How long will it take a 50w motor to lift a 20 kg object 5 m?

Initial speed of ball will be 2.7m/s.

When a body is rotating about an axis, then it has kinetic energy.

And this energy is called rotational kinetic energy.

It is given as -  R.K.E. = 1/2 Iω²

And if a ball is rolling without slipping.

Then the moment of inertia of the solid ball is written as -

I = 25MR²

Vi = Rω

Here it is given in the problem that-

height(h) = 0.53m

Now by the conservation of energy we can write the equation as -

1/2MVi² + 1/2Iω² = Mgh

so that -

(1/2)MVi² + (1/2)×(2/5MR²) ×(Vi/R)² = Mgh(1/2)Vi² + (1/5)Vi²

= gh(7/10)Vi² = 9.8 × 0.53

Vi = 2.7 m/s

So that the initial velocity of ball came out to be 2.7m/s after applying all concepts or rotational motion.

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Answer:

A

Explanation:

As it begins to fall

F = ma        a = 9.81

F = .21 * 9.81 = 2.06 N

The definition of force in physics is the push or pull on a massed object that changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. The correct option is A.

Force is a physical factor that alters or has the potential to alter an object's state of rest or motion as well as its shape. Newton is the SI unit of force.

One of the most fundamental types of physical entities is the force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction. Frictional force and contact force are the two categories under which all forces can be classified.

F = m × a

F = 0.21 × 9.81

F = 2.06 N

Thus the correct option is A.

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Answer:

Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.

Explanation:

Answer: Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.

Explanation:

Answer:

1. Information Gathering

2. Analysis and Planning.

3. Implementation of a solution.

4. Assessment of the effectiveness of the solution.

5. Documentation of the incident.

The horizontal force required to move the crate at a steady speed across the floor will be 58.86 N.

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).

The given data in the problem is;

Mass,m = 24 kg

The coefficient of kinetic friction, μ=0.25

The friction force is defined as the product of the coefficient of friction and normal reaction.

f=μN

If there is no other force, the horizontal force is equal to the friction force;

F=f

The horizontal force required to move the crate at a steady speed across the floor is found as;

F = μN

The normal reaction force is;

N=W

N=mg

F = μmg

F=0.25 ×  24-kg × 9.81 m/s²

F = 58.86 N

Hence horizontal force is required to move the crate at a steady speed across the floor will be 58.86 N.

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Their velocity (in m/s) just after impact if they cling together is -0.9 m/s.

Apply the principle of conservation of linear momentum;

Pi = Pf

where;

(92.5)(5.5)  +  (116)(-6)  =  v(92.5 + 116)

508.75 - 696 = v(208.5)

-187.25 = 208.5v

v = -187.25/208.5

v = -0.9 m/s

Thus, their velocity (in m/s) just after impact if they cling together is -0.9 m/s.

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Answer:

A. To determine jobs for people

The acceleration and time is mathematically given as

a=gsinθ

[tex]t=\frac{2d}{gsinθ}[/tex]

Generally, the equation for the Force of the car is mathematically given as

[tex]N = mg cos\theta[/tex]

Therefore

[tex]ma=mgsin\theta[/tex]

a=gsinθ

The acceleration of the car is mathematically given as

a=gsinθ

Generally, the equation for the final velocity of the car is mathematically given as

[tex]v^2−u^2=2as[/tex]

Therefore

[tex]v 2 −(0) 2 =2(gsin\theta)d[/tex]

[tex]v= \sqrt {2gsin\theta.d}[/tex]

​

Generally, the equation for the motion of the car is mathematically given as

v−u=at

[tex]2gsin\thetad =(gsin\theta)t[/tex]

[tex]t=\frac{2d}{gsinθ}[/tex]

In conclusion, the time it takes the front bumper to reach the bottom of the hill is

[tex]t=\frac{2d}{gsinθ}[/tex]

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The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R[/tex]

where [tex]m=4.10\,\rm kg[/tex], [tex]v=2.85\frac{\rm m}{\rm s}[/tex], and [tex]R[/tex] is the radius of the circular path.

