A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid

Answer:

[tex]d=0.019m[/tex]

Explanation:

From the question we are told that:

Far point [tex]x=0.759m[/tex]

Refractive power [tex]P=-1.35 D.[/tex]

Generally, the equation for Focal length is mathematically given by

[tex]F=\frac{1}{P}[/tex]

[tex]F=\frac{1}{-1.35}[/tex]

[tex]F=-0.74m[/tex]

Therefore

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

Where

[tex]u=o[/tex]

[tex]\frac{1}{-0.74}=\frac{1}{0}+\frac{1}{v}[/tex]

[tex]v=-0.74m[/tex]

Therefore,The between her glasses and eyes

[tex]d=x-v[/tex]

[tex]d=0.759-0.74m[/tex]

[tex]d=0.019m[/tex]

Answer:

Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)

Explanation:

Time=Distance/Speed

=14.5/6.63

=2.19s(to 3s.f.)

Acceleration=Final Velocity(v)-Initial Velocity(u)/Time

9.81=v-6.63/2.19

v-6.63=21.5

v=28.1m/s

Answer:

D) 1003 N​

Explanation:

Given the following data;

Mass of man = 85 kg

Acceleration of elevator = 2 m/s²

Acceleration due to gravity, g = 9.8 m/s²

To find the force exerted by the man on the floor;

Force = mg + ma

Answer:

The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Explanation:

From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h

The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,

Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s

First, convert 118 km/h to m/s

118 km/h = (118 × 1000) /3600 = 32.7778 m/s

∴ u = 32.7778 m/s

Now, to determine the deceleration, a, required to stop,

From one of the equations of motion for linear motion,

v² = u² + 2as

Then

0² = (32.7778)² + 2×a×85

0 = 1074.3841 + 170a

∴ 170a = - 1074.3841

a = - 1074.3841 / 170

a = - 6.3199

a ≅ - 6.32 m/s²

Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Answer:

simple is rumple a daily ok I'll be

Answer:

[tex]F=133N[/tex]

Explanation:

From the question we are told that:

Length [tex]l=3.0m[/tex]

Mass [tex]m=24kg[/tex]

Distance from Tip [tex]d=35cm[/tex]

Generally, the equation for Torque Balance is mathematically given by

[tex]mg(l/2)=F(l-d)[/tex]

[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]

Therefore

[tex]F=133N[/tex]

Answer:

the distance of the fish (as  seen by the bird) is greater than the actual distance.

Reason-

it is due to the apparent depth and differences between the refractive indices.

Have a nice day!

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

[tex]x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}[/tex] (1)

Where:

[tex]x_{o}[/tex] - Initial x-position, in meters.

[tex]v_{o,x}[/tex] - Initial x-velocity, in meters per second.

[tex]t[/tex] - Time, in seconds.

[tex]a_{x}[/tex] - x-acceleration, in meters per second.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:

[tex]x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}[/tex]

[tex]x(t) = 0.324\,m[/tex]

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

[tex]y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}[/tex] (2)

Where:

[tex]y_{o}[/tex] - Initial y-position, in meters.

[tex]v_{o,y}[/tex] - Initial y-velocity, in meters per second.

[tex]t[/tex] - Time, in seconds.

[tex]a_{y}[/tex] - y-acceleration, in meters per second.

If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o,y} = -3.6\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{y} = 0\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:

[tex]y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}[/tex]

[tex]y(t) = -2.16\,m[/tex]

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ([tex]v_{x}[/tex]), in meters per second, is calculated by this kinematic formula:

[tex]v_{x}(t) = v_{o,x} + a_{x}\cdot t[/tex] (3)

If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-velocity of the skateboarder is:

[tex]v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)[/tex]

[tex]v_{x}(t) = 1.08\,\frac{m}{s}[/tex]

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

[tex]v_{y} = -3.6\,\frac{m}{s}[/tex]

The y-velocity of the skateboard is -3.6 meters per second.

