what is the radius of the 5th orbital in hydrogen?

Answer:

54.0 x0.1=5.4 x0.03=0.162

kinetic force

If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,

The force required to get the block moving = μMg

                                                                          = 0.01×54×9.81

                                                                         = 5.2974 Newtons

Thus, the minimum force required to move the block would be 5.2974 Newtons.

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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

The time for which the baseball remains in the air will be 9.37 sec. Option C is correct.

It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.

From Newton's first equation of motion;

[tex]\rm v_y= u_y+gt \\\\ 0 = usin \theta - gt \\\\ t = \frac{u sin\theta }{g}\\\\t = 4.68 \ sec[/tex]

The time the baseball remains in the air will be twice the time he travels for one side motion;

T' = 2t

T'=2 × 4.68 sec

T'= 9.37 sec

Hence for 9.37 sec, the baseball willl remains in the air. Option C is correct.

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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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[tex]\textsf {C. Force}[/tex]

[tex]\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}[/tex]

Answer:

the answer to the question is a reflector

To deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.

The power generated in a circuit is directly proportional to the current flowing in the circuit.

P = IV

where;

Voltage in the circuit also increases with increase in the current flowing in the circuit.

Thus, to deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.

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Answer:equal mass

Explanation:the mass of both is 5kg

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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Answer:

Explanation:

Hope I give all correct answer please mark as brainlest answer

C. The Densities are equal.

Density is mass per unit volume or mass of a unit volume of a material substance.

If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]

Therefore,

[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]

Hence, D1 = D2

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The net force on the crate will be 99 N.Option B is correct.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration.Mathematically in the different components and balancing the equation gets.Components in the x-direction.

When all the forces get resolved the y-direction of forces are;

300 sin 10° +N = 445

N = 445-295.4

N = 392.9 N

The normal force is 392.9 N.

The net force on the crate is found by resolving the force in the x-direction;

F = 300 cos 10° - μN

F= 295.44-196.45

F= 98.99

F=99

Hence, the net force on the crate will be 99 N.Option B is correct.

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(a) The time spent in the air by the ball is 1.5 seconds.

(b) The horizontal distance of the ball is 9.83 m.

(c) The impact speed of the ball is 20.37 m/s.

(2) The magnitude of the force is 253.5 N in horizontal direction.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

18 = 8sin(35)t + (0.5)(9.8)t²

18 = 4.589t + 4.9t²

4.9t² + 4.589t - 18 = 0

solve the quadratic equation using formula method,

t = 1.5 s

X = vcosθ(t)

X = 8cos(35) x 1.5

X = 9.83 m

vyf = vyi + gt

vyf = 8sin(35) + 9.8(1.5)

vyf = 19.289 m/s

vxf = vxi

vxf = 8 x cos(35)

vxf = 6.55 m/s

Resultant speed = √(19.289² + 6.55²) = 20.37 m/s

F = ma

F = 65 x 3.9

F = 253.5 N

Since the motion is horizontal, the angle of the motion is zero.

Fx = 253.5cos(0) = 253.5 N

Fy = 253.5sin(0) = 0 N

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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

Explanation:

this is the answer 276 volts

The Total assets based on the given table is $88,800.

An asset refers to a resource owned by an individual or company which has tangible value and whose value is expected to rise in the future.

The Total assets based on the table = Cash + Supplies + Prepaid rent + Equipment

Total assets = 13,100 + 6,000 + 3,800 + 65,900

Total assets = $88,800

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By using an electric field, it is feasible to differentiate between these different forms of radiation.

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.

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The charge on each sphere is 742 n

In physics, Charge, also known as electric charge is a electrical charge, or electrostatic charge and symbolized q, is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.

According to the question,

Distance = 15cm or 0.15m

Force =  0.22N

[tex]K=9*10^{9} kgm^2/c^2[/tex] ( this is a constant value of K)

By using the formula,

F = [tex]\frac{kQq}{d^{2} }[/tex]

0.220 N = 8.99

N·m²/C² * Q² / (0.15m)²

Q = 7.42e-7

C = 742 n

C is the charge of each sphere .

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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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The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u= 0 m/s

Final velocity, v= 4.2 m/s

Time elapsed, t = 7.3 seconds.

To find ;

Acceleration, a

The acceleration when the change in velocity is observed by the formula as:

a= (v-u)/(t)

Substitute the given values:

a= (4.2-0)/(7.3)

a=(4.2)/(7.3)

a= 0.575 m/s²

Hence, the car's acceleration will be 0.575 m/s².

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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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(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 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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The acceleration of this sled can be calculated by doing: option A, B, E and F.

The acceleration of this sled can be calculated by using Newton's Second law of motion. Mathematically, the acceleration of an object is given by this formula:

Net force = Mass × acceleration

Deductively, the acceleration of this sled can be calculated by doing the following:

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Complete Question:

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.

A. The force of gravity must be broken into its parallel and perpendicular components.

B. Acceleration can be found by dividing the net force by mass.

C. Acceleration can be found by multiplying the net force by mass.

D. The magnitude of the force components must be multiplied by gravity.

E. Trigonometry can be used to solve for the magnitude of the force components.

F. Net force must be found before acceleration can be found.

Answer:

To find the acceleration, you do 20N/50kg = 0.4 m/s^2

Explanation:

example: a = F/m = 10/2 = 5 m/s2

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 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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(a) The acceleration of the two block is determined as 0.93 m/s².

(b) The tension in the string is 5.78 N.

The net force on the aluminum block is calculated as follows;

[tex]T - \mu_a m_a g = m_a a \ --- \ (1)[/tex]

The net force on the copper block is calculated as follows;

[tex]m_cg sin(30) - T -\mu _cm_c gcos(30) = m_c a --- (2)[/tex]

where;

Solve for T using (1) and (2)

[tex]a = g(\frac{m_c sin30\ - \ \mu _c m_c cos30 \ - \mu_ a m_a}{ma_a + m_c} )\\\\a = 9.8(\frac{6 sin30\ - \ 0.2 (6) cos30 \ - 0.2 (6)}{2 +6} )\\\\a = 0.93 \ m/s^2[/tex]

From equation (1);

T = μm_ag + m_aa

T = 0.2(2)(9.8) + 2(0.93)

T = 5.78 N

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Answer:

enable us to take in a wide variety of information, and process it based on our past experience and values.

Explanation:

Answer:

A 2kg block is moved a long a level floor by a horizontal force of 20N if the coefficient of sliding friction is 0.4

what is the acceleration of the blove

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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