As shown in the diagram, we can see that

[tex]\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)[/tex]

where [tex]r=1.69\,\rm m[/tex], so that

[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}[/tex]

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

[tex]F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}[/tex]

Solve for [tex]\theta[/tex] :

[tex]\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0[/tex]

Complete the square:

[tex]\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}[/tex]

Plugging in the known quantities, we end up with

[tex]\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27[/tex]

The second case has no real solution, since [tex]-1\le\cos(\theta)\le1[/tex] for all [tex]\theta[/tex]. This leaves us with

[tex]\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}[/tex]

(A)The total resistance of the parallel resistors is: [tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]

(B) The current in the overall circuit is: [tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]

(C) The current through each resistance is as follows:

For the three resistances connected in parallel, R1, R2, R3 , the total resistance, [tex]R_{T}[/tex] is calculated as follows:

[tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]

The current in the overall circuit is calculated using the formula:

[tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]

The current through each resistance is given as follows;

Current through R1, [tex]I_{1} = \frac{V}{R_{1}}[/tex]

Current through R2; [tex]I_{2} = \frac{V}{R_{2}}[/tex]

Current through R3; [tex]I_{3} = \frac{V}{R_{3}}[/tex]

In conclusion, the voltage across resistances in parallel is the same but the current varies.

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The acceleration is 2.2 m/s^2 and the net force is  62N towards the right

The net force is the effective force that acts on a body in a give direction.

1. The net force in the y axis is; ∑Fy = 0

2. The net force along the y axis is; ∑Fx =  277 N - 215 N

3. The normal reaction is given by;  28 kg * 9.8 m/s^2 = 274.4 N

4. The net force in the x axis is  277 N - 215 N = 62N towards the right

5. The acceleration of this force is obtained from;

F = ma

62 = 28 a

a = 62/28

a = 2.2 m/s^2

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(a) The peak voltage across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Vrms = 0.7071V₀

where;

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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a.

b.

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

Since velocity of spring v = ω√(A² - x²) where

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

Since velocity of spring v = ω√(A² - x²) where

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

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Answer:

dependent variable

Explanation:

The ball's height [tex]y[/tex] at time [tex]t[/tex] is given by

[tex]y = 29.2\,\mathrm m - \left(8.30\dfrac{\rm m}{\rm s}\right) t - \dfrac12 gt^2[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex].

Solve for [tex]t[/tex] when [tex]y=0[/tex]. Omitting the units, we have

[tex]29.2 - 8.30t - \dfrac g2 t^2 = 0[/tex]

I'll solve by completing the square.

[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t\right) = 0[/tex]

[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t + \dfrac{8.3^2}{g^2}\right) = -\dfracg2 \times \dfrac{8.3^2}{g^2}[/tex]

[tex]29.2 - \dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = -\dfrac{8.3^2}{2g}[/tex]

[tex]\dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = 29.2 + \dfrac{8.3^2}{2g}[/tex]

[tex]\left(t + \dfrac{8.3}g\right)^2 = \dfrac{58.4}g + \dfrac{8.3^2}{g^2}[/tex]

[tex]t + \dfrac{8.3}g = \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]

[tex]t = -\dfrac{8.3}g \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]

[tex]\implies t \approx -3.43 \text{ or } t \approx 1.74[/tex]

Ignore the negative solution; the ball hits the ground about 1.74 s after being thrown.

Answer:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

  The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

Explanation:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

  The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

(a) The angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.

(b) The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .

The angular acceleration of the rod is calculated as follows;

Apply the principle of conservation of angular momentum;

τ = Iα

where;

τ = Fr = mg(L/2) cosθ

I = mL²/3

mg(L/2) cosθ =  mL²/3(α)

g(1/2) cosθ =  L/3(α)

(³/₂)g cosθ  = L(α)

(³/₂ L)g cosθ  = α

1.5Lg cosθ  = α

1.5(2.1) cos (70.4) = α

1.06 rad/s² = α

ωf² = ω₁² + 2αθ

ωf² = 0 + 2(1.06)(70.4π/180)

ωf² = 2.605

ωf = √2.605

ωf = 1.61 rad/s

Thus, the angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.

The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .

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The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.

The given data in the problem is;  

Bending stress, σ = 120 N/mm2  

Moment of inertia, I = 8.5 × 106 mm⁴

Depth of beam, y = d/2 = 200/2 = 100 mm  

Length of beam, L = 8 m = 8000 mm  

Width of beam, W = 300 mm

The maximum bending moment of the beam with UDL;

[tex]\rm W = \frac{wL^2}{8}[/tex]

From the bending equation;

[tex]\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm[/tex]

The maximum bending moment of the beam with UDL;

[tex]\rm M = \frac{wL^2}{8} \\\\ 10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \ N/mm[/tex]

Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

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The frequency f of the AC source is determined as 0.446 Xc.

The frequency of the AC source is calculated as follows;

Xc ≡ 1/2fC

where;

fC = 2Xc

f = 2Xc / C

f = (2Xc)/4.48

f = 0.446 Xc

Thus, the frequency f of the AC source is determined as 0.446 Xc.

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If the metal ion has a 3+ charge and the nonmetal ion has a 4- charge, the formula of the compound is X4Z3.

