An elevator with its occupants weighs 2400 N and is supported by a vertical cable. What is the tension in the cable if the elevator is moving up with its speed decreasing at a rate of 1.7

Answer:

a

Explanation:

The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.

Answer:

1140 J, 6840 J, 10260 J

Explanation:

Lo x 2 Lo x 3 Lo, Lo = 0.2 m,  K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,

time, t = 6 s

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]

Answer:

the weight of the air in pound-force (lb-f) is 325.8 lbf

Explanation:

Given;

dimension of the room, = 15 ft by 15 ft by 20 ft

density of air in the room, ρ = 0.0724 lbm/ft³

The volume of air in the room is calculated as;

Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³

The mass of the air is calculated as;

mass = density x volume

mass = 0.0724 lbm/ft³  x  4,500 ft³

mass = 325.8 lb-m

The weight of the air is calculated as;

Weight = mass x gravity

Weight = 325.8 lb-m x 32.174 ft/s²

Weight = 10482.29 lbm.ft/s²

The weight of the air in pound-force (lb-f) is calculated as;

1 lbf = 32.174 lbm.ft/s²

[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]

Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf

Explanation:

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Answer:

The pressure is 2505 Pa.  

Explanation:

Height, h = 1.5 m

density of blood, d = 1000 kg/cubic meter

Gravity, g = 1.67 m/s^2

let the pressure is P.  

The pressure due to the fluid is given by

P = h d g

P = 1.5 x 1000 x 1.67

P = 2505 Pa

Answer: 11.36 m

Explanation:

Given

Mass of roller coaster is m=1200 kg

Initial speed of roller coaster is v=15 m/s

Energy at bottom and at the top is same i.e.

[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]

Thus, the highest point reach by the roller coaster is 11.36 m

Answer:

11.36m

Explanation:

Answer:

Explanation:

Answer:

D

Explanation:

This is a great way to manage health.

A would be avoiding everything which isnt good.

B. would be emotionally draining and damaging to bottle feelings and ignore them.

C. is unhealthy to not exercise and eat food while doing nothing.

Answer:

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

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

Answer:

[tex]n=3.8[/tex]

Explanation:

From the question we are told that:

Magnetic Field [tex]B=0.01T[/tex]

Current [tex]I=1.00[/tex]

Wire Diameter [tex]d_w=0.5*10^3m[/tex]

Layers Diameter [tex]d_l=1*10^2m[/tex]

Length [tex]l=0.2m[/tex]

Generally the equation for number of layers is mathematically given by

[tex]n=\frac{Bd_w}{\mu_o I}[/tex]

Where

[tex]Vacuum\ permeability=\mu_0[/tex]

[tex]n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }[/tex]

[tex]n=3.8[/tex]

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

Answer:

Explanation:

Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container

It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.

It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.

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Answer:

a) ΔT₁ = -4.68 N,   ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N

Explanation:

In this exercise we will use Newton's second law.

         ∑F = m a

Let's start with the set of three cars

         F_total = M a

         F_total = M 0.12

where the total mass is the sum of the mass of each charge

          M = m₁ + m₂ + m₃

This is the force with which the three cars are pulled.

Now let's write this law for each vehicle

car 1

         F_total - T₁ = m₁ a

         T₁ = F_total - m₁ a

car 2

         T₁ - T₂ = m₂ a

         T₂ = T₁ - m₂ a

car 3

         T₂ = m₃ a

note that tensions are forces of action and reaction

a) They tell us that 39 kg is removed from car 2 and placed on car 1

         m₂’= m₂ - 39

         m₁'= m₁ + 39

         m₃ ’= m₃

they ask how much each tension varies, let's rewrite Newton's equations

The total force does not change since the mass of the set is the same F_total ’= F_total

car 1

           F_total ’- T₁ ’= m₁’ a

           T₁ ’= F_total - m₁’ a

           T₁ ’= (F_total - m₁ a) - 39 a

           T₁ '= T₁ - 39 0.12

           ΔT₁ = -4.68 N

car 2

           T₁’- T₂ ’= m₂’ a

           T₂ ’= T₁’- m₂’ a

           T₂ '= (T₁'- m₂ a) + 39 a

           T₂ '= T₂ + 39 0.12

           ΔT₂ = 4.68 N

b) in this case the masses remain

            m₁ '= m₁

           m₂ ’= m₂ - 39

           m₃ ’= m₃ + 39

we write Newton's equations

car 3

          T₂ '= m₃' a

          T₂ ’= (m₃ + 39) a

          T₂ '= m₃ a + 39 a

          T₂ '= T₂ + 39 0.12

          ΔT₂ = 4.68 N

car 1

            F_total - T₁ ’= m₁’ a

            T₁ ’= F_total - m₁ a

car 2

            T₁' -T₂ '= m₂' a

            T₁ ’= T₂’- m₂’ a

            T₁ '= (T₂'- m₂ a) + 39 a

            T₁ '= T₁ + 39 0.12

            ΔT₁ = 4.68 N

The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

Tension is the pulling force carried by the flexible mediums like ropes, cables and string.

Tension in a body due to the weight of the hanging body is the net force acting on the body.

At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.

The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,

[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]

On solving the above 3 equation, we get the values of tension in each bar as,

[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]

The tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]

The tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]

Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

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Answer:

a) [tex] \alpha = 1.28 rad/s^{2} [/tex]  

b) Option ii. The angular acceleration would increase

Explanation:

a) The angular acceleration is given by:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular speed = v/r

v: is the tangential speed = 6.35 m/s

r: is the radius = 45.0 cm = 0.45 m

[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)

t: is the time = 11.0 s

α: is the angular acceleration

Hence, the angular acceleration is:

[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]  

b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).

Therefore, the correct option is ii. The angular acceleration would increase.

I hope it helps you!  

Answer:

elastic energy

Explanation:

When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.

During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.

The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:

elastic energy

Answer:

The force on the second piston is 5246.5 N .

Explanation:

Area of first piston, a = 3.002 cm^2

Area of second piston, A = 315 cm^2

Force on first piston, f = 50 N

let the force of the second piston is F.

According to the Pascal's law

[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

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].

Answer:

sup qwertyasdfghjk

Explanation:

Answer:

9.25 m/s

Explanation:

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

The time period of the spring is 2[tex]\pi[/tex][√(m/k)].

The spring constant of a spring is defined as the measurement of ratio of the force that is exerted on the spring to the displacement caused by it.

Here,

The mass of the block = m

Let F be the applied force on the spring and k be the spring constant.

When the mass attached to the spring is pulled to one side then released, it executes SHM.

Therefore we can write that, the applied force,

F = kx

Restoring force = -kx

According to Newton's law, we know that,

F = ma

So,

ma = -kx

Therefore, the acceleration,

a = (-k/m) x

For an SHM, the acceleration is given as,

a = -ω²x

Therefore, we can write that,

-ω²x = (-k/m) x

ω² = k/m

So, the time period of the spring,

T = 2[tex]\pi[/tex]/ω

T = 2[tex]\pi[/tex][√(m/k)]

Hence,

The time period of the spring is 2[tex]\pi[/tex][√(m/k)].

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Answer:

Decreases

Explanation:

because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again

The final speed of the ball is 91.72 m/s.

Given the following data:

To find the final speed of the ball, we would use the following formula:

[tex]F = \frac{M(V - U)}{t}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]

Final speed, V = 91.72 m/s

Therefore, the final speed of the ball is 91.72 m/s.

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Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad