A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na

Answers

Answer 1

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .


Related Questions

An electric point charge of Q = 22.5 nC is placed at the center of a cube with a side length of a = 16.3 cm. The cube in this question is only a mathematical object, it is not made out of any physical material. What is the electric flux through all six sides of the cube?

Answers

Answer:

The electric flux is  [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]

Explanation:

From the question we are told that

  The magnitude of the electric point charge  [tex]q = 22.5 nC = 22.5 *10^{-9} \ C[/tex]

   The length of the one side of the cube is [tex]l = 16.3 \ cm = 0.163 \ m[/tex]

   The number of  sides is  [tex]N= 6[/tex]  

The electric flux according to Gauss law is mathematically evaluated as

          [tex]\phi = \frac{q}{\epsilon_o}[/tex]

Where [tex]\epsilon _ o[/tex] is the permitivity of free space with value  [tex]\epsilon_o = 8.85*10^{-12}\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

  substituting values

        [tex]\phi = \frac{22.5 *10^{-9}}{8.85 *10^{-12}}[/tex]

         [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]

         

What's a line of best fit? Will give BRAINLIEST

Answers

A line of best fit expresses the relationship between the points.

Explanation:

It does not go through all the points but goes through most of them and it is like a hardrawn curve

What is the speed at which a spaceship shoots up from earth ?

Answers

Answer:

Once at a steady cruising speed of about 16,150mph (26,000kph

Explanation:

A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s

Answers

Answer:

B. 9.4 s

Explanation:

In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:

Vf₁ = Vi₁ + gt₁

where,

Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)

Vi = Initial Velocity in upward motion = 36.2 m/s

g = - 9.8 m/s² (negative due to upward motion)

t₁ = Time taken in upward motion = ?

Therefore,

0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)

t₁ = (36.2 m/s)/(9.8 m/s²)

t₁ = 3.7 s

Now, using 2nd equation of motion:

h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²

where,

h₁ = distance from top of building to highest point ball reaches = ?

Therefore,

h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²

h₁ = 133.58 - 66.86 m

h₁ = 66.72 m

No, considering downward motion and using subscript 2, for it.

Using 2nd equation of motion:

h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²

where,

h₂ = height of the highest point from ground = h₁ + height of building

h₂ = 66.72 m + 90 m = 156.72 m

Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)

t₂ = Time Taken in downward motion = ?

g = 9.8 m/s²

Therefore,

156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²

t₂² = (156.72 m)/(4.9 m/s²)

t₂ = √31.98 s²

t₂ = 5.7 s

Now, the total time taken by ball to reach the ground is"

Total Time = T = t₁ + t₂

T = 3.7 s + 5.7 s

T = 9.4 s

Therefore, the correct answer is:

B. 9.4 s

EASY! WILL REWARD BRAINLIEST!

Electrical current is defined as _____.

the capacity to store charge
the flow of electric charge per unit time
the amount of stored electric energy
the voltage of the battery

Answers

Electrical current is defined as the flow of electric charge per unit time.

the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s

Answers

Answer:

I dont know bro

Explanation:

Ask an expert

Answer:

Time=90s

Explanation:

Speed=distance /time

[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]

A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected across the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?

Answers

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]

The rms voltage (in V) measured across the secondary coil is calculated as;

[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

A force of 640 newtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.

Answers

Bhbbv h Gucci Janice)6225

The equation for distance is d= st. if a car has a speed of 20 m/s how long will it take to go 155m

Answers

Answer:

It will take 7.75 seconds for the car to go 155m

Explanation:

From the question, we can understand that the distance covered by the moving car is got by a product of its speed and the time it travels.

i.e distance = speed X time.

However, in this case, we have the distance travelled and the speed of the car, and we are looking for the time of travel

TO solve this, we will simply make the travel time the subject of the formula in the equation above.

i.e time = distance / speed

time = 155/20= 7.75 seconds.

Hence, it will take 7.75 seconds for the car to go 155m

Un levantador de pesas puede generar 3000 N de fuerza ¿Cuál es el peso máximo que puede levantar con una palanca que tiene un brazo de la fuerza de 2 m y un brazo de resistencia de 50 cm?

