Abdou was explaining to a classmate that graphite is a good lubricant
because it is bonded in layers that easily slip over each other. Which image
shows this quality?

Answer:

STATES OF MATTER The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.

Explanation:

Answer:

speed and velocity

Explanation:

A. The efficiency of the ramp is 54.6 %.

V.R = distance moved by effort/distance moved by load = L/h

V.R = (6.12)/(1.53) = 4

M.A = Load/Effort

M.A = (451 x 9.8)/(2025)

M.A = 2.183

E = (M.A / V.R) x 100%

E = (2.183 / 4) x 100%

E = 54.6 %

Thus, the efficiency of the ramp is 54.6 %.

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The reference point that should be used to describe Melanie motion is the pond (option C).

Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.

According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.

However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.

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The horizontal component of the force  is 336 N

The magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons

The system of the beam, hi_nge and cable are in equilibrium.

The horizontal component of the force exerted by the beam is given below as:

Horizontal component of the force = mg × cos 31°

Horizontal component of the force = 40 × 9.8 × cos 31°

Horizontal component of the force  = 336 Newtons

The magnitude of the force that the beam exerts on the hi_nge is given as:

F = mg × cos 31° + mg × sin 31°

F = 40 × 9.8 × cos 31° + 40 × 9.8 × sin 31°

F = 537.9 Newtons

Hence, the magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons.

In conclusion, the system of forces acting on the hi_nge and the beam are in equilibrium.

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The radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.

Using the formula:

r = [tex]\frac{mvsin \alpha }{Bq}[/tex]

m = 9.109 × 10-31 kg

α = 35°

B = 0. 040 T

q = 1.6 × 10-19 C

Substitute values into the formula

[tex]r = \frac{9. 109 * 10^-31 * 4.0 * 10^0 * sin 35}{0. 040 * 1.6 * 10^-19 }[/tex]

[tex]r = \frac{2. 089 * 10^-30}{6. 4* 10^-21}[/tex]

[tex]r = 3. 26 * 10^-10[/tex] m

b. [tex]T = \frac{2\pi m}{qB}[/tex]

[tex]T = \frac{2 * 3.142 * 9.109 *10^-31}{1.6* 10^-19* 0.040}[/tex]

[tex]T = \frac{5. 72* 10^-30}{6. 4* 10^-21}[/tex]

[tex]T = 8. 94* 10^-10[/tex]

Therefore, the radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.

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There should be a reinstatement of the National Maximum Speed Limit for the safety of road users because the minimum braking distance is 2,25m.

Hence, we can see that the minimum braking distance for the car would be:

d'/d= (v'1/v1)^2 = (1.5/1)^2

= 2.25m

The speed limit ensures that drivers do not exceed a set speed limit in order to reduce road accidents and keep road users safe and also the pedestrians would be able to navigate easier.

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Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

[tex]\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg[/tex]

We can simplify and rearrange the equation to solve for 'v'.

[tex]\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}[/tex]

Plugging in values:

[tex]v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}[/tex]

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

[tex]\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg[/tex]

Rearranging for 'T":
[tex]T = \frac{mv^2}{r} + mg\\\\[/tex]

Plugging in the appropriate values:
[tex]T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}[/tex]

Answer: 0.025 cm

Explanation:

The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm

The magnification of the lens is given by

m = v/u = h'/h

v/u = h'/ h

h' = ( v/u ) × h

h' = ( 2cm /-120cm ) × 1.5cm

h' = - 0.025cm

Therefore, the height of the image is 0.025 cm

The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

The average power produced by the rocket is calculated as follows;

P = FV

where;

Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

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The total work done is  5980 Joules and the power expended is 57 Watts.

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m  

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules  

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

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Answer:

P= phg

so:- the height is 10m

density of water 1000 kg/m3

gravity is 9.8m/s2

P=1000 kg/m3*10m*9.8m/s2

=98000Pa

=98KPa

The impact will last for 3.3 milliseconds, the average force applied to the nail will be [m (v² - u²)]/2s, and the average force applied to the nail will be 883.2 N, respectively.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Average force,F = ?

Mass,m = 0.46 kg

Acceleration,a

Final velocity,v = 0 m/sec

Initial velocity,u = 6.5 m/sec

The duration of the impact, assuming the acceleration is constant is found with the help of Newton's third equation of motion as;

v²=u²+2as

0  = (6.5)² + 2 ×a × 1.1 × 10⁻²

- 42.25 = 2 × a × 1.1 × 10⁻²

a = - 1920 m/s²

a = (v-u)/t

-1920 = (0-6.5)/t

t = 3 × 10⁻³ sec

t = 3.3 milli second

From Newton's third equation of motion;

v² = u² +2as

a = (v² - u² ) /2s

The average force exerted on the nail, in terms of the mass, initial velocity, and distance traveled is;

F = ma

F = [m (v² - u² )] /2s

The average force, in newtons, exerted on the nail is;

F = [m (v² - u² )] /2s

F= ma

Substitute the given value;

F = 0.469 × 1920

F = 883.2 N

Hence the duration of the impact, the average force exerted on the nail and the average force, in newtons, exerted on the nail will be 3.3 milliseconds, F = [m (v² - u² )] /2s and 883.2 N respectively.

