Someone help me with these questions please!

Answer:

Minimum Area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum Area of rectangle

Computation:

Minimum Area of rectangle = Length of rectangle x Width of rectangle

Minimum Area of rectangle = 6 x 4

Minimum Area of rectangle = 24 centimeter²

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;

[tex]F_{min}[/tex] = m·g - m·a

∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N

Answer:

[tex]T=8.1N[/tex]

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 [tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]

 [tex]\alpha=50 \textdegree[/tex]

Generally the equation for Tension is is mathematically given by

 [tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]

 [tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]

 [tex]T=8.1N[/tex]

Answer:no

Explanation:because 0.9*(30*60)=0.9*1800=1620

The turtle has already won the race

Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

It is given

[tex]V_{t} = 0.9 \frac{m}{s}[/tex]

[tex]V_{r} = 9 \frac{m}{s}[/tex]

[tex]D=1500 m[/tex]

Time taken by turtle  

 [tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]

[tex]T=1666 minutes= 27 hours[/tex]

Time taken by  rabbit

[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]

[tex]T=166 minutes[/tex]

since rabbit started 30 minutes after turtle then

[tex]T= 136+30=196 minutes[/tex]

[tex]T= 3.2 hours[/tex]

Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

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Answer:

f = 106.3 N

Explanation:

The force applied on the sled must be equal to the static frictional force to move the sled:

Tension Force Horizontal Component = Static Frictional Force

[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]

where,

T = Tension = 120 N

θ = angle of rope = 37°

μ = coefficient of static friction = ?

m = mass of children plus sled = 55 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]

Now, the static friction acting on the mother will be:

[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 106.3 N

Answer:

a) Light that passes through the floor to reveal yourself (not shadow).

b) 2 rays of light that bounce between 2 transparent media.

c) I don't know what is Diffraction?

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

The frictional force in unit of N that the floor must exert on the ladder is approximately 232.216 N

The known values are;

The weight of the painter = 1,071.628 N

The height to which the painter needs to climb along the ladder = 1.926 m

The length of the ladder = 12.014 m

The weight of the ladder = 250 N

The points where one of the ladder's ends is resting = On the ground

The points where the other end of the ladder is resting = A perfectly smooth wall

The angle with which the ladder rises above the horizontal floor = 51.96°

The acceleration due to gravity, g ≈ 10 m/s²

The unknown values include;

The friction force that the floor must exert on the ladder

The strategy to be used;

At equilibrium, the sum of moments about a point is zero

Finding the moments about the point of contact where the ladder rests on the wall, P, is given as follows;

At equilibrium, the sum of clockwise, [tex]M_{CW}[/tex], moment about P = The sum of the counterclockwise, [tex]M_{CCW}[/tex]moment about P

[tex]\mathbf{M_{CCW}}[/tex] = (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

[tex]\mathbf{M_{CW}}[/tex] = 12.014 × cos(51.96°) × [tex]\mathbf{F_N}[/tex]

Where;

[tex]\mathbf{F_N}[/tex] = The normal reaction of the of the ground on the end of the ladder that rests on the floor

[tex]\mathbf{M_{CCW}}[/tex] = [tex]\mathbf{M_{CW}}[/tex]

∴ (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250 = 12.014 × cos(51.96°) × [tex]F_N[/tex]

We get;

6,665.3068846 N·m =  7.40316448688 m × [tex]F_N[/tex]

[tex]\mathbf{F_N}[/tex] = 6,665.3068846 N·m/(7.40316448688 m) = 900.332135 N

The normal reaction of the floor on the ladder, [tex]\mathbf{F_N}[/tex] = 900.332135 N

Taking moment about the point the ladder rests on the floor, R, gives;

[tex]M_{CCW}[/tex] = 12.014 × sin(51.96°) × [tex]F_W[/tex]

Where;

[tex]\mathbf{F_W}[/tex] = The normal reaction at the wall

[tex]M_{CW}[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

At equilibrium, we have, [tex]M_{CCW}[/tex] = [tex]M_{CW}[/tex]

Therefore;

12.014 × sin(51.96°) × [tex]F_W[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

9.46199511627 m × [tex]F_W[/tex] = 2,197.22861125 N·m

[tex]F_W[/tex] = 2,197.22861125 N·m/(9.46199511627 m)

The reaction of the wall, [tex]\mathbf{F_W}[/tex] = 232.216206 N

We note that also at equilibrium, the sum horizontal forces = 0

The horizontal forces acting  on the ladder = The normal reaction on the, [tex]F_W[/tex] wall and the friction force on the ground, [tex]\mathbf{F_f}[/tex]

∴ At equilibrium; [tex]\mathbf{F_W}[/tex] + [tex]\mathbf{F_f}[/tex] = 0

[tex]\mathbf{F_f}[/tex] = -[tex]\mathbf{F_W}[/tex]

[tex]\mathbf{F_W}[/tex]  = 232.216206 N

Therefore;

The frictional force in unit of N that the floor must exert on the ladder, [tex]\mathbf{F_f}[/tex] = 232.216206 N ≈ 232.216 N.