Answers

Responder: 12000N

Explicación: Usando la fórmula para encontrar la eficiencia de una máquina. Eficiencia = ventaja mecánica / relación de velocidad × 100%

Dado MA = Carga / Esfuerzo

Relación de velocidad = distancia recorrida por esfuerzo (brazo de fuerza) / distancia recorrida por carga (brazo de resistencia)

MA = Carga / 3000

VR = 2 / 0.5 VR = 4

Asumiendo que la eficiencia es 100% 100% = (Carga / 3000) / 4 × 100%

1 = (Carga / 3000) / 4

4 = Carga / 3000

Carga = 4 × 3000

Carga = 12000N

Esto significa que el peso máximo que se puede levantar es 12000N

Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly atoms are assembled in this protein synthesis.

Answers

Answer:5.35nm

Explanation:

Consider that 1 inch is = 0.0254m

we have,

1m= 1x10^9 nm  

While:

0.0254m = 2.54x10^7nm  

1/55 (2.54x10^7) = 4.6181 x 10^5nm  

1 day= 24 hrs  

= (24x60) when calculating in min  

= (24x60x60) calculating in seconds we have:

= 8.64x10⁴sec  

In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm

Therefore, the amount by which the hair grows in 1 second  will be;

= (4.6181 x 10^5)/(8.64x10^4)  

= 5.35nm  

The rate of growth will be 5.35nm

A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine the tension of the wire.

Answers

Answer:

Tension of the wire(T) = 169 N

Explanation:

Given:

f = 65Hz

Length of the piano wire (L) = 2 m

Mass density = 5.0 g/m² = 0.005 kg/m²

Find:

Tension of the wire(T)

Computation:

f = v / λ

65 = v / 2L

65 = v /(2)(2)

v = 260 m/s

T = v² (m/l)

T = (260)²(0.005/2)

T = 169 N

Tension of the wire(T) = 169 N

5.

The solar system coalesced due to rotational forces and

gravity.

heat.

radioactivity.

solar wind.

Answers

Answer:

Gravity

Explanation:

The solar system is held together by rotational forces and gravity. This can be seen from billions of years ago when the solar system was a cloud of dust and gas. This cloud of dust and gas is known as the solar nebula. All of these dust and gas were brought together by the rotational movement as well as the action of gravity which brought all the particles together to form a larger one. This alone brought about the sun's formation in the center of the nebula as well the formation of other planetary bodies, etc.

Cheers.

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time

Answers

Answer:

Explanation:

The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force

impulse by net  force = change in momentum

Net force = R - mg , mg is weight of the man .

( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0

R - mg = 227.81 x 10³

Average net force = 227.81 x 10³ N .

Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?

Answers

Answer:

a. about 20 years younger

b. Malcolm sits around for 49.94 years

c. 2.268x[tex]10^{17}[/tex] m

Explanation:

light travels 3x[tex]10^{8}[/tex] m in one seconds

in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m

for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex]  = 3.78x[tex]10^{17}[/tex] m

Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex]  = 2.4x[tex]10^{8}[/tex]  m/s

time that will pass for Malcolm will be  distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]  

= 1575000000 s = 49.94 years

the relativistic time t' will be

t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]

t' = 49.94 x 0.6 = 29.96 years       this is the time that has passed for Flo

this means that Flo will be about 20 years younger than Malcolm when she returns

relativistic distance is

d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x 0.6

d' = 2.268x[tex]10^{17}[/tex] m     this is how far it is to Flo

A 888 kg car is driven clockwise around a flat circular track of radius 59 m. The speed of the car is a constant 7 m/s. Which factor, when doubled, would produce the greatest change in the centripetal force acting on the car? A. Radius of the track B. Weight of the car C. Mass of the car D. Velocity of the car

Answers

Answer:

D. Velocity of the car

Explanation:

The centripetal force acting on the car is given by the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]    (1)

m: mass of the car  = 888 kg

v: tangential speed of the car = 7 m/s

r: radius of the flat circular track = 59 m

By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:

[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]

Then, the greatest values of the centripetal force is:

[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]

The greatest change in Fc is obtained by changing the value of the speed

answer

D. Velocity of the car

Using the equation for the distance between fringes, Δy = xλ d , complete the following. (a) Calculate the distance (in cm) between fringes for 694 nm light falling on double slits separated by 0.0850 mm, located 4.00 m from a screen. cm (b) What would be the distance between fringes (in cm) if the entire apparatus were submersed in water, whose index of refraction is 1.333? cm

Answers

Answer:

Explanation:

Distance between fringe or fringe width =  xλ /  d

where x is location of screen and d is slit separation

Given x = 4 m

λ = 694 nm

d = .085 x 10⁻³ m

distance between fringes

= 4 x 694 x 10⁻⁹ / .085 x 10⁻³

= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶

= 32.66 x 10⁻³ m

= 32.66 mm .

3.267 cm

b )

when submerged in water , wavelength in water becomes as follows

wavelength in water = wave length / refractive index

= 694 / 1.333 nm

= 520.63 nm

new distance between fringes

3.267 / 1.333

= 2.45 cm .

Suppose a stone is thrown vertically upward from the edge of a cliff on Mars (where the acceleration due to gravity is only about 12 ft/s2 with an initial velocity of 64 ft/s from a height of 192 ft above the ground. The height s of the stone above the ground after t seconds is given by
s=−6t2+64t+192
a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground?

Answers

Answer:

a) v = -12t + 64

b) t = 5.33s

c) s = 362.66ft

d) t = 13.10s

e) v = 93.2ft/s

Explanation:

You have the following equation for the height of a stone thrown in Mars:

[tex]s(t)=-6t^2+64t+192[/tex]       (1)

a)  The velocity of the stone after t seconds is obtained with the derivative of s in time:

[tex]v=\frac{ds}{st}=-12t+64[/tex]   (2)

The equation for the speed of the stone is v = -12t + 64

b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:

[tex]-12t+64=0\\\\t=5.33s[/tex]

The time in which the stone is at the maximum height is 5.33s

c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):

[tex]s(1)=-6(5.33)^2+64(5.33)+192=362.66ft[/tex]

the maximum height is 362.66 ft

d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:

[tex]0=-6t^2+64t+192[/tex]

you use the quadratic formula:

[tex]t_{1,2}=\frac{-64\pm\sqrt{64^2-4(-6)(192)}}{2(-6)}\\\\t_{1,2}=\frac{-64\pm 93.29}{-12}\\\\t_1=13.10s\\\\t_2=-2.44s[/tex]

You use the result with positive values because is the onlyone with physical meaning.

The time for the stone hits the ground is 13.10 s

e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:

[tex]v=-12t+64=-12(13.10)+64=-93.2\frac{ft}{s}[/tex]

The minus sign is because the stone's direction is downward.

The speed of the stone just when it strikes the ground is 93.2ft/s

1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a turn b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.

Answers

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  [tex]\mu mg = \frac{mv^2}{r}[/tex]

    where [tex]\mu[/tex] is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

Q) Considering the value of ideal gas constant in S.I. unit, find the volume of 35g O2 at 27°C and 72
cm Hg pressure. Later, if we keep this pressure constant, the r.m.s velocity of this oxygen molecules
become double at a certain temperature. Calculate the value of this temperature.

Answers

Answer:

V = 0.0283 m³ = 28300 cm³

T₂ = 1200 K

Explanation:

The volume of the gas can be determined by using General Gas Equation:

PV = nRT

where,

P = Pressure of Gas = (72 cm of Hg)(1333.2239 Pa/cm of Hg) = 95992.12 Pa

V = Volume of Gas = ?

n = no. of moles = mass/molar mass = (35 g)/(32 g/mol) = 1.09 mol

R = General Gas Constant = 8.314 J/ mol.k

T = Temperature of Gas = 27°C + 273 = 300 k

Therefore,

(95992.12 Pa)(V) = (1.09 mol)(8.314 J/mol.k)(300 k)