The formula of an ionic compound is determined by the charges of the ions that make up the compound. Recall that when the compound is  formed, the ions exchange valences.

Hence, is the metal ion has a 3+ charge and the nonmetal ion has a 4- charge, the formula of the compound is X4Z3.

Missing parts:

A metal ion (X) with a charge of 3+ is attracted to a nonmetal ion (Z) with a charge of 4-. Which of these formulas represents the resulting compound.

A. 7XZ

B. X3Z4

C. X4Z3

D. 4X3Z

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The upward force exerted on the board by the support is mathematically given as

Fu= 764.8 N

Generally, the equation for is  mathematically given as

Considering that the Net Force on the system is null

The weight of the children plus the weight of the board equals the upward force imposed on the support.

The upward force

Fu= 440 + 272 + 52.8 N

Fu= 764.8 N

In conclusion, he upward force exerted on the board by the support

Fu= 764.8 N

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The bus will travel a further 20 m before coming to rest.

The term acceleration has to do with a change in velocity with time. Now we have;

u = 20 m/s

v = 10 m/s

s =  60 m

Now;

v^2 = u^2 -2as

v^2 -  u^2 = -2as

(10)^2 - (20)^2 = - 2 * a * 60

a = (10)^2 - (20)^2/ - 2 * 60

a = 100 - 400/ - 2 * 60

a = 2.5 m/s^2

At that time;

v = 0 m/s

u = 20 m/s

a = 2.5 m/s^2

s = ?

Hence;

u^2 = 2as

s = u^2/2a

s = (20)^2/ 2 *  2.5

s = 400/5

s = 80 m

Hence, the bus will travel a further 20 m before coming to rest.

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The change in kinetic energy of the proton is 4.8 x 10⁻²⁰.

The change in kinetic energy of the proton is calculated as follows;

ΔK.E = W = eV

where;

V = Ed

V = 3 V/m x 0.1 m = 0.3 V

ΔK.E  = 1.6 x 10⁻¹⁹ x 0.3 = 4.8 x 10⁻²⁰ J

Thus, the change in kinetic energy of the proton is 4.8 x 10⁻²⁰ J.

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The coasts of Africa and South America were touched  125 million years ago due to continental drift.

The time can be computed from the ratio of distance and speed.

Given:

Distance = 5, 000,000

Speed = 4 cm/year = 0.04 m/year

The time is computed as:

Time = Distance / Rate

Time = 5,000,000 meters / 0.04 meters per year

Time = 125,000,000 years

Hence, the coasts of Africa and South America were touched  125 million years ago due to continental drift.

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Boiling points are raised by hydrogen bonds because they make different molecules desire to "attach" to one another, which requires more energy to do so. In water, for instance, the hydrogen proton is in a state that resembles ionization because the connections between oxygen and hydrogen, while covalent, are strongly polar. The oxygen also receives a partial negative charge. Therefore, hydrogen bonds are formed when the electro-positive H in one molecule is strongly electrostatically attracted to the electro-negative O in nearby molecules. Despite being weak links, they are powerful enough to significantly alter the liquid's characteristics.  

Thanks!

>> ROR

Heat transfer in a closed system is the addition of changes in internal energy and the total amount of work done by it.  The final energy of the system is 35.5kJ.  

Heat transfer is the transfer of heat energy due to temperature differences.

The paddle-wheel paintings are quantities of workdone, 500 N.m or 0.5kJ.

The preliminary (initial) power of the device is 10 kJ.

Total warmness transferred in the course of the method is 30 kJ

Total warmness misplaced in the course of the method to the encompassing air is 5 kJ.

The energy of the system is given as:

The energy of the system = Energy in - Energy out

The energy of the system = Initial energy + Energy transferred + Work done - Energy lost

Energy of the system = 10 + 30 + 0.5 - 5 kJ

Energy of the system = 35.5 kJ

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Time taken by the ball to reach the ground is 1.8s.

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

A ball is thrown directly downward with an initial speed of 7.30 m/s, from a height of 29.0 m

Using the second equation of motion,

s = ut + 1/2 at²

Plug the values, we get

29 = 7.3t + 1/2 x9.81t²

4.905t² + 7.3t -29 =0

t =1.79871 or −3.28699

As time can't be negative, time taken to strike the ground is  approximately 1.8 s.

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The  energy was lost to air friction is determined as 42.56 J.

The energy lost due to friction is calculated as follows;

ΔE = ΔK.E + ΔP.E

where;

ΔE = ¹/₂(0.45)(19.89² - 15.1²) + (0.45)(9.8)(12 - 30.2)

ΔE = 37.7 - 80.26

ΔE = -42.56 J

Thus, the  energy was lost to air friction is determined as 42.56 J.

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