That depends on where YOU are when you measure it.
If you're motionless in the laboratory, then you measure the particles flying apart at 1.6c .
If you're riding on one of the particles, you measure the other one flying away from you at less than c
What is the estimated density of the golf ball? Record your answer to the nearest hundreth.
g/cm3
The estimated density of the golf ball is 170 kg/m³
What is density?
Density is the ratio as Mass per unit Volume.
In displacement method,
First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.
Then calculate the density by dividing the mass by the volume.
Here, D = m/V
By displacement method , the estimated volume of golf ball is 100 cm³ and estimated mass is 600 g
Then ,
Density = 100 cm³ / 600 g = 0. 17 g/ cm³ = 170 kg/m³
So the estimated density of golf ball is 170 kg/m³
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Psychology is a combination of the Greek words psyche, which means “soul” or “mind,” and logos, which means “the study of.”
Please select the best answer from the choices provided
T
F
Greek terms psyche, which means "soul" or "mind," and logos, which means "the study of," are combined to form the english word psychology. The assertion is true.
What is Psychology?Psychology is the name given to the scientific study of the mind and behavior. Psychologists are actively interested in researching and comprehending how the mind, the brain, and behavior work.
Psychology is a combination of the Greek words psyche, which means “soul” or “mind,” and logos, which means “the study of.”The given statement is corect.
Hence the best answer from the choices provided will be true.
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A uniform plate of height 1.870 m is cut in the form of a parabolic section.
The lower boundary of the plate is defined by: y = 0.500[tex]x^{2}[/tex]. Find the distance from the rounded tip of the plate to the center of mass.
The distance from the rounded tip of the plate to the center of mass is 1.87 m.
What is center of mass?The center of mass is a point inside or outside the mass where all of the mass is concentrated.
The y-coordinate of the centroid is given by the ratio of two definite integrals;
Yc = ∫ydm/∫dm,
where dm is a density function
For the uniform plate, δ does not change with position in the plate.
Yc = ∫yδdA/∫δdA
Yc = ∫ydA/∫dA.
dA is a horizontal slice of the plate with dimensions xdy.
Solving the parabola for x,
y = 0.5x²
x = ± √(y/0.50), where the negative value corresponds to the left half of the parabola and the positive to the right half.
dA = (√(y/0.50)
= √(y/0.50))dy
= 2(√(y/0.50))dy
The limits of integration are from zero to 1.870, the top of the plate.
∫ydA = ∫2y√(y/0.50)dy = 7.232 m³
∫dA = ∫2√(y/0.50)dy = 3.868 m²
∫ydA/∫dA = 7.232 m³/3.868 m²
∫ydA/∫dA = 1.869700 m
∫ydA/∫dA = 1.87 m
Thus, the distance from the rounded tip of the plate to the center of mass is 1.87 m.
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A piece of steel expands 10 cm when ur is heated from 20 to 50 degrees Celsius. How much would it expand if it was heated from 20 to 60 degrees Celsius
The steel would expand by 4. 8 * 10^-3 cm
How to determine the linear expansionThe change in length ΔL is proportional to length L. It is dependent on the temperature, substance, and length.
Using the formula:
ΔL= α LΔT
where ΔL is the change in length L = 10cm
ΔT is the change in temperature = 60° - 20° = 40° C
α is the coefficient of linear expansion = 1.2 x 10^-5 °C
Substitute into the formula
ΔL = [tex]1.2 * 10^-5 * 10 * 40[/tex]
ΔL = [tex]4.8 * 10 ^-3[/tex] cm
Therefore, the steel would expand by 4. 8 * 10^-3 cm
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If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____, and ____, and all planets will orbit the sun successfully.
Pleas help this is Flvs comprehensive science class
6.01 please please help
So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.
When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.
When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.
When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.
So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.
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a. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by how much is the spring is compressed.
b. A 0.010 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position.
c. The same dart is now fired horizontally from a height of 4.70 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.
d. Find the horizontal distance from the equilibrium position at which the dart hits the ground.
a. The spring is compressed by 0.174 m.
b. The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.
c. The horizontal velocity of the dart at that time is 13.74 m/s.
d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.
What is potential energy?The energy by virtue of its position is called the potential energy.
a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.