V = 2718.678 J/95992.12 Pa

V = 0.0283 m³ = 28300 cm³

The Kinetic Energy of gas molecule is given as:

K.E = (3/2)(KT)

Also,

K.E = (1/2)(mv²)

Comparing both equations, we get:

(3/2)(KT) = (1/2)(mv²)

v² = 3KT/m

v = √(3KT/m)

where,

v = r.m.s velocity

K = Boltzamn Constant

T = Absolute Temperature

m = mass of gas molecule

At T₁ = 300 K, v = v₁

v₁ = √(3K*300/m)

v₁ = √(900 K/m)

Now, for v₂ = 2v₁ (double r.m.s velocity), T₂ = ?

v₂ = 2v₁ = √(3KT₂/m)

using value of v₁:

2√(900 K/m) = √(3KT₂/m)

4(900) = 3 T₂

T₂ = 1200 K

Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.

Answers

Answer:

a) v = -30 - 32 t ,  s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s

c) v (1) = -62 ft / s,  v (3) = -126 ft / s , d) t = 7.13 s , e)  v = -258.16 ft / s

Explanation:

a) For this exercise they give us the function of the position of the ball

          s (t) = s (o) + v_o t - 16 t²

notice that you forgot to write the super index

indicate the initial position of the ball

        s (o) = 600 ft

also indicates initial speed

        v_o = - 30 ft / s

let's substitute in the equation

        s (t) = 600 - 30 t -16 t²

to find the speed we use

       v = ds / dt

       v = v_o - 32 t

       v = -30 - 32 t

b) To find the average speed, look for the speed at the beginning and end of the time interval

t = 1 s

     v (1) = -30 -32 1

     v (1) = - 62 ft / s

t = 3 s

     v (3) = -30 -32 3

     v (3) = -126 ft / s

the average speed is

    v = (v (3) -v (1)) / (3-1)

    v = (-126 +62) / 2

    v = -32 ft / s

c) instantaneous speeds, we already calculated them

    v (1) = -62 ft / s

    v (3) = -126 ft / s

d) the time to reach the ground

in this case s = 0

    0 = 600 - 30 t -16 t²

     t² + 1,875 t - 37.5 = 0

we solve the quadratic equation

     t = [-1,875 ±√ (1,875² + 4 37.5)] / 2

     t = [1,875 ± 12.39] / 2

     t₁ = 7.13 s

     t₂ = negative

Since the time must be positive, the correct answer is t = 7.13 s

e) the speed of the ball on reaching the ground

     v = -30 - 32 t

     v = -30 - 32 7.13

      v = -258.16 ft / s

What is the velocity of a car that travels 556km northwest in 3.2 hours

Answers

Answer:

173.75 km/hr in the NW direction.

Explanation:

Velocity is the time rate of change in displacement of a body. Mathematically:

v = d / t

where d = displacement

t = time

Therefore, the velocity of the car is:

v = 556 / 3.2 = 173.75 km/hr

The velocity of the car is 173.75 km/hr in the NW direction.

The velocity of a car will be "173.75 km/hr".

Displacement and Velocity,

The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.

Displacement, d = 556 km

Time, t = 3.2 hours

We know the relation,

→ Velocity = [tex]\frac{Displacement}{Time}[/tex]

or,

→ V = [tex]\frac{d}{t}[/tex]

By substituting the values, we get

      = [tex]\frac{556}{3.2}[/tex]

      = [tex]173.75[/tex] km/hr

Thus the response above is right.

Find out more information about velocity here:

https://brainly.com/question/6504879

A 328-kg car moving at 19.1 m/s in the x direction hits from behind a second car moving at 13.0 m/s in the same direction. If the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision

Answers

Answer:

14.04 m/s

Explanation:

To find the velocity of the first car after the collision, we can use the equation of conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

We have the following data:

m1 = m1' = 328,

m2 = m2' = 790,

v1 = 19.1,

v2 = 13,

v2' = 15.1.

