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:

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

Answer:

1. 92, 138, 92

2. 7, 7, 7

3. 17, 18, 17

4. 29, 34, 29

Explanation:

the number on the bottom is always going to be the atomic number, or the number of protons.

the number on the top is going to be the atomic mass number, which is the sum of the number of protons and neutrons.

there is always going to be the same number of electrons as protons, because that was the overall charge of each atom is 0.

for example, for the first one:

there are 92 protons

there are 230 - 92 = 138 neutrons

there are 92 electrons

Answer:

i think the answer is 20.......

(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.

An equilibrant force is a single force that will bring other bodies into equilibrium.

Vector A: mass = 0.2 kg, θ = 20⁰

Vector B: mass = 0.15 kg, θ = 80⁰

Fx = mg cosθ

Fy = mg sinθ

where;

Force 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

R = √(0.67² + 1.84²)

R = 1.96 N

Force 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

R = √(0.26² + 1.45²)

R= 1.47 N

Equilibrant 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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The amount of current in the blood sample is determined as 0.12 A.

The current in the blood is calculated by applying Ohm's law as shown below;

V = IR

where;

From resistivity chart corresponding to 0.98 g/L, resistance, R = 7.4

I = V/R

I = 0.9 /7.4

I = 0.12 A

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It is important to remember "association", instead of "causation" when discussing correlations because associations can be used to make predictions.

The associations among variables have a fundamental value in science because they can determine how such variables behave in hypothetical situations.

Moreover, correlation only indicates a type of association between variables and therefore it is less informative.

In conclusion, it is important to remember "association", instead of "causation" when discussing correlations because associations can be used to make predictions.

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One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.

The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.

The complete question is;

"Choose the correct star life cycle.

A. Supernova, star as well as red giant, the nebula is incorrect.

B. Nebula, white dwarf, planetary nebula That is incorrect.

C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!

D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."

Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.

Hence option C is correct.

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the answer is in the picture ok

good luck don't worry it's a correct

#carry on learning

(a) A graph of u as a function of position (r) along the line between the two planets is attached below.

(b) The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]

If an object is lifted, work is done against gravitational force. The object gains energy.

Given are two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as

u(r) = − (Gm /R1 −r) − (Gm / R2 −r) .

This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets

(a) A graph of u (r)  versus position (r) along the line between the two planets is attached in answer.

(b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.

The range of the projectile is given by R =  v²sin2θ / g

g = gravitational acceleration of Earth

If g = g(p) for planet , range  R =  v²sin2θ / g(p)..................(1)

The gravitational force of attraction = weight force

Gm² /d² = m g(p)

g(p) = Gm/d².........................(2)

For R = d/3, from equation (1), we have

d/3 =  v²sin2θ / g(p)

Plug the expression for g(p) , we get

v = √ [Gm/3dsin2θ ]

For velocity to be minimum, sin2θ =1

So, the minimum velocity will be

v = √ [Gm/3d]

Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]

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The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1  and q2 = charges  =  1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F =  [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]

F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]

F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]

F = [tex]6. 426 * 10^-11[/tex] Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

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The work done in pulling the sled is 445,930.2 Joules.

The acceleration in the x direction, assuming that friction is negligible, is  20.85 m/s².

Time taken to pull the sled 185 m is 4.21s.

Work done is equal to product of force applied and distance moved.

Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.

Work = Force x Distance x cos(angle)

W= 1225 x 385 x cos 19°

W = 445,930.2 Joules

Thus, the  work done in pulling the sled is 445,930.2 Joules

From the Newton's second law of motion, we have'

F = ma

acceleration, a = 1225cos19° / ( 545 /9.81)

a = 20.85 m/s²

Thus, the acceleration in the x direction is  20.85 m/s²

Using the second equation of motion, we get

s = ut+ 1/2 at²

Substitute the values, we have

185 m = 0 +1/2 x 20.85 x t²

t = 4.21 s

Thus, the time taken to pull the sled 185 is 4.21s.

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Answer:

Explanation:

Solution

In physics, a period is a measurement of time. The answer must be either B or C. It is a direct measurement of time, not its reciprocal. So the answer is B. A and D is a measurement of the actual cycle that it takes before the motion begins to repeat itself.

Answer

B

A. The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 4 i m/s.

B. The velocity and acceleration at any time t is  4t i +1 j and 4i.

C. The average acceleration in the time interval given in part (a) is 4

Acceleration can be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given the following equation shows the position of a particle in time t, x=at² i + bt j

where t is in second and x is in meter. a=2m/s², b=1m/s.