PE of spring = 1/2 kx²
Put the values, we get
P.E = 0.940 = 1/2 x 62 x²
x = 0.174 m
Thus, the spring is compressed by 0.174 m.
b. Given is a 0.010 kg dart is fired straight up.
The vertical height is find out by
0.940 J = (0.010 kg) (9.8 m/s²) h
h = 9.6 m
Thus, the vertical distance the dart travels from its position is 9.6 m
c. From the conservation of energy principle, total mechanical energy is conserved.
1/2 mv² =mgh
v = √2gh
Plug the values, we get
v = √2 x 9.8x 9.6
v = 13.74 m/s
Thus, the horizontal velocity is 13.74 m/s.
d. Time that dart spends in air, t = √2h/g
t = √(2x9.6)/9.81
t = 1.4 s
The horizontal distance from the equilibrium position at which the dart hits the ground.
Horizontal distance = (Velocity on x direction) x time
Horizontal distance = 13.74 m/s x 1.4s
Horizontal distance = 19.236 m
Thus, the horizontal distance is 19.236 m.
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In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is .......... a) 300 J c) 600 J (b) 450 J d) 750 J F=100 N CamScanner Wa 3m In the opposite figure , the work done by the force F to push the box upwards from the ground to the top of the inclined plane is .......... a ) 300 J c ) 600 J ( b ) 450 J d ) 750 J F = 100 N CamScanner Wa 3m
In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is 600 J.Option c is correct.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
The displacement value from the triangle is;
[tex]\rm sin 30^0 = \frac{P}{H} \\\\ H=S =\frac{3 \ m}{sin 30^0} \\\\ S = 6\ m[/tex]
The work done is found as ;
[tex]\rm W= Fd \\\\ W = 100 \ N \times 6 \\\\ W= 600 \ J[/tex]
The force F in the opposing figure does 600 J of effort to lift the box from the ground to the top of the inclined plane.
Hence option c is correct.
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5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) How much charge is stored in the 5.00-μF capacitor?
(b) What is the potential difference across the 10.0-μF capacitor?
(a) The charge stored in the 5.00-μF capacitor is 37.2 μC.
(b) The potential difference across the 10.0-μF capacitor is 3.72 V.
What is capacitor?The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.
The capacitance of a capacitor is denoted by C and expressed as
C = Q/V
Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) The equivalent capacitance is
1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃
Substitute the values, we get
Ceq = 3.1 μF
The charge stored in 5.00-μF capacitor is
Q = Ceq x V
Q = 3.1 μF x 12 V
Q = 37.2 μC
Thus, the charge stored in the 5.00-μF capacitor is 37.2 μC
(b) The potential difference across the 10.0-μF capacitor is given by
V = Q/C₂
Put the values, we get
V = 37.2 / 10
V = 3.72 V
Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.
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two rods one aluminum and one brass are is clamped at one end. At zero degrees celsius, the roads are each 50 cm long and separated by 0.024 CM at their unfastened ends. At what temperature will the rod just come into contact.
At 11.3°C the rod will just come into contact
Coefficient of linear expansion of aluminium [tex]\alpha_{Al}[/tex] = [tex]23*10^-^6[/tex] °[tex]C[/tex]
Coefficient of linear expansion of Brass [tex]\alpha_{B} =19*10^-^6[/tex] °[tex]C[/tex]
[tex]\alpha = \Delta{L}/Lo(T2-T1)[/tex]
For aluminium
[tex]\alpha_{Al} = \Delta{L}/Lo(T-0)[/tex]
[tex]\Delta{L_{1}} = (23*10^-^6*50*T)[/tex]
For Brass
[tex]\alphaB = \Delta{L_{2}/Lo(T-0)[/tex]
[tex]\Delta{L_{1} +\Delta{L_{2} =0.024cm(23*10^-^6*50*T)+(19*10-6*50*T) =0.024[/tex]
T =11.43 °C
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What is temperature?
A. A type of heat transfer
ANUN
B. The measure of an object's "hotness"
APATIN
HERRE
C. Electromagnetic waves
Digita
D. The energy transferred between objects
H
The term temperature has to do with the measure of an object's "hotness".
