A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?

Answer:

9. (B) ¼ Mv²

10. (A) √(3gL)

11. 20 N

12. 5 m/s²

Explanation:

9. The rotational kinetic energy is:

RE = ½ Iω²

RE = ½ (½ MR²) (v/R)²

RE = ¼ Mv²

10. Energy is conserved.

Initial potential energy = rotational energy

mgh = ½ Iω²

Mg(L/2) = ½ (⅓ ML²) ω²

g(L/2) = ½ (⅓ L²) ω²

gL = ⅓ L² ω²

g = ⅓ L ω²

ω² = 3g / L

ω = √(3g / L)

The velocity of the top end is:

v = ωL

v = √(3gL)

11. Sum of torques about the hinge:

∑τ = Iα

-(Mg) (L/2) + (T) (r) = 0

T = MgL / (2r)

T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)

T = 20 N

12. Sum of forces on the block in the -y direction:

∑F = ma

mg − T = ma

Sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a / R)

T = ½ Ma

Substitute:

mg − ½ Ma = ma

mg = (m + ½ M) a

a = mg / (m + ½ M)

Plug in values:

a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))

a = 5 m/s²

Answer:

Explanation:

As we've seen, humans have on average a heart rate of around 60 to 70 beats per minute, give or take. We live roughly 70 or so years, giving us just over 2 billion beats all up.Apr

Answer:

explained

Explanation:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.

Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.

Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at  5  m/s

straight toward the goal and drives the ball in the same direction with a velocity of  30 m/s

relative to her body, then the velocity of the ball is  35  m/s

relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (

The figure shows components of velocity v in horizontal  vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.

Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors  and  

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.

Answer:

I. light house 1 will be warmer during the day ii. light house 2 will be warmer at night.

Explanation:

Because the land conducts heat better than water the light house farthest away from the water will get hotter during as the ground will heat up faster than the water. But this also means that the ground will lose heat faster at night where the water won't making the light house closest to the water hotter at night.

Answer:

Energy is the strength and vitality required for sustained physical or mental activity.

Matter occupies space and possesses rest mass, especially as distinct from energy.

Hope this helps! （づ￣3￣）づ╭❤～

Answer:

a

  [tex]Q = 5.34 *10^{19} \ J[/tex]

b

   [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Explanation:

From the question we are told that

     The  area of  Manhattan is  [tex]a_k = 87.46 *10^{6} \ m^2[/tex]

      The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]

        The  thickness is  [tex]t = 500 \ m \\[/tex]

Generally the volume of the ice is mathematically represented is

         [tex]V = a_i * t[/tex]

substituting value

         [tex]V = 500 * 3.498*10^{8}[/tex]

         [tex]V = 1.75 *10^{11}\ m^3[/tex]

Generally the mass of the ice is

       [tex]m_i = \rho_i * V[/tex]

Here [tex]\rho_i[/tex] is the density of ice the value is  [tex]\rho _i = 916.7 \ kg/m^3[/tex]

=>   [tex]m_i = 916.7 * 1.75*10^{11}[/tex]

=>    [tex]m_i = 1.60 *10^{14} \ kg[/tex]

Generally the energy needed for the ice to melt is mathematically represented as

        [tex]Q = m _i * H_f[/tex]

Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is  [tex]H_f = 3.33*10^{5} \ J/kg[/tex]

=>    [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]

=>    [tex]Q = 5.34 *10^{19} \ J[/tex]

Considering part b

  We are told that the annual energy consumption is  [tex]G = 1.2*10^{20 } \ J / year[/tex]

So  the time taken to melt the ice is

      [tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]

        [tex]T = 0.445 \ years[/tex]

converting to days

      [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Answer;

A)S(t)=96t-16t² +432

B)it will take 9 seconds for the ball to reach the ground.

C)864feet

Explanation:

We were given an initial height of 432 feet.

And v(t)= 96-32t

A) we are to Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

S(t)= ∫(96-32)dt

S(t)=96t-16t² +K

S(t)=96t-16t² +432

In which the constant of integration K is the initial height, so K= 432

b) we need to know how long will the ball take to reach the​ ground

This is t when S(t)= 0

S(t)=96t-16t² +432

-16t² +96t +432=0

This is quadratic equation, if you solve using factorization method we have

t= -3 or t= 9

Therefore, , t is the instant of time and it must be a positive value.

So it will take 9 seconds for the ball to reach the ground.

C)V=s/t

Velocity= distance/ time

=96=s/9sec

S=96×9

=864feet

By applying the integrations,

(a) [tex]S = 96t-16t^2+432[/tex]

(b) Time will be "t = 9".

(c) Height will be "576"

Given:

Height,

Initial velocity,

According to the question,

(a)

Integrate v:

Initial Condition,

→ [tex]S = 96t-16t^2+432[/tex]

(b)

Hits the ground when,

→ [tex]0=96t-16t^2+432[/tex]

→ [tex]t =9[/tex]

(c)

Maximum height when,

→ [tex]0 = 96-32 t[/tex]

→ [tex]t = 3[/tex]

Now,

→ [tex]S = 96\times 3-16\times 3^2+432[/tex]

      [tex]= 576[/tex]

Thus the answer above is correct.

