The driver provides a constant force on the engine through the foot pedal. Eventually the van stops accelerating and reaches a constant speed.
c Explain why the van reaches a constant speed if the driver provides a constant driving force to the van.

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed,  v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]

Therefore, the maximum speed is 0.20 m/s

Answer:

a)  [tex]F=475.7Hz[/tex]

b)  [tex]F'=410.899Hz[/tex]

Explanation:

From the question we are told that:

Velocity of eagle [tex]V_1=35m/s[/tex]

Frequency of eagle [tex]F_1=440Hz[/tex]

Velocity of Black bird [tex]V_2=10m/s[/tex]

Speed of sound [tex]s=343m/s[/tex]

a)

Generally the equation for Frequency is mathematically given by

 [tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]

 [tex]F=440(\frac{343-10}{343-35})[/tex]

 [tex]F=475.7Hz[/tex]

b)

Generally the equation for Frequency is mathematically given by

 [tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]

 [tex]F'=440(\frac{343+10}{343+35})[/tex]

 [tex]F'=410.899Hz[/tex]

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.

Answer:

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

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:

[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:

[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:

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]

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

Answer:

imma try and fail again and again

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!

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]

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:

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:

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:

hehe

Explanation:

I dont know because I am a noob ant study

Answer:

Quantum theory gives the concept of

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.

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:

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.

Answer:

38.33°C

Explanation:

Applying,

180/100 = (F-32)/C............. Equation 1

Where F = Temperature of the hot day in fehrenheit, C = Temperature of the hot day in centigrade.

make C the subject of the equation

C = 100(F-32)/180.............. Equation 2

From the question,

Given: F = 101°F

Substitute into equation 2

C = 100(101-32)/180

C = 38.33°C

Answer:

34.64 cm

Explanation:

Given that:

The depth of the hole h = 10 cm

height of the water holding in the tank H = 40 cm

For a stream of flowing water, the distance (x) at which the stream strikes the floor can be  computed by using the formula;

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

[tex]x = 2 \sqrt{10(40-10)}[/tex]

[tex]x = 2 \sqrt{10(30)}[/tex]

[tex]x = 2 \sqrt{300}[/tex]

[tex]x = 2 \times 17.32[/tex]

x = 34.64 cm

Answer:

See the explanation below

Explanation:

Explain how solar radiation is converted into thermal energy

From the concept of the particulate nature of matter, when the gases(air) receives a sufficient amount of radiation from the sun they tend to move in a constant rapid random motion as they move they hit one another and hence heat in form of thermal energy

Answer:

η = 0.5 = 50%

Explanation:

The efficiency of the power cycle is given by the following formula:

[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]

where,

where,

η = efficiency = ?

Q₁ = heat received from hot reservoir = 1000 KJ

Q₂ = heat discharged to cold reservoir = 500 KJ

Therefore,

[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]

η = 0.5 = 50%

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

Answer:

a)  I = 3.75 10⁴  kg  m², b)  w = 0.6 rad / s, c)  K = 6.75 10³ J, d) P = 6.75 10² W

Explanation:

This is a rotations exercise

a) the proper shape for a wheel is that of a rim where most of the weight is in the circumference plus the point weights of the people sitting on its periphery.

We are going to approximate the reda with a thin ring

           I = M r²

           I = 1500 5²

           I = 3.75 10⁴  kg  m²

b) angular and linear velocity are related

         v = w r

         w = v / r

         w = 3/5

         w = 0.6 rad / s

c) the expression for kinetic energy is

         K = ½ I w²

         K = ½ 3.75 10⁴   0.6²

         K = 6.75 10³ J

d) the power is

         P = W / t

to find the work we use the relationship between work and the variation of kinetic energy

         W = ΔK = K_f - K₀

the system part of rest wo = 0

         W = K_f

         W = 6.75 10³ J

we calculate

         P = 6.75 10³/10

         P = 6.75 10² W

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022

Answer:

5g/cm

Explanation:

denisty=mass/volume

100/20

5g/cm

Answer:

speaker64

--------

34x

Explanation:

64-34

x

speaker

4

2

4

788

- circuit

voltage

100000

x.34

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