What is temperature?The term temperature has to do with how hot or cold a body is. In other words, the word temperature brings us to call to mind the degree of hotness or coldness of a body.
Succinctly put, the term temperature has to do with the measure of an object's "hotness".
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which items can be classified as matter? check all that apply?
-pencil
-oxygen
-idea
-horse
-dream
Answer:horse
Explanation:
A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top when standing straight. The stress produced is 75000kN/m2. Assume Young's modulus for cast iron as 1.5x 108 kN/m^2 and find (i) magnitude of the load, (ii) longitudinal strain produced, and (iii) total decrease in length.
Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.
Given,
Length, L = 4 m
Outer diameter, D = 300mm, D= 0.3 m
Thickness, t = 50 mm, t = 0.05 m
Stress produced, σ = 75000 kN/m²
Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²
Calculating the diameter of the cylinder,
Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)
d= 0.2 m
(i) Magnitude of the load P:
Using the relation, σ =P/A
P = σ × A = 75000 × π /4 (D² – d² )
P= 75000 × π/4 (0.3² – 0.2²)
P= 75000 × π/4 (0.09 – 0.04)
P = 2945.2 kN
Hence, Magnitude of the load P is 2945.2 kN.
(ii) Longitudinal strain produced, e :
Using the relation, Strain, (e) = stress/E
e= 75000/(1.5 x 10⁸)= 0.0005
Hence, the Longitudinal strain produced is 0.0005.
(iii)Total decrease in length, dL:
The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.
Strain = change in length/original length
e= dL/L
0.0005 = dL/4
dL = 0.0005 × 4m = 0.002m=2mm
Hence,the decrease in length is 2 mm.
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A 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up and slide together with a speed of 15.0 km/h. What was the speed of the first car just before the collision?
The speed of the first car just before the collision is 26.73 km/h.
What is conservation of momentum principle?When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.
The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.
m₁u₁ +m₂u₂ =(m₁ +m₂) v
Given a 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up and slide together with a speed of 15.0 km/h.
Second car is parked, so its velocity will be zero.
808 x u +632 x 0 = (808 +632 ) x 15
u = 26.73 km/h
Thus, the speed of the first car just before the collision is 26.73 km/h
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A plane leaves with an acceleration of 6.34 m/s squared and takes 1.5 hours to stop. What is the speed of the plane? What was the distance it traveled?
The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.
Three equation of motion are:-
v = u + ats = ut + (1/2)at²v² - u² = 2asWhere v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.
In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.
Applying equation 1 to find the initial speed of plane
v = u + at
0 = u + (-6.34 × 5400) {v=0 as plane will stop after 5400 sec}
u = 6.34 × 5400
u = 34236 m/s
Initial velocity of plane is 34236 m/s
Applying equation 2 to find the displacement of plane in that time period
s = ut + (1/2)at²
s = ( 34236 × 5400 ) - ( (1/2) × 6.34 × 5400² )
s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )
s = 5400 × ( 34236 - 17118 )
s = 5400 × 17118 metres
s = 5.4 × 17118 Km
s = 92437.2 Km
Distance travelled by plane is 92437.2 Km
So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.
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A resistor has a resistance of 8.2 kΩ. If a voltage of 15.9 V were placed across it, what would be the current, in mA? Give the answer to two decimal places; don't worry if the computer adds zeroes.
The value of the electric current for the given conditions willl be 1939.02 A.
What is ohm’s law?Ohm's law claims that the voltage across a conductor is directly proportional to the current flowing through it.
Ohm's law claims that the voltage across a conductor is direct to the current flowing through it. This current-voltage connection may be expressed mathematically as,
V=IR
15.9 V = I × 8.2 × 10³ Ω
I = 1939.02 A
Hence, the value of the electric current for the given conditions willl be 1939.02 A.
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the 3kg object in figure is released from rest at height of 5m on curved frictionless ramp.at the foot of the ramp is a spring of force constant 400N/m.the object slide down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. a) find x
b)describe the motion of the object (if any) after the block momentarily comes to rest?