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Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

Answer:

692 J/kg/°C

Explanation:

Electric energy added = amount of heat

Power × time = mass × SHC × increase in temperature

Pt = mCΔT

(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)

C = 692 J/kg/°C

Answer:

56°

Explanation:

Brewsters angle can be simply derived from

n1sin theta1= n2sintheta2= n2costheta1

because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is

Tan theta= n2/n1

1.66/1.11= 1.495

Theta = 56°

Explanation:

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

Answer:gamma ray

Explanation:

Answer:

Explanation:

A machine is which no part of the work done on the machine is wasted, is called an ideal or perfect machine

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

Answer:

True

Explanation:

Because the resistors in series is the sum of the two resistors given as

R= R1+R2

While that of resistors in parallel is the sum of the reciprocal of the resistance given as

1/R = 1/ R1+ 1/R2

So that of series connection will be greater

One of the two requirements of electric current is there must be an electric potential between two bodies

For electric current to flow, there must be an electric potential between two bodies.

This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.

The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.

This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.

So, one of the two requirements of electric current is there must be an electric potential between two bodies.

So, the answer is A

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Answer:

See attached file

Explanation:

Answer:

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Explanation:

Given that:

the time period T = 88.5 min = 88.5 × 60  = 5310 sec

The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg

if  the radius of orbit is r,

Then,

[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e }{r}[/tex]

[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Similarly :

[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]

where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Then:

[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]

[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]

[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]

[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]

[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]

[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]

[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]

[tex]r= 2565.38^2[/tex]

r = 6579225 m

The  circumference of the satellites  orbit can now be determined by using the formula:

 circumference = 2π r

 circumference = 2π  × 6579225 m

 circumference = 41338489.85 m

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Answer:

1.227 m

Explanation:

Given that

Minimum wavelength is 410 nm

Maximum wavelength is 750 nm

Grating is 7800 slits/cm

Distance is 3.2 m

To solve this question, we would use the formula

sin θ = λ/d

sin θ = (410*10^-9) / (0.01/7800)

Sin θ = 410*10^-9 / 1.282*10^-6

Sin θ = 0.32 and θ = 18.67 degrees

For the second wavelength = 750 nm

sin θ = [(0.32x750)/410]

sin θ = (240 / 410)

sin θ = 0.5853 or

θ = 35.8 degrees

And finally, the width of spectrum would be

3.2[tan 35.8 - tan 18.67]

3.2 * 0.3833

= 1.227 m

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

Explanation:

When an ideal gas undergoes a slow isothermal expansion, following phenomenon occur

1. Work done bu the gas = Energy absorbed as heat.

2. Work done by environment = Energy absorbed as heat.

3. Increase in internal energy= Heat absorbed= work done by gas = work done by environment.

Hence all option are correct.

Increase in internal energy is equal to the heat absorbed or work done by gas or environment. All the statements are correct.

If an ideal gas undergoes a slow isothermal expansion,

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Answer:

a) 6.8--5.10 thats equal 11.9

b) m=ris/run +10 equal 0.06/8 =7.5*10^-3

Answer:

T_object = 380.35 K

Explanation:

From Stefan–Boltzmann law, the power output is given by the formula:

P = σAT⁴

where;

σ is Stefan-Boltzmann constant

A is area of the radiating surface.

T is temperature of the body

Now, we are told that the power the object emitted is 3 times the power absorbed from the room.

Thus, we have;

P_e = 3P_a

Where P_e is power emitted and P_a is power absorbed.

So, we have;

σA(T_object)⁴ = 3σA (T_room)⁴

σA will cancel out to give;

(T_object)⁴ = 3(T_room)⁴

We are given T_room = 289 K

Thus;

(T_object)⁴ = 3 × 289⁴

(T_object) = ∜(3 × 289⁴)

T_object = 380.35 K

Answer:

Magnetic fields exist near a magnet, farther away from a magnet, and within a magnet.

So, the answer is D. All of the above.

Let me know if this helps!

Answer:

[tex]\theta=34 \textdegree[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=55kg[/tex]

Angle [tex]\theta =28.0[/tex]

Coefficient of static friction [tex]\alpha =0.680[/tex]

Generally, the equation for Newtons second Law is mathematically given by

For

[tex]\sum_y=0[/tex]

[tex]N=mgcos \theta[/tex]

for

[tex]\sum_x=0[/tex]

[tex]F_{s}=mgsin\theta[/tex]

Where

[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]

[tex]F_{s}=0.68*55*9.8*cos 28[/tex]

[tex]F_{s}=323.62N[/tex]

Therefore

[tex]\alpha mgcos \theta=mg sin \theta[/tex]

[tex]\theta=tan^{-1}(0.68)[/tex]

[tex]\theta=34 \textdegree[/tex]

(a) The static frictional force which holds the box in place is 323.62 N.

(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.

The net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.

∑F = 0

The static frictional force is calculated as follows;

Fs = μFncosθ

Fs = 0.68 x (55 x 9.8) x cos28

Fs = 323.62 N

Fn(sinθ) - μFn(cosθ) = 0

mg(sinθ) - μmg(cosθ) = 0

μmg(cosθ) = mg(sinθ)

μ(cosθ) = (sinθ)

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(μ)

θ = tan⁻¹(0.68)

θ = 34.2⁰

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Answer:

 I = 7.96 W / m²

Explanation:

The light bulb emits a power of P = 100W, this power is distributed over the surface of a sphere, thus the emission is in all directions.

Intensity is defined by power per unit area

            I = P / A

The area of ​​a sphere is

         A = 4π r²

we substitute

         I = P / (4π r²)

in this case it tells us that the distance is r = 1 m

let's calculate

        I = 100 / (4π 1²)

        I = 7.96 W / m²

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.