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

W = mg

W = 0.15 x 9.8 = 1.47 N

T = mg = 1.47 N

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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Answer:

Approximately [tex]4.61\times 10^{3}\; {\rm N}[/tex] upwards (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

External forces on this astronaut:

Let [tex](\text{normal force})[/tex] denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

[tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex].

Let [tex]m[/tex] denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be [tex](\text{weight}) = m\, g[/tex].

Let [tex]a[/tex] denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be [tex](\text{net force}) = m\, a[/tex].

Rearrange [tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex] to obtain an expression for the magnitude of the normal force on this astronaut:

[tex]\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}[/tex].

The work done is 8427 J while the energy expended is  80.25 W.

Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;

W = mgh

Total mass of the water bucket and chain = 13.9 kg +  19.3 kg = 33.2Kg

Distance covered =  25.9 m

W = 33.2Kg * 9.8 m/s^2 * 25.9 m

W = 8427 J

Recall that the work done = Energy expended

Power = Energy expended/ Time

Power = 8427 J/ 1.75 * 60 seconds

Power = 80.25 W

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The magnitude of the tension in the string marked A and B is mathematically given as

A = 52.5 N

Generally, the equation for is mathematically given as

The angle at A is...

[tex]tan\theta = 3/8[/tex]

When below the negative x

B= tan\phi

[tex]tan\phi = 5/4[/tex]

When below the negative x

C=

[tex]tan \rho = 1/6[/tex]

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Hence, considering the Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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The theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.

The theory of uniform accelerated motion is a scientific statement regarding the acceleration of an object in steady conditions.

This theory (uniform accelerated motion) is fundamentally applied in physic and astrophysics.

In conclusion, the theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.

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The sound of a lamb is grave due to its high pitch depending on its low frequency. Option B is correct.

A sound pressure wave's frequency is the number of times it repeats itself every second.

The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.

Pitch and the frequency of sound are inverse to each other. A lamb's sound is grave because of its high pitch and low frequency.

The sound of a lamb is grave due to its high pitch depending on its low frequency.

Hence, option B is correct.

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Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that

[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]

Solve for [tex]a[/tex] :

[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]

The tension in the rope B is determined as 10.9 N.

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

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A 520 Hz tone is sounded at the same time as a 516 Hz tone now the beat frequency is 4 Hz.

The pace of direction changes in current per second is known as frequency. It is expressed in hertz (Hz), a unit of measurement that is used internationally. One hertz is equal to one cycle per second. One hertz (Hz) is equivalent to one cycle per second. A complete alternating current or voltage wave is referred to as a cycle.

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The minimum coefficient of static friction is  0.807

Static friction is friction between two or more solid objects that are not moving relative to each other.

According to the question,

When an object turns on a curve path at a specific speed, two forces will be acting on the object which balance each other:

Centripetal force and friction force.

The friction coefficient between the object and the path is independent of the object's mass.

Given,

The radius of a corner is, r=4.16m

The turning speed of the car is, v

[tex]20km/h * (\frac{\frac{5}{18}m/s }{1km/h} )[/tex]

= 5.72 m/s

To avoid slipping a coffee cup on the car's dashboard, the centripetal force would balance the friction force.

So,  Fc=Ff

[tex]\frac{mv^2}{r}[/tex]=μmg

μ=[tex]\frac{v^2}{rg}[/tex]

Here,

Fc represents the centripetal force.

Ff represents the friction force.

μ represents the coefficient of static friction.

g represents the gravitational acceleration whose value is 9.8m/s²

By Substituting the values in the above formula, we get:

μ =[tex]\frac{(5.72m/s)^2}{(4m)(9.8m/s^2)} \\[/tex]  ≈ 0.807

Therefore,

The minimum coefficient of static friction is  0.807

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Answer:

The purpose of an experiment is to test out your hypothesis. If your hypothesis is correct, then it is a theory that could work every single time the experiment has been performed by scientists.

Answer:

The value of work being done on the object is 958J.

Explanation:

Work done is equal to force multiply by distance, but when the angle is between the force and distance work done=Force (cos theta) × distance

Work done = Force(cos theta) × distance

Work done = 25N(cos 40.0°) × 50m

Work done = 25N(0.7660) × 50m

Work done = 19.15N × 50m

Work done = 957.5J = 958J

Therefore the value of work being done on the object is 958J.

Answer:

it's 6a - 18

Explanation:

because 6 is multiplied witk a and -3

Answer:

True

Explanation:

it is true because blabla bla

Answer:

The centripetal acceleration (ac)=314m/s²

Explanation:

look at the attachment ☝️

Answer:

Power = 2000J/s

Explanation:

Power= Total Work done/Time

Given

mass = 100kg

distance(height) = 10m

time = 5s

acceleration due to gravity = 10m/s²

Work done = force × distance

Finding Force

Force = mass × acceleration due to gravity

Force = 100kg × 10m/s²

Force = 1,000N

Finding work done

Work done = Force × distance

Work done = 1,000N × 10m

Work done = 10,000J

Finding Power

Power = work done/time

Power = 10,000J/5s

Power = 2,000J/s

Therefore Power is 2,000J/s