The value of x given the data from the question is 0.86 m
How to determine the energyFrom the data given above, we can s determine the energy as follow:
Mass (m) = 3 KgHeight (h) = 5 mAcceleration due to gravity (g) = 9.8 m/s² Energy (E) = ?E = mgh
E = 3 × 9.8 × 5
E = 147 J
How to determine the value of xSpring constant (K) = 400 N/mEnergy (E) = 147 JExtention (e) = x = ?E = ½Ke²
147 = ½ × 400 × x²
147 = 200 × x²
Divide both side by 200
x² = 147 / 200
Take the square root of both sides
x = √( 147 / 200)
x = 0.86 m
Since the block came to rest, there is no motion
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Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?
This is a uniform rectilinear motion (MRU) exercise.
To start solving this exercise, we obtain the following data:
Data:v = 4.6 m/sd = ¿?t = 10 secTo calculate distance, speed is multiplied by time.
We apply the following formula: d = v * t.
We substitute the data in the formula: the speed is equal to 4.6 m/s, the time is equal to 10 s, which is left as follows:
[tex]\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }[/tex]
[tex]\bf{d=46 \ m}[/tex]
Therefore, the speed at 10 seconds is 46 meters.
[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}[/tex]
The road from city A to city B is described by a car with Vm 40 km / h. When the car turns (from B to A) the average speed is 60 km / h. Find the average round trip speed.
Answer:
i think the answer is 20.......
Q3, A ball of mass 5.0 kg moving with a Velocity of 10.0 ms collides
with a 15.0 kg ball moves with a Velocity of 4 ms! If both balls
Stick together after Collision, Calculate their Common Velocity after Impact if they initially moves in The Same direction, and Opposite direction.
Answer:
Their common velocity after the collision will be 5.5m/s
Explanation:
look at the attachment above ☝️
28. An electron with a speed of 4.0 x 10° m/s enters a uniform magnetic field of magnitude 0.040 T at an angle of 35 degrees to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?
Answer:
r= 1.09×10^-4
Explanation:
Given
V=speed=4.0×10^5 m/s
B= magnetic field= 0.040 T
©=angle= 35°
m= mass of electron= 9.11×10^-31
q= charge of electron= 1.60×10^-19
solution
qv×B= mv²/r
qvBsin©=mv²/r
qBsin©=mv/r
r=mv/qBsin©
r=9.11×10^-31× 4.0×10^5/1.064×10^-19×0.04T(sin35°)
r= 1.09×10^-4 m
a) r = 1.09 * [tex]10^{-4}[/tex] m
b) Distance travelled : 6.845 * [tex]10^{-4}[/tex] m
What is an electron ?
An electron is a stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids.
given
charge of electron : 1.6 * [tex]10^{-19}[/tex] C
mass of electron = 9.11 * [tex]10^{-31}[/tex] kg
v = 4.0 x 10 m/s
B = 0.040 T
theta = 35 degrees
since ,
force in magnetic field on electron = centripetal force
a) q(v*B) = m [tex]v^{2}[/tex] / r
q v B sin(theta) = m [tex]v^{2}[/tex] / r
r =m v /q B sin(theta)
r = 9.11 * [tex]10^{-31}[/tex] * 4.0 x [tex]10^{5}[/tex]/ 1.6 * [tex]10^{-19}[/tex] sin (35)
r = 1.09 * [tex]10^{-4}[/tex] m
b) far forward will the electron have moved after completing one circle will be equal to circumference of the circle = 2πr
= 2 * 3.14 * 1.09 * [tex]10^{-4}[/tex] m = 6.845 * [tex]10^{-4}[/tex] m
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1) 40 coloumbs of charge flow in 10 seconds. What is the current?
[tex]\large\boxed{Formula: I= \frac{Q}{t}}[/tex]
Note that:I: CurrentQ: Charge t: TimeSolution:Current is the flow of electrons around a circuit.SI unit of current is Amperes or Amps Symbol is ( A )Let's solve!
The formula is given above.
Let's substitute the values according to the formula.
We'll have to divide the measure of charge by the given time.
[tex]I= \frac{40}{10}[/tex]
Final answer:
[tex]\large\boxed{= 4 \: Amps \: (A)}[/tex]
Hence, the current is 4 amps.
What does the first law of thermodynamics state about energy?
The first law states that the internal energy change of that system is given by Q − W . Since added heat increases the internal energy of a system, Q is positive when added to the system and negative when removed from the system.
an object is held 33.5 cm from a convex mirror. it creates a virtual image of magnification 0.253. what is the focal length of the mirror?
