Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answers

Answer 1

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

Approximately What Applied Force Is Needed To Keep The Box Moving With A Constant Velocity That Is Twice

Related Questions

When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?

Answers

Answer: If we have equilibrium, the magnitude must be zero.

Explanation:

If the charges are in equilibrium, this means that the total charge is equal to zero.

And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero

A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2

Answers

Answer:

ML²/6

Explanation:

Pls see attached file

The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,

Moment of inertia:

The length of the rod is L. It has a non-uniform distribution of mass given by:

dm/dx = Cx

where C has units kg/m²

dm = Cxdx

the total mass M of the rod can be calculated by integrating the above relation over the length:

[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]

Thus,

C = 2M/L²

Now, the moment of inertia of the small element dx of the rod is given by:

dI = dm.x²

dI = Cx.x²dx

[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]

I = ML²/2

Learn more about moment of inertia:

https://brainly.com/question/6953943?referrer=searchResults

Bromine, a liquid at room temperature, has a boiling point

Answers

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

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A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.

Answers

Answer:

Yes, work has been done on the mud.

Explanation:

Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:

Answers

Answer:

The ratio  is  [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

Explanation:

Generally  the Moment of inertia of a spherical object (shell) is mathematically represented as

              [tex]I = \frac{2}{3} * m r^2[/tex]

Where m is  the mass of the spherical object

       and   r is the radius  

Now the the rotational kinetic energy can be mathematically represented as

       [tex]RE = \frac{1}{2}* I * w^2[/tex]

Where  [tex]w[/tex] is the angular velocity which is mathematically represented as

             [tex]w = \frac{v}{r}[/tex]

=>           [tex]w^2 = [\frac{v}{r}] ^2[/tex]

So

             [tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]

            [tex]RE = \frac{1}{3} * mv^2[/tex]

Generally the transnational  kinetic energy of this motion is  mathematically represented as

                [tex]TE = \frac{1}{2} mv^2[/tex]

So  

      [tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]

       [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?

Answers

Answer:

The sun will be located near the Gemini constellation at sunset

A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice​

Answers

Answer:

nvbnncbmkghbbbvvvvvvbvbhgggghhhhb

When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly

Answers

Answer:

Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force,  air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.

5) What is the weight of a body in earth. if its weight is 5Newton
in moon?​

Answers

Answer:

8.167

Explanation:

Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop

Answers

Answer:

v = 1.7 m/s

Explanation:

By applying conservation of energy principle in this situation, we know that:

Loss in Potential Energy of Car = Gain in Kinetic Energy of Car

mgΔh = (1/2)mv²

2gΔh = v²

v = √(2gΔh)

where,

v = velocity of car at top of the loop = ?

g = 9.8 m/s²

Δh = change in height = 45 cm - Diameter of Loop

Δh = 45 cm - 30 cm = 15 cm = 0.15 m

Therefore,

v = √(2)(9.8 m/s²)(0.15 m)

v = 1.7 m/s


An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.

(ii)Power dissipated

Answers

Plz check attachment for answer.

Hope it's helpful

A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.

Answers

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:

[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons

Answers

Answer:

The  tension on the rope  is  T  =  900 N

Explanation:

From the question we are told that  

     The mass of the person on the left is  [tex]m_l = 100 \ kg[/tex]

      The force of the person on the left is  [tex]F_l = 1000 \ N[/tex]

       The mass of the person on the right  is  [tex]m_r = 70 \ kg[/tex]

       The force of the person on the right is  [tex]F_r = 830 \ N[/tex]

     

Generally the net force is  mathematically represented as

         [tex]F_{Net} = F_l - F_r[/tex]

substituting  values

        [tex]F_{Net} = 1000-830[/tex]

       [tex]F_{Net} = 170 \ N[/tex]

Now the acceleration net acceleration of the rope is mathematically evaluated as

        [tex]a = \frac{F_{net}}{m_I + m_r }[/tex]

substituting  values

     [tex]a = \frac{170}{100 + 70 }[/tex]

     [tex]a = 1 \ m/s ^2[/tex]

The  force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by  a  is  mathematically represented as

        [tex]m_l * a = F_r -T[/tex]

Where T  is  the tension on the rope  

      substituting values

        [tex]100 * 1 = 1000 - T[/tex]

=>      T  =  900 N

         

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you at 13.0 m/s. The maximum force you apply in catching the water balloon is twice the average force. How long must the interaction time of your catch be to make sure the water balloon doesn't soak you

Answers

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

[tex]|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s[/tex]

The interaction time to avoid that the water balloon breaks is 0.029s

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices

Answers

Answer:

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Explanation:

The power dissipation by an electrical circuit is given by the following formula:

Power Dissipation = (Voltage)(Current)

P = VI

but, from Ohm's Law, we know that:

Voltage = (Current)(Resistance)

V = IR

Substituting this in formula of power:

P = (IR)(I)

P = I²R   ---------------- equation 1

Now, if we double the current , then the power dissipated by that circuit will be:

P' = I'²R

where,

I' = 2 I

Therefore,

P' = (2 I)²R

P' = 4 I²R

using equation 1

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?

Answers

Answer:

v = 8.5 m/s

Explanation:

In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,

Weight = Centripetal Force

but,

Weight = mg

Centripetal Force = mv²/r

Therefore,

mg = mv²/r

g = v²/r

v² = gr

v = √gr

where,

v = minimum speed required = ?

g = 9.8 m/s²

r = radius = 7.4 m

Therefore,

v = √(9.8 m/s²)(7.4 m)

v = 8.5 m/s

Minimum speed for a roller coaster while travelling upside down  so that the person will not fall out = 8.5 m/s

For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.

In the given condition the weight of the person must be balanced by the centrifugal force.

and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person

According to the Newton's Second Law of motion we can write force balance

[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]

Given Radius of loop = r = 7.4 m

Putting the value  of r = 7.4 m  in equation (1) we get

[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]

For more information please refer to the link below

https://brainly.com/question/13259103

A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?

Answers

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

From the question;

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

Substitute these values into equation (i) as follows;

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z

Answers

The question is not complete, the value of z is not given.

Assuming the value of z = 4.0m

Answer:

the magnitude of the electric field at any point having z(4.0 m)  =

E = 5.65 N/C

Explanation:

given

σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²

z = 4.0 m

Recall

E =F/q (coulumb's law)

E = kQ/r²

σ = Q/A

A = 4πr²

∴ The electric field at point z =

E = σ/zε₀

E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)

E = 5.65 N/C

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