Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.

Answers

Answer 1

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) g

• the upward normal force of the surface, magnitude n₁

• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction

• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction

• the applied force, mag. F, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) g, pointing downward

• normal force, mag. n₂, pointing upward

• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction

• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

n₂ - (3.0 kg) g = 0   ==>   n₂ = (3.0 kg) g   ==>   f₂ = 0.30 (3.0 kg) g

• net horizontal force:

c₂ - f₂ = 0   ==>   c₂ = 0.30 (3.0 kg) g8.8 N


Related Questions

Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.

Diagram A
Diagram B
Diagram C
Diagram D

Answers

Answer:Diagram A

Explanation:

Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.

describe the cause of earth's magnetism ?​

Answers

The magnetic field of Earth is caused by currents of electricity that flow in the molten core. ... The currents flow in the outer core, and the lines of force shown in yellow, travel outwards through the rest of the earth's interior. If the earth rotated faster, it would have a stronger magnetic field.

0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b. 300 K c. 250 K d. 125 K​

Answers

Answer:

c. 250k

Explanation:

The temperature of the sink is approximately 250 K.

To find the temperature of the sink, we can use the formula for the efficiency of a heat engine:

Efficiency = 1 - (Temperature of Sink / Temperature of Source)

Given that the temperature of the source (T_source) is 500 K and the source energy (Q_source) is 2003 J, and the sink energy (Q_sink) is 100 J, we can rearrange the formula to solve for the temperature of the sink (T_sink):

Efficiency = (Q_source - Q_sink) / Q_source

Efficiency = (2003 J - 100 J) / 2003 J

Efficiency = 1903 J / 2003 J

Efficiency = 0.9497

Now, plug the efficiency back into the first equation to solve for T_sink:

0.9497 = 1 - (T_sink / 500 K)

T_sink / 500 K = 1 - 0.9497

T_sink / 500 K = 0.0503

Now, isolate T_sink:

T_sink = 0.0503 * 500 K

T_sink = 25.15 K

Since the temperature should be in Kelvin, we round down to the nearest whole number, which is 25 K. Thus, the temperature of the sink is approximately 250 K.

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A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?​

Answers

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

[tex]h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5 \ m[/tex]

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

Answers

Answer:

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Explanation:

We are given that

[tex]PV^{1.4}=C[/tex]

Where C=Constant

[tex]\frac{dP}{dt}=-7KPa/minute[/tex]

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]

Substitute the values

[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]

[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]

[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Answer:

[tex]\dot V=2786.52~cm^3/min[/tex]

Explanation:

Given:

initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]

initial volume during the process, [tex]V_1=420~cm^3[/tex]

The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.

Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]

Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:

[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]

[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]

[tex]\dot V=2786.52~cm^3/min[/tex]

[tex]\because P\propto\frac{1}{V}[/tex]

[tex]\therefore[/tex] The rate of change in volume will be increasing.

Which is the most difficult subject?​

Answers

Answer:

Quantum Mechanics

Explanation:

Well, that's what I think personally.

Physical quantities expresed only by their magnitude is​

Answers

Answer:

Scalar quantity is the Physical quantity expresed only by their magnitude.

Given a 64.0 V battery and 30.0 Ω and 88.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W (b) Repeat when the resistances are in parallel. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W

Answers

Answer:

a. i. 0.542 A ii. 8.813 W iii. 0.542 A iv. 25.85 W

b. i. 2.13 A ii. 136.53 W iii. 0.727 A iv. 46.55 W

Explanation:

a. Find the current (in A) and power (in W) for each when connected in series.

Since the resistors are connected in series, their combined resistance is R = R₁ + R₂ where R₁ = 30.0 Ω and R₂ = 88.0 Ω.

So, substituting the values of the variables into the equation, we have

R = R₁ + R₂

R =  30.0 Ω +  88.0 Ω

R =  118.0 Ω

Since from Ohm's law, V = IR where V = voltage across circuit = battery voltage = 64.0 V, I = current in circuit and R = total resistance of circuit = 118.0 Ω

So, I = V/R = 64.0V/118.0 Ω = 0.542 A

Since the resistors are in series, the same current flows through them

i. Current in 30.0 Ω

Current in 30.0 Ω is I = 0.542 A since the resistors are in series.

ii Power in the 30.0 Ω

The power in the 30.0 Ω is P₁ = I²R₁ where I = current = 0.542 A and R₁ = resistance = 30.0 Ω

So, P₁ = I²R₁

= (0.542 A)² × 30.0 Ω

= 0.293764  A² × 30.0 Ω

= 8.8129 W

≅ 8.813 W

iii. Current in 88.0 Ω

Current in 88.0 Ω is I = 0.542 A since the resistors are in series.

iv. Power in the 88.0 Ω

The power in the 88.0 Ω is P = I²R₂ where I = current = 0.542 A and R₂ = resistance = 88.0 Ω

So, P₂ = I²R₂

= (0.542 A)² × 88.0 Ω

= 0.293764  A² × 88.0 Ω

= 25.8512 W

≅ 25.85 W

(b) Repeat when the resistances are in parallel.

