The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?

Answer:

sup qwertyasdfghjk

Explanation:

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

Explanation:

hope this will help u

mark me as brinalist or vote me

Thankyou

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]

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,

Learn more about the tension here;

https://brainly.com/question/25743940

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

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:

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:

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

Answer:

The current is 0.67 A.

Explanation:

Density, J = 1 A/m^2

Area, A = 1 m^2

Let the radius is r. And outer is R.

Use the formula of current density

[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]

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:

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

Answer:

parametric representation: x = u, y = v - u ,  z = - v

Explanation:

Given vectors :

i - j ,  j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + ua + vb

where:  r₀ = position vector

            u and v = real numbers

             a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v,  z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u,  y₀ - u + v,   z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

∴ parametric representation: x = u, y = v - u ,  z = - v

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:

[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:

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

Answer:

a)[tex]V=1.067\: m/s[/tex]

b)[tex]v=434.65\: m/s [/tex]  

Explanation:

a)

Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.

[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]

Where:

Then, let's find the initial speed of the bullet-block system.

[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]

[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]

[tex]V=1.067\: m/s[/tex]

b)

Using the conservation of momentum we can find the velocity of the bullet.

[tex]mv=MV[/tex]

[tex]v=\frac{MV}{m}[/tex]

[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]

[tex]v=434.65\: m/s [/tex]  

I hope it helps you!

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.

Read more here: https://brainly.com/question/24029674

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.

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ2

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]

[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]

[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)

Where:

[tex]h[/tex] - Maximum height of the wood block, in meters.

[tex]v[/tex] - Initial speed of the block, in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]m[/tex] - Mass, in kilograms.

[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.

[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:

[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]h = 4.249\,m[/tex]

The wood block reaches a height of 4.249 meters above its starting point.

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

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:

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:

Explanation:

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

Answer:

9.25 m/s

Explanation:

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:

[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:

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