What is the acceleration of the object?
m/s²

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move will be 81.03 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).

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

On resolving the given force and acceleration. Mathematically aion in the different components and balancing the equation gets. Components in the x-direction.

The normal force is balanced by weight;

N = mg

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move is;

F = μN

F= μmg

F=0.14 ×  59 kg × 9.81 m/s²

F = 81.03 N

Hence, the magnitude of the maximum force will be 81.03 N,

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Explanation:

Power= 40 W

cost=21¢/kWh

t=1 month=30 days = 720h

Energy= power*time

E= 40W* 720h = 28800kWh

Cost= Energy*rate

cost= 28800kWh*21¢/kWh

cost=604800¢

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

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(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (2ati + bj) m/s and a = 2ai m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 4 m/s².

x = at²i + btj

x = 2t²i + tj

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

x = at²i + btj

v = dx/dt

v =  (2ati + bj) m/s

a = dv/dt

a = 2ai m/s²

a = 2ai m/s²

a = 2(2)(1)

a = 4 m/s²

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The x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

There are three little objects that are densely spaced out in the x-y plane.;

The value of the masses

m₁=1.41 kg

m₂=2.55 kg

m₃=3.19 kg

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3x_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 6 +3.19 \times 4}{1.41+2.55+3.19} \\\\ X_{cm} =4.31 \ m[/tex]

[tex]\rm Y_{cm} =\frac{ (W_1y_1 + W_2y_2 + W_3y_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 4 +3.19 \times 7}{1.41+2.55+3.19} \\\\ X_{cm} =3.12 \ m[/tex]

Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.

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Compared with the car, the truck will travel four times as far.

The initial velocity of car and truck, u = 0

The acceleration of both the truck and car = a

The length of time for the acceleration = t

Let the time the truck accelerated be 2t

The distance traveled by car is calculated as;

s = ut + ¹/₂at²

s₁ = 0(t) + ¹/₂at²

s₁ = ¹/₂at²

The distance traveled by truck

s = ut + ¹/₂at²

s₂ = 0(2t) + ¹/₂a (2t)²

s₂ =  ¹/₂a x 4t²

s₂ = 4 (¹/₂at²)

s₂ = 4(s₁)

We can conclude that the Truck distance is four times the car distance.

Therefore, Compared with the car, the truck will travel four times as far.

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Answer:

42.9°C

Explanation:

Resistance of a conducting material is directly proportion to Temperature, that is, when being heated, their is more of collisions between the molecules of the material which as a result offers more opposition to the flow of current through it.

therefore R=kT

where R=resistance of the wire

T=Temperature

k= proportionality constant

when R=1.05 T=30°C

so k=1.05/30

k=0.035

when R=1.5 k=0.035

so T=1.5/0.035

T=42.9°C. or 43°C approximated

Answer:

A

Explanation:

F = m * a

100 = 5 * a

a = 100/5 = 20 m/s^2

The magnitude of the net force that the atmosphere applies to the roof when the outside pressure drops suddenly is  1.9 x 10⁵ N.

The pressure is the amount of force applied per unit area.

Pressure p = Force/Area

Given is a house has a roof (colored gray) with the dimensions shown in the drawing. The outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can adjust.

The pressure difference ΔP = 13.2 mm of Hg

The length of the roof l = 14.5m

the breadth of the roof h = 4.21m

The force exerted by pressure is

Force, F = P x A

         = (13.2 mm of Hg) [(133 N/m²) /1 mm of Hg ](14.5 x 4.21)

         = 107,170.6 N

Then the net vertical force

    Fnet = 2F cos30

    Fnet = 2 ( 107,170.6) cos30

    Fnet = 185625 N

    Fnet =1.9 x 10⁵ N

The direction of the force is downwards, since the horizontal components of the forces cancel each other.

Hence,  magnitude of the net force is 1.9 x 10⁵ N

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(a) The compression of the spring is 7.4 cm.

(b) When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

The compression of the spring is calculated by applying Hooke's law as follows;

F = kx

x = F/k

x = (mg)/k

x = (3 x 9.8)/400

x = 0.074 m

x = 7.4 cm

When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

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The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

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Answer:

1680 seconds

Explanation:

[tex]28 hrs * \frac{60 s}{1 hr} =1680s[/tex]

This planet's mass must be four times that of Earth in order for the gravitational pull at its surface to be comparable to Earth's.

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

The planet's mass is directly related to its mass at the surface and inversely proportional to the square of its radius;

When mass increases and distance reduces, gravity rises. Gravity also lowers when the distance between two points grows and the mass decreases.

The astronomers discovered a planet twice the size of Earth that resembles Earth.

This planet's mass must be four times that of Earth in order for the gravitational pull at its surface to be comparable to Earth's.

Hence the masses of this planet will be 4 times that of Earth.

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Answer:

v(t)=s'(t)=5t⁴-sin(4).4, for velocity function at time t

a(t)=v'(t)=20t³, for acceleration function at time t

Answer:

588J

Explanation:

Work done = Fd

=wh=mgh

=6×9.8×10

=558J

Answer:

Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].

Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)

Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].

Explanation:

Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.

Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.

Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.

Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].

Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].

