:
the highr and jebad to kilogram mizans
Answer:
w=m×g
Explanation:
w is given and take g as 9.8m/s^2
The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newtons on Earth weigh on Pluto? Note that the value of acceleration due to gravity on Earth is 9.8 meters/second2.
The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.
What is gravitation?Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.
By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.
m = 250 N / 9.8 m/s²
m = 25.51 kg
Multiply the acquired mass by Pluto's gravitational acceleration (g);
W = (25.51 kg) x (0.61 m/s²)
W = 15.56 N
Thus, the item will only be 15.56 N in weight.
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Answer: B.
15.6 newtons
Explanation: edmentum
Particles q1, q2, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C,
and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.250 m. What is the net force on q2?
The net force on q2 will be 1.35 N
A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.
Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.
We have to find the net force on q2
At first we will find Force due to q1
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²
F = 450 × 10⁻³
F₁ = 0.45 N (+)
Now we will find Force due to q2
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²
F = 1800 × 10⁻³
F₂ = 1.8 N (-)
So net force (F) will be
F = F₂ - F₁
F = 1.8 - 0.45
F = 1.35 N
Hence the net force on q2 will be 1.35 N
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velocity and distance of free falling object dropped from rest
The diagram shows the speed and distance of a free-falling item dropped from rest. In the attachment, there is a graph.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The velocity and the time at the different points are;
A(0 sec ,0 m/sec)
B(1,9.8)
C(2,19.6)
D(3,29.4)
E(4,39.2)
F(5,49.0)
The point is plotted on the graph and get the graph.
The graph is attached in the attachment.
The complete question is;
Velocity and distance of free-falling object dropped from rest are given in the digrame. Draw the velocity-time graph for the given conditions.
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A spaceprobe in outer space is flying with a constant speed of 1.795 km/s. The probe has a payload of 1635.0 kg and it carries 4092.0 kg of rocket fuel. The rocket engines of the probe are capable of expelling propellant at a speed of 4.161 km/s. Then the rocket engines are fired up. How fast will the spaceprobe travel when all the rocket fuel is used up?
The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
What is the law of conservation of linear momentum?
According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.
Unit conversion;
1 km/sec = 1000 m/sec
Given data;
Spaceprobe speed = 1.795 km/s = 1795 m /sec
Probe mass = 635.0 kg
Fuel mass = 4092.0 kg
Expelled propellent velocity = 4.161 km/s = 41461 m/sec
From the momentum conservation principle;
[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]
Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
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state two factors that affect current carrying capacity of an accumulator
The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature
What is the function of the accumulator?Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.
These key deciding elements are:
1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.
2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.
3. Accumulator identification:
4. Conditions for Installation:
Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature
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Wavelength is determined by dividing a wave's speed by its frequency. If a wave has a frequency of 1 hertz and a speed of 20 m/s, what is its wavelength?
Select one:
a. 80 m
b. 5 m
c. 0.05 m
d. 20 m
Answer:
D. 20m
Explanation:
this answer is 20 m.
wavelength = velocity / frequency
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant then changes to k2 = 250 N/m as the spring is stretched to x2 = 65 cm. From x2 = 65 cm to x3 = 89 cm the spring force is constant at F3 = 105 N.
Write an equation for the work done in stretching the spring from x1 to x2.
Calculate the work done, in joules, in stretching the spring from x1 to x2.
Calculate the work, in joules, necessary to stretch the spring from x = 0 to x3.
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
Work done in the springThe work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
where;
k is spring constantΔx is compression of the springW(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
where;
k₁ is first spring constantk₂ is second spring constantF₃ is third force applied to the springW(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
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A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.
The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground. At what distance x will the ball land?
The distance x will the ball land after flies off with a horizontal initial velocity is 3.0635 m.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.
M.E = KE +PE
A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.
The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground.
The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.
Kinetic energy = Potential energy
1/2 mv² =mgh₁
The velocity at the bottom, when the height h = 5m, is
v= √2gh₁...................(1)
The vertical height h₁ = l- lcosθ
h₁ = l- lcosθ
h₁ = 1.85 - 1.85cos48.5°
h₁ =0.6241 m
Putting the values in equation (1), we get
v = √2x 9.81 x0.6241
v = 3.499 m/s
The horizontal distance traveled is
x = vt
x = v x √2h/g
Plug the values, we get
x = 3.499 x √2x3.76 / 9.81
x = 3.0635 m
Thus, the horizontal distance ball travels is 3.0635 m.
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A truck accelerating at 0.0083 meters/second2 covers a distance of 5.8 × 104 meters. If the truck's mass is 7,000 kilograms, what is the work done to reach this distance?
Answer:
Explanation:
Force:
⇒ Mass × Acceleration
Work Done:
⇒ Force × Distance
Calculations:
⇒ Force = 0.0083 × 7000 N
⇒ Work done = 0.0083 × 7000 × 5.8 × 10^4 Joules
Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a Young's modulus of 2.10 × 1011 N/m2. Calculate the force a piano tuner applies to stretch the steel piano wire in Newtons.
The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.
What is force?Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.
Given data;
Young modulus, E=2.10 × 10¹¹ N/m²
Cross-sectional area,A
Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]
Initial length,[tex]\rm L_i = 7.5 mm[/tex]
[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]
Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.
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A metallic spoon is placed in a hot cup of coffee. If the coffee gives away 190 calories to the spoon to cool down by 0.75°C, what is the mass of the coffee? (Assume that ccoffee = 1.0 cal/gC°.)
