A wheel accelerates from rest to 20 rad/s at a uniform rate of 3.5 rad/s2. Through what angle (in radians) did the wheel turn while accelerating

Answer:

Initial velocity, u = 60 km/h = 16.7 m/s

Final velocity, v = 72 km/h = 20 m/s

time, t = 2 sec

From first equation of motion:

[tex]{ \bf{v = u + at}}[/tex]

Substitute the variables:

[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

All the energy in oil, gas, and coal originally came from the sun captured through photosynthesis. In the same way that we burn wood to release energy that trees capture from the sun, we burn fossil fuels to release the energy that ancient plants captured from the sun.

Answer:

Everything on the planet will move twice faster than on Earth.

Explanation:

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

Answer:

false?

Explanation:

The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.

Answer:

I think the answer is False.

Answer:

If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.

Explanation:

Answer:

   x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]

Explanation:

This is an exercise in fluid mechanics, let's find the speed of the water in the hole

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

where the subscript 1 is for the tank surface and the subscript 2 is for the depth of the hole

the pressure inside and outside the tank is the same

       P₁ = P₂

we must measure the distance from the same reference point, let's locate it on the surface of the water

       y₁ = 0

       y₂ = h

Suppose the gap is small compared to the diameter of the tank

        v₁ «v₂

       v₂² = 2 g (0-h)

This is the speed of the outlet water in the tank, as the force is horizontal this speed is horizontal.

Let's use the projectile launch ratios

         vₓ = [tex]\sqrt{2g |h|}[/tex]

         y = y₀ + v₀ t - ½ g t²

the height when reaching the floor is y = 0,

the initial height is measured from the floor therefore y₀ = H-h

        0 = (H-h) + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2(H-h)}{g} }[/tex]

we look for the distance x traveled

        x = vₓ t

        x = [tex]\sqrt{2g |h| } \ \sqrt{ 2(H-h)/g}[/tex]

        x = [tex]\sqrt{4 h (H-h)}[/tex]

        x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]

Answer:

The acceleration is in 2 D as in between east and south.

Explanation:

mass, m = 50 kg

acceleration, a = 0.25 m/s^2 horizontal

acceleration of elevator, a' = 1 m/s^2 downwards

When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:

Answer:

In the given circuit, R

2

,R

6

and R

4

are in series. So,

R

1

′

=7+5+12=24Ω

Now R

1

′

and R

5

′

are in parallel. So,

R

2

′

1

=

8

1

+

24

1

=

24

3+1

=

24

4

=

6

1

R

2

′

=6ohm.

Now R

2

′

,R

1

and R

3

are in series. So,

R=R

2

′

+R

1

+R

3

=6+3+2=11ohm.

We know i=

R+r

E

=

11+1

6

=

12

6

=

2

1

i=0.5amp.

Answer:

imma try and fail again and again

Explanation:

It isn't possible to learn an entire language fast. For an examination, memorise some of the easy words. Practice mew minutes speaking French and keep speaking it till you know it.

Answer:

[tex]D'=2.933*10^{18}m[/tex]

Explanation:

From the question we are told that:

Distance [tex]D=310ly[/tex]

Where

[tex]1ly=9.46*10^{15}[/tex]

Generally the equation for  meters will you traverse to reach your vacation destination is mathematically given by

[tex]D'=D*1ly[/tex]

[tex]D'=310*9.46*10^{15}[/tex]

[tex]D'=2.933*10^{18}m[/tex]

Answer:

Explanation:

Changing the resistance of a resistor means the resistance is either increased or decreased.

a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.

Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;

V = IR

6 = I x 30

I = 0.2 A

If the resistance is changed to 50 Ohm's, then:

I = 0.12 A

(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.

In the previous example, if the resistance is changed to 5 Ohm's, then:

V = IR

6 = I x 5

I = 1.2 A

(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.

In the Homogenous equation with respect to its unit if the constants has been removed, then the equation is still homogenous but it is incorrect.

         Homogenous equation is one of the type of the differential equations. These equation have zero on the right hand side of the equality sign. If the equation has the independent variable on the right side of equal sign then it is said to be non - homogenous equation. The homogenous and non - homogenous equations are the two types of linear equations.