Since the resistors are connected in parallel, the same voltage is applied across them.

i. Current in 30.0 Ω

Using Ohm's law, V = I₁R₁ where V = voltage = 64.0 V, I₁ = current in 30.0 Ω resistor and R₁ = resistance = 30.0 Ω

So, I₁ = V/R₁ = 64.0 V/30.0 Ω = 2.13 A

ii Power in the 30.0 Ω

The power in the 30.0 Ω resistor is P₁ = V²/R₁ where V = voltage across resistor = 64.0 V and R₁ = resistance = 30.0 Ω

So, P₁ = V²/R₁

P₁ = (64.0 V)²/30.0 Ω

P₁ = 4096 V²/30.0 Ω

P₁ = 136.53 W

iii. Current in 88.0 Ω

Using Ohm's law, V = I₂R₂ where V = voltage = 64.0 V, I₂ = current in 88.0 Ω resistor and R₂ = resistance = 88.0 Ω

So, I₂ = V/R₂ = 64.0 V/88.0 Ω = 0.727 A

iv. Power in the 88.0 Ω

The power in the 30.0 Ω resistor is P₂ = V²/R₂ where V = voltage across resistor = 64.0 V and R₂ = resistance = 88.0 Ω

So, P₂ = V²/R₂

P₂ = (64.0 V)²/88.0 Ω

P₂ = 4096 V²/88.0 Ω

P₂ = 46.55 W

Please show steps as to how to solve this problem
Thank you!

Answers

Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.

Answers

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T

Explain the following observations:
a) A balloon filled with hydrogen gas floats in air;
B) A ship made of steel floats on water.

Answers

Answer and Explanation:

a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.

b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.

A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?

Answers

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?

Answers

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.

Answers

Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

[tex]q_{1} = q_{2} = +q[/tex]

[tex]q_{3} = -q[/tex]

As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).

Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r

Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]

where,

k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]

Substitute the values into above formula as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)

Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex]   ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

What is the pH of a solution with a hydrogen ion concentration of 2.0x10^3.(Use 3 digits)

Answers

Answer:

2.70

Explanation:

pH = -log[H+]

pH = -log[2.0x10^-3]

pH = 2.70

need help pleaseee,question is in the pic​

Answers

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a
distance of 20 m before bouncing.
(i) Calculate how long it takes the ball to travel this distance.
(1 mark)

Answers

Answer:

t=0.417s

Explanation:

After the ball hits the racket it is in freefall(assume air resistance as negligible)

so a=-g

use

x-x0=v0t+1/2at^2

Plug in givens

20=50t-4.9t^2

Solve quadratic equation using quadratic formula

t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)

How does the theory of relativity explain the gravity exerted by massive objects?
A. More massive objects create stronger forces of gravity.
B. More massive objects create shallower curves of space-time.
C. More massive objects pull objects from farther away.
D. More massive objects create larger curves of space-time.

Answers

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

The theory of relativity explain the gravity exerted by massive objects is

more massive objects create larger curves of space-time (option-d).

Do bigger objects exert more gravity?

The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.

Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.

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An aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft dears the building by 12 m, how far from the building does the aircraft touch down on the runway

Answers

The aircraft is 12 meters higher than the building so it is at 45 + 12 = 57 meters high.

For every 12 meters it travels it drops 1 m.

Divide the height by 12 to find the distance it travels:

57 / 12 = 4.75

It touches down 4.75 meters from the building.

The building is 684 meters away from the aircraft touching down on the runway.

What are trigonometric functions?

A right-angled triangle's side ratios are the easiest way to express a function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant. These functions are known as trigonometric functions.

As given in the problem an aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft clears the building by 12 m,

the total height of the aircraft when it clears the building = 45 +12

the total height of the aircraft when it clears the building is 57 meters

It is given that the Glide ratio is 12:1,

The distance of the building from touch down on the runway = 12 ×57

The distance of the building from the touch-down on the runway is 684 meters.

Thus, the building is 684 meters away from the aircraft touching down on the runway.

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NEED HELP ASAP. Please show all work.

A point on a rotating wheel (thin hoop) having a constant angular velocity of 200 rev/min, the wheel has a radius of 1.2 m and a mass of 30 kg. ( I = mr2 ).
(a) (5 points) Determine the linear acceleration.
(b) (4 points) At this given angular velocity, what is the rotational kinetic energy?

Answers

Answer:

Look at work

Explanation:

a) I am not sure if you want tangential or centripetal but I will give both

Centripetal acceleration = r*α

Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2

Tangential acceleration = ω^2*r

convert 200rev/min into rev/s

200/60= 10/3 rev/s

a= 100/9*1.2= 120/9= 40/3 m/s^2

b) Rotational Kinetic Energy = 1/2Iω^2

I= mr^2

Plug in givens

I= 43.2kgm^2

K= 1/2*43.2*100/9=2160/9=240J

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

Investigators measure the size of fog droplets using the diffraction of light. A camera records the diffraction pattern on a screen as the droplets pass in front of a laser, and a measurement of the size of the central maximum gives the droplet size. In one test, a 690 nm laser creates a pattern on a screen 30 cm from the droplets. If the central maximum of the pattern is 0.24 cm in diameter, how large is the droplet?

