A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d

Answers

Answer 1

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

A Uniform Horizontal Bar Of Mass M1 And Length L Is Supported By Two Identical Massless Strings. String

Related Questions

Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.

Required:
What is the total work performed on the toolbox?

Answers

If both forces are measured in Newtons, then the net force is

F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N

The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector

d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m

The total work done by the astronauts on the toolbox is then

F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J

The work done by the two astronauts is equal to 96 J.

What is work done?

work done?Work done is defined as the product of force applied and the distance moved by the force.

Work done = Force × Distance

The forces applied = 18+16 N, 7+ -10 N, and -12 + 16N

Forces = 34 N, -3 N, and 4N

Distances = (17 - 15, 14 - 14, -1 - - 8) m

Distances = 2, 0, 7

Work done = 34 × 2 + -3 × 0 + 4 × 7

Work done = 96 J

Therefore, the work done by the two astronauts is equal to 96 J.

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Based on the information in the table, what
is the acceleration of this object?

t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2

Answers

Answer:

Option A. –5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (v₁) = 9 m/s

Initial time (t₁) = 0 s

Final velocity (v₂) = –6 m/s

Final time (t₂) = 3 s

Acceleration (a) =?

Next, we shall determine the change in the velocity and time. This can be obtained as follow:

For velocity:

Initial velocity (v₁) = 9 m/s

Final velocity (v₂) = –6 m/s

Change in velocity (Δv) =?

ΔV = v₂ – v₁

ΔV = –6 – 9

ΔV = –15 m/s

For time:

Initial time (t₁) = 0 s

Final time (t₂) = 3 s

Change in time (Δt) =?

Δt = t₂ – t₁

Δt = 3 – 0

Δt = 3 s

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Change in velocity (Δv) = –15 m/s

Change in time (Δt) = 3 s

Acceleration (a) =?

a = Δv / Δt

a = –15 / 3

a = –5 m/s²

Thus, the acceleration of the object is

–5 m/s².

Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system

Answers

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

Which indicates the first law of thermodynamics ​

Answers

Answer:

(d)

Explanation:

because dU = Q -W so ,that the option d(D) is correct

Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)

Answers

Answer:

[tex]M_b=6kg[/tex]

Explanation:

From the question we are told that:

Coefficient of restitution [tex]\mu=1.00[/tex]

Mass A [tex]M_a=6kg[/tex]

Initial Velocity of A [tex]U_a=6m/s[/tex]

Initial Velocity of B [tex]U_b=0m/s[/tex]

Generally the equation for Coefficient of restitution is mathematically given by

 [tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]

 [tex]1=\frac{v_B}{6}[/tex]

 [tex]V_b=6*1[/tex]

 [tex]V_b=6m/s[/tex]

Generally the equation for conservation of linear momentum  is mathematically given by

 [tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]

 [tex]6*6+=M_b*6[/tex]

 [tex]M_b=6kg[/tex]

What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5

Answers

Answer:

2) when acceleration triples force triples,  3) a diagram with dynamic friction force in the opposite direction of movement of the car

4)  a = 2.44 ft / s², 5)  fr = 894.3 N

Explanation:

In this exercise you are asked to answer some short questions

2)  Newton's second law is

         F = m a

when acceleration triples force triples

3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.

The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car

4) let's use the scientific expressions

          v = v₀ - a t

as the car stops v = 0

          a = v₀ / t

let's reduce the magnitudes

          v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s

          a = 58.667 / 24

          a = 2.44 ft / s²

5) let's use Newton's second law

           fr = m a

We must be careful not to mix the units, we will reduce the acceleration to the system Yes

           a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²

           fr = 1200  0.745

           fr = 894.3 N

190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?

Answers

Answer:

W = 4.75 KW

Explanation:

First, we will calculate the heat to be removed:

Q = (No. of students)(Metabolic Power of Each Student)

Q = (190)(125 W)

Q = 23750 W = 23.75 KW

Now the formula of COP is:

[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]

W = 4.75 KW

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.

Answers

Answer:

[tex]\lambda=6.83\times 10^{-5}\ m[/tex]

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]

So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?

Answers

Answer:

1. 588 N

2. 738 N

3. 588 N

Explanation:

time, t = 4 s

initial velocity, u = 0

final velocity, v = 10 m/s

mass, m= 60 kg

1.

Weight of passenger before starts

W =m g = 60 x 9.8 = 588 N

2.

When the elevator is speeding up

v = u + a t

10 = 0 + a x 4

a = 2.5 m/s2

Now the weight is

W' = m (a + g) = 60 (9.8 + 2.5) = 738 N

3.

When he reaches the cruising speed, the weight is

W = 588 N

A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km

Answers

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.​

Answers

Answer:

1568J

Explanation:

Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.

Use conservation of Energy

ΔUg+ΔKE=0

ΔUg= mgΔh=2*9.8*(20-100)=-1568J

ΔKE-1568J=0

ΔKE=1568J

since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J

Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.​

Answers

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI  

Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results

Answers

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

how did kepler discoveries contribute to astronomy

Answers

Answer:

They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.