An example of Homogenous equation is,

The formula for kinetic energy can be represented as,

K. E = 1/2 (mv²)

Here, 1/2 can be removed as it was a constant.

K. E = mv².

This equation can be homogenous but still it is incorrect.

Learn more about homogenous equation,

https://brainly.com/question/12884496

#SPJ2

Answer:

[tex]F=202650N[/tex]

Explanation:

From the question we are told that:

Area [tex]a=50m^2[/tex]

Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]

Generally the equation for Force is mathematically given by

[tex]F=dP*A[/tex]

[tex]F=4053*50[/tex]

[tex]F=202650N[/tex]

Answer:

[tex]Emf=24.4V[/tex]

Explanation:

From the question we are told that:

Angular speed [tex]\omega=40*\omega'[/tex]

Radius [tex]r=0.33m[/tex]

No. Turns [tex]N=163[/tex]

Area [tex]A= 3.23 * 10^{-3} m^2[/tex]

Magnetic field. [tex]B=0.0948T[/tex]

Acceleration [tex]a=0.60m/s^2[/tex]

Time [tex]t=6.72s[/tex]

Generally the equation for momentum is mathematically given by

[tex]\omega'=\omega_o+\frac[a}{r}t[/tex]

[tex]\omega'=0+\frac{0.60}{0.33}*6.72[/tex]

[tex]W=12.22rad/s^2[/tex]

Therefore

[tex]\omega=Aw[/tex]

[tex]\omega=12.22*40[/tex]

[tex]\omega=488.7rads/s[/tex]

Generally the equation for Peak emf is mathematically given by

[tex]Emf=NBA \omega[/tex]

[tex]Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s[/tex]

[tex]Emf=24.4V[/tex]

Answer:

Quantum theory gives the concept of

Answer:

[tex]l=12916.5m[/tex]

Explanation:

Distance [tex]d=3600km[/tex]

Since

Density of steel [tex]\rho=7900kg/m^3[/tex]

Stress of steel [tex]\mu= 1*10^9[/tex]

Generally the equation for Stress on Cable is mathematically given by

[tex]S=\frac{F}{A}[/tex]

[tex]S=\frac{\rho Alg}{A}[/tex]

Therefore

[tex]l=\frac{s}{\rhog}[/tex]

[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]

[tex]l=12916.5m[/tex]

Answer:

The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.

It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is

∑ F = (force from engine) - (resistive forces) = 0

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{\rm total}=\Delta K[/tex]

or

[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]

Compute the work performed by friction:

[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]

[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]

[tex]\implies \boxed{\mu\approx0.45}[/tex]

The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{total}=\Delta K[/tex]

or

[tex]W_{friction}+W_{spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]

Compute the work performed by friction:

[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{friction}=-f(0.20)[/tex]

[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]

[tex]\mu =0.45[/tex]

Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

To know more about Coefficient of friction follow

https://brainly.com/question/136431

Answer:

The correct answer is "121.36 meters and 2.40 MHz".

Explanation:

Given:

Red shift,

[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]

Wavelength,

[tex]\lambda = 125 \ meters[/tex]

The observed frequency will be:

⇒ [tex]f = \frac{c}{\lambda}[/tex]

      [tex]=\frac{3\times 10^8}{125}[/tex]

      [tex]=24\times 10^5[/tex]

      [tex]= 2.40\times 10^6[/tex]

      [tex]=2.40 \ MHz[/tex]

hence,

The Emitted wavelength will be:

⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]

   [tex]\frac{125}{\lambda_0}-1 =0.03[/tex]

         [tex]125=1.03 \ \lambda_0[/tex]

           [tex]\lambda_0=\frac{125}{1.03}[/tex]

                [tex]=121.36 \ m[/tex]

Answer:

In heterogeneous mixture do the physically distinct component parts each have distinct properties.

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

Consulta los dibujos adjuntos

Answer:

A

Explanation:

So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.

Answer:

Len's law

Explanation:

We can explain this exercise using Len's law

when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.