Answer:
Magnification= -image distance/object distance
.253=image distance/33.5
image distance= 8.48 cm
Mark is diving to the bottom of a pool to pick up a penny. Which of the
following describes the fluid pressure on his ears as he dives?
A. The fluid pressure increases as he dives.
B. After he dives, the fluid pressure increases and then decreases as
he dives deeper.
C. The fluid pressure on his ears doesn't change as he dives.
D. The fluid pressure decreases as he dives.
Answer:
A
Explanation:
As you go deeper, the pressure increases because the water is going farther away from the air and atmosphere
a. Recent studies have raised concern about 'heading' in youth soccer (i.e., hitting the ball with the head). A soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg. What is the magnitude of the impulse which the player must impart to the ball?
b. If the player's head has a mass of 2.90 kg, what is the magnitude of the average acceleration of the player's head during the impact. Assume that over the brief time of the impact, 30.40 ms, the player's head can be treated separately from the player's body.
a. The magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s.
b. The magnitude of average acceleration of the player's head during the impact is 195.66 m/s².
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, a soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg.
a. Substitute the values into the expression, we get
Impulse (x) = 2 x (13.20 - 13.20cos40°)
Impulse (x) = 3.088 kg.m/s
Impulse in y direction, is
Impulse (y) = 2 x 13.20sin40°
Impulse (y) = 16.97 kg.m/s
so, the magnitude of impulse is
I = sq rt(Ix² + Iy²)
Put the values, we get
I = 17.25 kg.m/s
Thus, the magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s
b. The player's head has a mass of 2.90 kg. The brief time of the impact, 30.40 ms,
F.t = I
Put the given values, we have
F = 17.25 / 30.40 x 10⁻³
F = 567.434 N
From the Newtons second law of motion,
F = ma
Plug the values from the question, we have
567.434 N = 2.90 x a
a = 195.66 m/s²
Thus the magnitude of average acceleration of the player's head during the impact is 195.66 m/s².
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Consider the baby being weighed in Figure 4.25.
Figure 4.25
(a) What is the mass of the child and basket if a scale reading of 104 N is observed?
kg
(b) What is the tension T in the cord attaching the child to the scale?
N
(c) What is the tension T' in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?
N
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. (Do this on paper. Your instructor may ask you to turn in this work.)
The mass and tension due to the system are as follows:
The mass of the child and scale = 10.6 kgThe tension T, in the cord attaching the child to the scale = 104N The tension T', in the cord attaching the scale to the ceiling T' = 108.9 NWhat is tension?Tension is a type of pulling force due transmitted by means of a string or cable.
Force = mass * acceleration due to gravitya) The mass of the child and scale = 104/9.81 = 10.6 kg
b) The tension T, in the cord attaching the child to the scale = scale reading = 104N
c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale
T' = 104 + (0.5 * 9.81)
T' = 108.9 N
d) The sketch is attached in the picture
In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.
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143°C = _____
416 K
-130 K
0 K
143 K
During a NASCAR race a car goes 58 m/s around a curved section of track that has a radius of 260 m. What is the car's acceleration?
Answer:
12.9m/s^2
Explanation:
As, a=(v^2)/r
=(58^3)/260
=12.9
There are several ways to model a compound one type of model is shown ?what is the chemical formula for the molecule modeled?
The is organic compound with the correct chemical formula C4H9O2.
What is a model?A model is a representation of reality. A model serves the purpose of prediction as well as explanation.
Looking at the model of the molecule we can see that it is the organic compound with the correct chemical formula C4H9O2. The molecule is shown in the image attached to this answer.
Missing parts:
There are several ways to model a compound. One type of model is shown.
What is the chemical formula for the molecule represented by the model?
CHO
C4H9O2
C4H8O
C3H8O2
Learn more about molecular models:https://brainly.com/question/156574?
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Two force of 20N and 40N act at a
Point and the angle between them is 50 degrees.
Find the resultant force
Find the direction using sine rule
R=√20²+40²+2.20.40 cos 50
R=55 N
R/sin β = F/sin α
55/sin 50 = 40/sin α
α = 33°