Using this data, we can find v1' (final velocity of the first car):

328 * 19.1 + 790 * 13 = 328 * v1' + 790 * 15.1

16534.8 = 328 * v1' + 11929

328 * v1' = 4605.8

v1' = 14.04 m/s

The brakes of a car are applied to give it an acceleration of -2m/s^2. The car comes to a stop in 3s. What was its speed when the brakes were applied?

Answers

Answer:

So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s

ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick

Answers

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

A mechanic applies a force of 60N at a distance of 80 cm from the pivot on a wheel wrench. What is the size of the moment?

Answers

Answer:

48 Nm

Explanation:

Moment, or torque, is the cross product of radius and force vectors.

τ = r × F

τ = (0.80 m) (60 N)

τ = 48 Nm

How much work is done by 0.30 m of gas if its pressure increases by 8.0 x105 Pa and the volume remains constant
Salerno

Answers

Answer:

0

Explanation:

if the volume remains constans, the works is 0 because the equation

W = P . ∆V

P = pressure

∆V = change in volume

Someone plzzz helpppppp with this last question

Answers

Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

A corpse is discovered in a room that has its temperature held steady at 25oC. The CSI ocers ar- rive at 2pm and the temperature of the body is 33oC. at 3pm the body's temperature is 31oC. Assuming Newton's law of cooling and that the temperature of the living person was 37oC, what was the approximate time of death

Answers

Answer: Around 0:35 Pm or 12:35 Am

Explanation:

The equation that describes the cooling of objects can be written as:

T(t) = Ta + (Ti - Ta)*e^(k*t)

Where Ta is the ambient temperature, here Ta = 25°C.

Ti is the initial temperature of the body, we have Ti = 37°C.

t is the time.

k is a constant.

So our equation is:

T(t) = 25°C +12°C*e^(k*t)

at 2pm, the temperature was 33°C

at 3pm, the temperature was 31°C.

we want to find the hour where we have our t = 0, suppose this hour is X.

then we can write our times as:

2pm ---> 2 - X

3pm ----> 3 - X

and our equations are:

33°C = 25°C + 12°C*e^(k2 - k*X)

31° = 25°C + 12°C*e^(k3 - k*X)

So we have two equations and two variables, let's solve the system.

first, simplify it a bit, for the first eq:

33 - 25 = 12*e^(k2 - k*X)

8/12 = e^(k2 - k*X)

ln(8/12) = k*2 - k*X

for the second equation we have:

31 - 25 = 12*e^(k3 - k*X)

6/12 = e^(k3 - k*X)

ln(6/12) = k*3 - k*X

So our equations are:

1) ln(2/3) = 2*k - X*k

2) ln(1/2) = 3*k - X*k

First, let's isolate one of the variables in one of the equations. let's isolate k in the first equation.

ln(2/3)/(2-X) = k

now we can replace it in the second equation:

ln(1/2) = 3*ln(2/3)/(2 - X) - X*ln(2/3)/(2-X)

now let's solve it for X, i will take a = ln(1/2) and b = ln(2/3) so it is easier to read.

a = 3*b/(2 - X) - X*b/(2 - X)

a*(2 - X) = 3*b - X*b

2a - aX = 3b - Xb

X(a - b) = 2a - 3b

X = (2*ln(1/2) - 3*ln(2/3))/(ln(1/2) - ln(2/3)) = 0.590

now, knowing that one hour has 60 minutes, then this is:

0.59*60m = 35 minutes

So the hour of death is 0:35 Pm or 12:35 Am

You are on a train traveling east at speed of 19 m/s with respect to the ground. 1)If you walk east toward the front of the train, with a speed of 1.5 m/s with respect to the train, what is your velocity with respect to the ground

Answers

Answer:

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Explanation:

Relative velocity with respect to the ground is;

Vbg = velocity of train with respect to the ground + your velocity with respect to the train

Vbg = Vtg + Vbt ......1

Given;

velocity of train with respect to the ground;

Vtg = 19 m/s

your velocity with respect to the train;

Vbt = 1.5 m/s

Substituting the given values into the equation 1;

Vbg = 19 m/s + 1.5 m/s

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

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