Answers

Answer:

the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm

Explanation:

Given the data in the question;

Diameter of bright central maxima;

⇒ 2 × ( 1.22 × (λD/d) ) ⇒ 2.44( λD/d )

where D is the distance from the the droplet to the screen ( 30 cm )

d is the diameter of the droplet

λ is the wavelength of light ( 690 nm  = 690 × 10⁻⁷ cm )

since the central maximum of the pattern is 0.24 cm in diameter,

we substitute

0.24 cm = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / d )

solve for d

d = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / 0.24 cm

d = 0.0050508 cm² / 0.24 cm

d = 0.021045 cm or 2.1 × 10⁻² cm

Therefore, the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm

Action and reaction are equal in magnitude and opposite in direction.Then Why do not balance each other

Answers

Answer:

Action and reaction are equal in magnitude and opposite in direction but they do not balance each other because they act on different objects so they don't cancel each other out.

hope this will help you more

Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080 arcsec,as measured by the Hipparcos satellite. What is the distance to Betelgeuse, and what is the uncertainty in that measurement?

Answers

Betelgeuse, the bright red star in the constellation of Orion the Hunter, is in the end stage of its stellar life. Astronomers have long thought it will someday explode to become a supernova. In late 2019 and early 2020, Betelgeuse generated a lot of

We have that  the distance to Betelgeuse, and the uncertainty in that measurement is

[tex]d=(221.7\pm39.33)pc[/tex]Uncertainty U = 0.00080

From the Question we are told that

Betelgeuse (in Orion)  has a parallax of 0.00451 + 0.00080

Generally

[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]

Where

Parallax [tex]P =0.00451[/tex]

Uncertainty [tex]U = 0.00080[/tex]

Generally the equation for the distance  is mathematically given as

[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]

Therefore

[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]

[tex]d=(221.7\pm39.33)pc[/tex]

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A wheel accelerates so that it's angular speed increases uniformly from 150 rads/s to 580 rads/s in 16 revolutions.Cakcjlate its angular acceleration. ​

Answers

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

A is the angular acceleration.t is the timeVw is the final angular speed.Uw is the initial angular speed.

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

A 5 kg object is moving in a straight-line with an initial speed of v m/s. It takes 13 s for the speed of the object to increase to 13 m/s and it kinetic energy increases at a rate of 15 J/s. What is the initial speed v (in m/s)?

Answers

The object's kinetic energy changes according to

dK/dt = 15 J/s

If v is the object's initial speed, then its initial kinetic energy is

K (0) = 1/2 (5 kg) v ²

Use the fundamental theorem of calculus to solve for K as a function of time t :

[tex]K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t[/tex]

After t = 13 s, the object's kinetic energy is

K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J

Put this as the left side in the equation above for K(t) and solve for v :

[tex]422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)[/tex]

==>   v9.5 m/s

A child with a weight of 230 N swings on a playground swing attached to 2.20-m-long chains. What is the gravitational potential energy of the child-Earth system relative to the child's lowest position at the following times?
(a) when the chains are horizontal (in J)
(b) when the chains make an angle of 33.0° with respect to the vertical (in J)
(c) when the child is at his lowest position (in J)

Answers

Answer:

a)  U = 506 J, b)  U = 37.11 J, c) U = 0

Explanation:

The gravitational power energy is given by the expression

         U = m g (y -y₀)

In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0

Let's use trigonometry to find the child's height

          h = y = L - L cos θ

         

we substitute

           U = m g L (1 - cos θ)

a) when the chain is horizontal θ = 90 and cos 90 = 0

           U = mg L

weight and mass are related

            W = mg

            m = W / g

           

           

           U = 230 2.20

           U = 506 J

b) θ = 33.0º

           cos 33 = 0.83867

           U = 230 (1 - 0.83867)

           U = 37.11 J

c) in this case θ = 0 cos 0 = 1

            U = 0

MCQ
................ ​

Answers

Answer:

I think it would be (-7 C )..

What power (in kW) is supplied to the starter motor of a large truck that draws 260 A of current from a 25.5 V battery hookup

Answers

Answer:

P = 6.63 kW

Explanation:

Given that,

Current, I = 260 A

Voltage of the battery, V = 25.5 V

We need to find the power supplied to the starter motor. We know that,

P = VI

Put all the values,

P = 25.5 × 260

P = 6630 W

or

P = 6.63 kW

So, the power supplied to the motor is 6.63 kW.

Answer:

The power is 6.63 kW.

Explanation:

Current, I = 260  A

Voltage, V = 25.5 V

Power of an electrical appliance is given by

P = V I

P = 25.5 x 260

P = 6630 W

1 kW = 1000 W

So, the power is

P = 6.63 kW

The unit of kinetic energy is the _______. The unit of kinetic energy is the _______. hertz meter watt joule radian

Answers

Answer:

joule

Explanation:

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