Explanation:

How do you find the product of gamma decay?

Answers

Answer:

The mass and atomic numbers don't change

Explanation:

An excited atom relaxes to the ground state emitting a photon...called a gamma ray.

The answer is that the mass and atomic numbers don't change.

In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.

To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.

During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.

The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).

For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.

It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).

Thus, the product nucleus remains unchanged in terms of atomic number and mass number.

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The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

Answers

Answer:

Explanation:

From the information given:

The angular frequency ω = 430 π rad/s

The wavenumber k = 4.00π which can be expressed by the equation:

k = ω/v

4.00 =  430 /v

v = 430/4.00

v = 107.5 m/s

Similarly: k  = ω/v = 2πf/fλ

We can say that:

k = 2π/λ

4.00 π = 2π/λ

wavelength λ = 2π/4.00 π

wavelength λ = 0.5 m

frequency of the wave can now be calculated by using the formula:

f = v/λ

f = 107.5/0.5

f = 215 Hz

Also, the Period(T) = 1/215 secs

The time at which particle proceeds from point A  to its maximum upward displacement  and to its maximum downward displacement  can be computed as t = T/2;

Thus, the distance(x) covered by each wave during this time interval(T/2) will be:

x = v * t

x = v * T/2

x = λ/2

x = 0.5/2

x =  0.25 m

what is time taken by radio wave to go and return back from communication satellite to earth??​

Answers

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

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~~~~NEED HELP ASAP~~~~

Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.

Block A= 3kg

Block B= 4kg

Block A velocity before collision =30m/s.

Block B velocity before collision = 15 m/s

The velocity of block B after the collision is 20m/s.


a.) What is the velocity of block A after collision?

b.) Is the collision elastic? Show work to explain answer why or why not.

Answers

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic

water contracts on freezing is it incorrect or conrrect

Answers

Answer:

hope it helps

much as you can

Which phase of matter makes up stars?
O liquid
O gas
O plasma

Answers

Answer:

The answer to this question is plasma

Answer:

Plasma

Explanation:

The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answers

Complete question:

A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answer:

The tension on the rope is 1.95 N

Explanation:

The general equation of a progressive wave is given as;

[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]

Compare the given equation with the general equation of wave, the following parameters will be deduced.

A = 0.75 cm

k = 0.400π cm⁻¹

ω = 250π s⁻¹

The frequency of the wave is calculated as;

ω = 2πf

2πf = 250π

2f = 250

f = 250/2

f = 125 Hz

The wavelength of the wave is calculated as;

[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]

The velocity of the wave is calculated as;

v = fλ

v = 125 x 0.05

v = 6.25 m/s

The tension on the rope is calculated as;

[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]

Therefore, the tension on the rope is 1.95 N

vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?​

Answers

Answer:

45 × 8 units = A + B as formular

* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)​

Answers

Explanation:

Given

Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]

time taken to reach ground is given by

[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]

(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]

From the figure, it can be written

[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]

Considering horizontal motion

[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]

(c) The vertical velocity with which it strikes the ground is given by

[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]

Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]

Explanation:

Given

Ball is projected horizontally from a building of height  

time taken to reach ground is given by

(b) Line joining the point of projection and the point where it hits the ground makes an angle of  

From the figure, it can be written

Considering horizontal motion

(c) The vertical velocity with which it strikes the ground is given by

Thus, the ball strikes with a vertical velocity of

A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips

Answers

Answer:

I could not find the answer or do it myself if I did find it I would defenetly share

Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.

Answers

Answer:

I = 0.25 A

Explanation:

Given that,

Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.

In series combination, the equivalent resistance is given by :

[tex]R=R_1+R_2+R_3+....[/tex]

So,

[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]

The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.

V = IR

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]

So, the current of 0.25 A passes through each bulb.

what is conservation energy?

Answers

Explanation:

Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant

hope it is helpful to you

The pressure exerted at the bottom of a column of liquid is 30 kPa. The height of the
column is 3,875 m. What type of liquid is used?

Answers

Answer:

For example, the pressure acting on a dam at the bottom of a reservoir is ... pressure = height of column × density of the liquid × gravitational field ... The density of water is 1,000 kg/m 3.

a. Give an example of the conversion of light energy to electrical energy.

b. Give an example of chemical energy converting to heat energy.

c. Give an example of mechanical energy converting to heat energy.

Answers

Explanation:

a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy

b) burning of coal converts chemical energy to heat energy

c) rubbing of both hands against each other converts mechanical to heat energy

Answer:

a. solar cells

b.coal,wood,petroleum

c.rubbing ours palms

Which level of government relies the most on income tax?
OA.
federal
state
OC.
local

Answers

Answer:

Its the Federal government

Federal government tax

Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.

Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?

Answers

Answer:

you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q

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