For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Answer:

a

Explanation:

The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1  and q2 = charges  =  1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F =  [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]

F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]

F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]

F = [tex]6. 426 * 10^-11[/tex] Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

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Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

The acceleration of this sled can be calculated by doing: option A, B, E and F.

The acceleration of this sled can be calculated by using Newton's Second law of motion. Mathematically, the acceleration of an object is given by this formula:

Net force = Mass × acceleration

Deductively, the acceleration of this sled can be calculated by doing the following:

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Complete Question:

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.

A. The force of gravity must be broken into its parallel and perpendicular components.

B. Acceleration can be found by dividing the net force by mass.

C. Acceleration can be found by multiplying the net force by mass.

D. The magnitude of the force components must be multiplied by gravity.

E. Trigonometry can be used to solve for the magnitude of the force components.

F. Net force must be found before acceleration can be found.

Answer:

To find the acceleration, you do 20N/50kg = 0.4 m/s^2

Explanation:

example: a = F/m = 10/2 = 5 m/s2

Answer:

See below

Explanation:

The work done equals the increase in Potential Energy

PE = mgh

    = 150 * 9.81 * 1 = 1471.5 j

3 sets of 10 is 30 times this = 44145 j

The amount of work done by a football player in the weight room when he squats 150 kg will be equal to 1471.5 J.

Potential energy is a form of stored energy that is dependent on the relationship among different components. When a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the floor as opposed to falling to the ground, it has more potential energy. It is capable of carrying out additional work when raised.

Potential energy is a feature of systems rather than of particular bodies or particles; for instance, the system created up of Earth and the elevated ball has more energy stored as they become further apart.

Potential energy develops in systems components whose configurations, or spatial arrangement, determine the amount of the forces they apply to one another.

As per the instructions given in the question,

Work performed is equivalent to Potential Energy Increase,

PE = mgh

= 150 × 9.81 × 1

PE = 1471.5 J

Also,

3 sets of 10 are 30 times this = 44145 J

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Answer:

0.5 m/s²

Explanation:

according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:

F = m×a

F for force

m for mass

a for acceleration

we use the given data and get:

20 = 40×a

we find a=20/40=0.5m/s²

acceleration of the box is 20m/s^2

given

mass (m) = 40 kg

force (f) = 20n

acceleration (a) =?

we know,

f = m × a

20 = 40 × a

40 - 20 = a

20m/s^2 = a

Answer:

1. 92, 138, 92

2. 7, 7, 7

3. 17, 18, 17

4. 29, 34, 29

Explanation:

the number on the bottom is always going to be the atomic number, or the number of protons.

the number on the top is going to be the atomic mass number, which is the sum of the number of protons and neutrons.

there is always going to be the same number of electrons as protons, because that was the overall charge of each atom is 0.

for example, for the first one:

there are 92 protons

there are 230 - 92 = 138 neutrons

there are 92 electrons

Answer:

1.163 seconds

Explanation:

see the attached. solve the quadratic equation.

The boat’s resultant velocity with respect to North is determined as 17 m/s.

The boat’s resultant velocity with respect to North is calculated as follows;

R² = Va² + Vb²

where;

R = √(Va² + Vb²)

R = √(15² + 8²)

R = 17 m/s

Thus, the boat’s resultant velocity with respect to North is determined as 17 m/s.

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Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.

The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.

The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.

The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.

Hence option B is corect.

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The allowable ampacity of a No. 12 copper wire with type THHN insulation installed in a raceway with 35 wires is approximately 12 amperes. This is best represented by option D.

According to the National Electrical Code Table (NEC), a derating factor must be applied for any raceway with more than three current-carrying conductors (in our case, copper wires). In a raceway with 31-40 current-carrying conductors, the derating factor is 50%. Because the ampacity of a No. 12 copper wire usually is 25 amperes (A), we can multiply this by our derating factor to find that:

25 • 0.50 = 12.5

The exact allowable ampacity of this copper wire is 12.5 amperes; however, we can round down to find that the closest answer is 12 A.

Hence, option D represents the best answer.

Answer:

  D)  12 A

Explanation:

You want to know the allowable ampacity of AWG 12 type THHN copper wire in a raceway with 35 wires.

Table 310.15(B)(16) of the 2017 National Electrical Code gives the nominal rating of AWG 12 type THHN copper wire as 30 A when installed in an environment with a temperature of 30 °C or less.

Table 310.15(B)(3)(a) gives the adjustment factor when there are more than three conductors in a raceway or cable. For 31–40 conductors, the ampacity must be derated to 40% of its nominal value.

In a raceway with 35 wires, the allowable ampacity of THHN AWG 12 wire is ...

  (30 A)(0.40) = 12 A

__

Additional comment

For environmental temperatures above 30 °C, Table 310.15(B)(2)(a) gives additional derating factors. The adjustment factor for temperatures up to 50 °C (122 °F) is 0.82, which would bring the allowable ampacity down to about 9.8 A.

The cost to use all lamps is $62.

           power=energy/time

          (60+ 100+150)=energy/2.5

          energy=775 joule

cost for 1 hour=$0.08

                      = 775 * 0.08

                     = $62

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Answer:

6.2 cents

Explanation:

(60+100+150/1000)*2.5*8=6.2 cents