(The coefficient of friction, μ = [tex]\mathbf{F_N}[/tex]/[tex]\mathbf{F_W}[/tex] = 900.332135/232.216206 ≈ 3.877).

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what

what

what

what

sorry

sorrry

sorry

Answer:

i. -4m

ii. 20m

Explanation:

The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)

Total distance = 8m going east + 8m back to origin + 4m west = 20m

Answer:

The gauge pressure is equal to 147 kPa.

Explanation:

The pressure exerted by fluid is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is density of water

h is height

So, put all the values,

[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]

or

P = 147 kPa

So, the gauge pressure is equal to 147 kPa.

Answer:

The gauge pressure is 147000 Pa.

Explanation:

Height, h = 15 m

density of water, d= 1000 kg/m^3

gravity, g = 9.8 m/s^2

The gauge pressure is the pressure exerted by the fluid.

The pressure exerted by the fluid is given by

P  = h d g

P = 15 x 1000 x 9.8 =  147000 Pa

Answer:

The required fraction is 0.023.

Explanation:

Given that

Mass of a car, m = 1030 kg

Mass of 4 wheels = 12 kg

We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.

The rotational kinetic energy due to four wheel is

[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]

Linear kinetic Energy of the car is:

[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]

Fraction,

[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]

So, the required fraction is 0.023.

Answer:

The speed of car A before collision is 3.5 km/h.

Explanation:

Mass of car A = 690 kg eastwards

Mass of car B = 520 kg at 74 km/h west wards

Distance, s = 6 m

coefficient of friction = 0.75

Let the speed after collision is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]

Let the initial speed of car A is v'.

Use conservation of momentum

690 x v' - 520 x 74 = (690 + 520) x 33.8

690 v' + 38480 = 40898

v' = 3.5 km/h

Answer:

  f = 276.6 Hz

Explanation:

This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.

In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is

          L = λ/ 4

speed is related to wavelength and frequency

          v = λ f

          λ = v / f

we substitute

          L = v / 4f

          f = v / 4L

the speed of sound at 20ºC is

          v = 343 m / s

let's calculate

          f = [tex]\frac{343 }{4 \ 0.31}[/tex]

          f = 276.6 Hz

In vacuum, going at 2.99×10^8 m/s.

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)

[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

[tex]v_B=+5\ m/s[/tex]

Therefore, Carlos velocity in Bill's reference frame will be

[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]

So, the magnitude is 20 m/s and the direction is westward (negative sign).

Answer:

44 N

Explanation:

Given that,

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.

It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.

Answer:

Explanation:

The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.

From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.

Answer:

t = 2.26 x 10⁵ s

Explanation:

The energy supplied to the water will be equal to the heat required for the boiling of water:

E = ΔQ

Pt = mL

where,

P = Power = 10 W

t = time = ?

m = mass of water = 1 kg

L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]

t = 2.26 x 10⁵ s

This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.

Answer:

the instantaneous velocity is 51 m/s

Explanation:

Given;

acceleration, a = 2 + 5t²

Acceleration is the change in velocity with time.

[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]

Therefore, the instantaneous velocity is 51 m/s

Answer:

Where is the figure ?????

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

Answer:

220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.

Answer:

Fq = k q Q / R^2

Fp = k q Q / (6 R)^2

The force exerted between S and p is 1 / 36 of that between S and q

or the force between S and q is 36 times that between S and p.

Hooke's law gives the relationship between the force applied and observed compression and expansion

(a) Hooke's law states that force applied is proportional to the deformation of an object

(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre

The reason for the above explanation are as follows;

(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F

Mathematically, we get;

F = k × ΔL

[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]

Given that the material cross sectional area = A, and the original length of the material = L, we get;

F/A = Stress = σ, ΔL/L = strain = ε

[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]

Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material

(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation

Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve

As the tyre straightens, the path of the stress change curve is given by the lower curve

The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening

Energy absorbed < Energy released

The balance energy is transformed into other forms of energy, including  heat energy, which is observed by the raising in temperature, warming, of the tyre

Energy absorbed = Energy released + Heat energy

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Answer:

The velocity becomes [tex]v\sqrt 2[/tex].

Explanation:

The force acting on the bobber is centripetal  force.

The centripetal force is given by

[tex]F =\frac{mv^2}{r}[/tex]

when mass remains same, radius is doubled and the force is same, so the velocity is v'.

[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]

Explanation:

Given that,

Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.

The rms speed for this collection is as follows :

[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]

The average speed of these molecules is :

[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]

So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.

Answer:

 I = - m 16  the two impulses are the same,

Explanation:

The impulse is given by the relationship

         I = Δp

         I = p_f - p₀

in this case the final velocity is zero therefore p_f = 0

        I = -p₀

For driver A the steering wheel impulse is

        I = - m v₀

        I = - m 16

For driver B, the airbag gives an impulse

        I = - m 16

We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less