A. Velocity is time rate of change of displacement.

V = dx/dt

V = d/dt (2t² i + 1t j)

V = 4t i +1 j

Velocity in time interval,

At  t₁=2sec , V₂= 4x2 i + 1 j = 8 i + 1 j

At  t₂=3sec, V₃ = 4x3 i + 1 j = 12 i + 1 j

Average velocity in time interval t₁=2sec and t₂=3sec is

, V₃ - V₂ = (12 i + 1 j) - ( 8 i + 1 j )

Average velocity =  4 i

B. Velocity at time t, is

V(t) = V = 4t i +1 j

Acceleration a = dV/dt

a(t) = d/dt (4t i +1 j)

a(t) = 4 i +0 j

C. Average acceleration in time interval t₁=2sec and t₂=3sec is

a = 4 m/s²

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Answer:(note that values will be on top in small)

cobalt :- 60Co

potassium :- 40K

neon :- 24Ne

lead :- 208Pb

An object, which weighs 98 N on the surface of Earth, weigh on this point is 2.23 x 10²⁴ kg.

The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.

Given, the Earth has the acceleration due to gravity, g =  9.81 m/s².

Force of gravity W = mass x acceleration due to gravity

98N = m x 9.8m/s²

m = 10 kg

The value of acceleration due to gravity (g) on a point 10,000 km above sea level is about 1.49 m/s².

The acceleration due to gravity and mass is related as

g = GM/R²

where G = gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg² and R is the distance between two masses.

Substituting the values, we get

1.49 m/s² =  6.67 x 10⁻¹¹ N.m²/kg² x M/ (10000 x 10³ m)²

M = 2.23 x 10²⁴ kg

Therefore, an object will weigh on this point is 2.23 x 10²⁴ kg

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Answer:  Solar Photovoltaic (PV) cells generate electricity by absorbing sunlight and using that light energy to create an electrical current. There are many PV cells within a single solar panel, and the current created by all of the cells together adds up to enough electricity to help power your school, home and businesses.

                              --Cited from Solar Schools

The magnitude of the force exerted by the left cube on the right cube is 17.64N.

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.

Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.

From the equilibrium of forces in vertical direction

Normal force N= 2m x g

friction force f = μN =μ(2m)g

From the equilibrium of forces in horizontal direction

F₁ =ma =0

using Newton's third law of motion, we get

F₁  - f =0

F₁  =f = μ(2m)g

Put the values, we get

F₁  = 17.64N

Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.

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The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

The given data in the problem is given by ;

u is the initial speed =  0 m/sec

v is the final speed= 35  m/sec

t is the time interval= 1.2 second

a is the acceleration=? m/sec²

The formula for acceleration is;

[tex]\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2[/tex]

Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

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The value of the electric current for the given conditions willl be  1939.02 A.

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 distance from the rounded tip of the plate to the center of mass is 1.87 m.

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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Answer:

The last one - each  vector pointing towards the center of the circle must be the same length for uniform circular motion

(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.

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

where;

W(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;

W(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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The final speed of the nickel at the given quantity of heat is determined as 202.1  m/s.

Apply the principle of conservation of energy.

Q = mcΔθ

Q = (18)(0.444)(66 - 20)

Q = 367.63 J

Q = K.E = ¹/₂mv²

2K.E = mv²

v = √(2K.E/m)

where;

v = √(2 x 367.63)/(0.018))

v = 202.1 m/s

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Answer:

the answer is ambivert have a good fourth of July

Answer:

i believe the answer is ambivert

Answer:

0.6192

Explanation:

1.44 x 0.430=0.6192

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 term for a pair of forces described by Newton's third law is action-reaction pair. Details about Newton's law can be found below.

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

This law proposed that when an object 1 acts on object 2 with a force of a particular magnitude, object 2 also acts on object 1 with an opposite force of same magnitude.

Therefore, it can be said that the term for a pair of forces described by Newton's third law is action-reaction pair.

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The average force exerted on the ball by the glove will be 129 N.

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.

Given data;

Mass,(m)=0.15-kilogram

Initial speed,u=43 meters per second

Time period,t= 0.050 seconds

Average force exerted,F=?

The average force exerted on the ball by the glove is;

[tex]\rm F = m\frac{v-u}{t} \\\\ F=0.15( \frac{43-0}{0.050} )\\\\ F=129 N[/tex]

Hence, the average force exerted on the ball by the glove will be 129 N.

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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°