Taking into account the definition of calorimetry, the mass of the coffee is 253.33 g.
CalorimetryCalorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m.c is the specific heat sudstance.ΔT is the temperature variation.Mass of coffeeIn this case, you know:
Q= 190 caloriesc= 1 [tex]\frac{cal}{gC}[/tex]m= ?ΔT= 0.75 CReplacing in the equation that allows to calculate heat exchanges is:
190 cal = 1 [tex]\frac{cal}{gC}[/tex]× m× 0.75 C
Solving:
m= 190 cal÷ (1 [tex]\frac{cal}{gC}[/tex]× 0.75 C)
m=253.33 g
Finally, the mass of the coffee is 253.33 g.
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1. List some environmental challenges identified by the Brundtland Commission's overview, as well as a key strategy for addressing them. Identify urgencies that the Commission sees as necessary for an optimistic environmental future.
The Brundtland Commission spotted as a matter of urgency, the interconnection between natural resource use and the economy.
What is the Brundtland Commission?The Brundtland Commission was set up with the intention to achieve sustainable development. The tenure of the commission lasted from 1984 to 1987.
The commission identified the pillars of sustainable growth as economic growth, environmental protection, and social equality. The committee opined that environmental problem emanates from poverty in the southern region and reckless consumption of resources in the northern region.
The Brundtland Commission spotted as a matter of urgency, the interconnection between natural resource use and the economy.
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Find the current flowing out of the battery in the circuit. I = [?] A 9.0 V 30 ww 40 Ω www 50 Ω 20Ω 10Ω
We need Net resistance
solve from right words
[tex]\\ \rm\Rrightarrow R_1=20+10=30\Omega[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{1}{R_2}=\dfrac{1}{30}+\dfrac{1}{50}=\dfrac{5+3}{150}[/tex]
[tex]\\ \rm\Rrightarrow R_2=\dfrac{150}{8}=18.75\Omega[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{1}{R_3}=\dfrac{1}{30}+\dfrac{1}{40}=\dfrac{4+3}{120}[/tex]
[tex]\\ \rm\Rrightarrow R_3=\dfrac{120}{7}=17.14\Omega[/tex]
[tex]\\ \rm\Rrightarrow R_{net}=17.14+18.75=35.89\Omega[/tex]
Use ohm's law
[tex]\\ \rm\Rrightarrow I=\dfrac{V}{R}[/tex]
[tex]\\ \rm\Rrightarrow I=\dfrac{9}{35.89}[/tex]
[tex]\\ \rm\Rrightarrow I\approx 0.25A[/tex]
HELP!! ASAP!!
2. Describe the difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl. Which molecule contains a polar covalent bond and which molecule contains a nonpolar covalent bond? Explain your reasoning alongside describing the differences between the types of bonds.
3. How is the metallic bonding different than ionic or covalent bonding? What are some properties of metals that result from this type of bonding? Explain/connect how the nature of the bonding leads to the properties of metallic substances.
Answer:
The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.
Why HCl is polar and H2 is now no longer?HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .
Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.
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how do I calculate equilibrant and fx and fy. I don't understand what they are asking
(a) The equilibrant C for force of vector A and B is 3.43 N.
(b) The equilibrant C for fx of vector A and B is 2.1 N.
(c) The equilibrant C, for fy of vector A and B is 2.12 N.
What is equilibrant force?An equilibrant force is a single force that will bring other bodies into equilibrium.
From configuration 1:Vector A: mass = 0.2 kg, θ = 20⁰
Vector B: mass = 0.15 kg, θ = 80⁰
Fx = mg cosθ
Fy = mg sinθ
where;
m is mass g is acceleration due to gravityVector AForce of A due to its weight
F(A) = mg
F(A) = 0.2 x 9.8 = 1.96 N
Fx = (0.2 x 9.8) cos(20) = 1.84 N
Fy = (0.2 x 9.8) sin(20) = 0.67 N
Resultant forceR = √(0.67² + 1.84²)
R = 1.96 N
Vector BForce of B due to its weight
F(B) = mg
F(B) = 0.15 x 9.8
F(B) = 1.47 N
Fx = (0.15 x 9.8) cos(80) = 0.26 N
Fy = (0.15 x 9.8) sin(80) = 1.45 N
Resultant forceR = √(0.26² + 1.45²)
R= 1.47 N
Equilibrant C of vector A and BEquilibrant force:
Force, C = 1.96 N + 1.47 N
Force, C = 3.43 N
Equilibrant FX:
Fx, C = Fx(A) + Fx(B)
Fx, C = 1.84 N + 0.26 N = 2.1 N
Equilibrant FY:
Fy, C = Fy(A) + Fy(B)
Fy, C =0.67 N + 1.45 N = 2.12 N
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What is the frequency of light with a wavelength of 2.10 × 10-7 m?
Answer:
V = ?
C = 3 × 10^8
wavelength = 2.10 × 10^-7
wavelength = C/V
V = C/wavelength
V = 3 × 10^8/2.10 × 10^-7
V = 1.43 × 10^ 15
Water flows through a pipe of diameter 8.0 cm with a speed of 10.0 m/s. It then enters a smaller pipe of diameter 3.0 cm. What is the speed of the water as it flows through the smaller pipe? (4) Water flows through a pipe of diameter 8.0 cm with a speed of 10.0 m / s . It then enters a smaller pipe of diameter 3.0 cm . What is the speed of the water as it flows through the smaller pipe ? ( 4 )
Answer:
71m/s
Explanation:
when you convert it you get 158.822 miles per hour