Answer:

  Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

        a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

   a = 2 lines 1/600

   a = 2/600

    a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

      tan θ = y / L

as the measured angles are small

      tan θ = sin θ / cos θ sin θ

      sin θ = y / L

we substitute

     a  y/L = λ

     y = λ L / a

for λ = 400 10-9 m

      I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

      i = 346.89 10⁻⁶ m

f

or λ = 700 nm

        y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

        y_f = 609.609 10⁻⁶ m

the separation of this spectrum

        Δr = v_f - i

        Dr = (609.609 - 346)  10 ⁻⁶

        Dr = 263 10⁻⁶ m

Answer:

56.1 kg

Explanation:

Given

[tex]T = 2200Nm[/tex] -- torque

[tex]d_1 = 1.0m[/tex]

[tex]d_2 = 3.0m[/tex]

Required

The mass of the diver

From the question, we understand that the diver is at the extreme of the board.

So, we make use of the following torque equation

[tex]T = F * (d_1 + d_2)[/tex]

Where:

[tex]F \to Force[/tex]

So, we have:

[tex]2200 = F * (1.0 + 3.0)[/tex]

[tex]2200 = F * 4.0[/tex]

Divide both sides by 4.0

[tex]550 = F[/tex]

[tex]F = 550 N[/tex] --- This is the force exerted by the diver (in other words, the weight).

To calculate the mass, we use:

[tex]F = mg[/tex]

Make m the subject

[tex]m = \frac{F}{g}[/tex]

This gives:

[tex]m = \frac{550}{9.8}[/tex]

[tex]m = 56.1kg[/tex]

Answer:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

Explanation:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.

If the hypothesis and the model used is correct, the relationship to be tested is

              T² =(4π² /g)   L  

by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.

Answer:

velocity is the answer of this question.

Answer:

Velocity is the right answer ok

Answer:

speed = 6.71 mph and angle is 71.2 degree.

Explanation:

speed of person, u = 3 miles per hour

speed of water, v = 6 miles per hour

Resultant speed

[tex]V =\sqrt{v^2 + u^2}\\\\V = \sqrt{3^2 + 6^2}\\\\V = 6.71 mph[/tex]

The angle from the west is

tan A = 6/2 = 3

A = 71.6 degree

Answer:

Option (a) is correct.

Explanation:

The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.

When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.

Answer:

4.56×10¯⁷¹ N

Explanation:

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

Distance apart (r) = 1.10 m

Force (F) =?

NOTE:

Gravitational constant (G) = 6.67×10¯¹¹ Nm² /Kg²

Mass of electron = 9.1×10¯³¹ Kg

Mass of the two elections = M₁ = M₂ = 9.1×10¯³¹ Kg

Thus, we can obtain the force of attraction between the two elections as illustrated below:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × (9.1×10¯³¹)² / (1.1)²

F = 4.56×10¯⁷¹ N

Thus, the force of attraction between the two elections is 4.56×10¯⁷¹ N

Answer:

(A) Series and Parallel

Explanation:

Circuit Component: These are electrical devices that makes up the circuit. They include, resistors, capacitors, inductors, voltmeters, ammeters, cell/batteries, earth connection, bulb, switch, connecting wire etc.

These component can either be connected in series or in parallel.

(1) Series Connection: This can be refered as end to end connection of electric component.

(2) Paralel Connection: This can be refered as the side to side connection of electric component.

From the question above,

A electric component in a circuit can be combined in series and in parallel.

The right option is (A) Series and Parallel

When the rock is suspended in the air, the net force on it is

∑ F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

∑ F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

∑ F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³

Answer:

The correct answer is - c.  Spaceship will have a velocity to the East and will be slowing down.

Explanation:

In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.

As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.

Answer:

The current is more in the parallel combination than in the series combination.

Explanation:

two resistances, R1 and R2 are connected to a battery of voltage V.

When they are in series,

R = R1 + R2

In series combination, the current is same in both the resistors, and it is given by Ohm's law.

V = I (R1 + R2)

[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)

When they are connected in parallel.

the voltage is same in each resistor.

The effective resistance is R.

[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]

So, the current is

[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)

So, the current is more is the parallel combination.

Answer:

metal

Explanation:

sound can travel best in materials with higher elastic properties like metal than it can through other solids like plastic or rubber which have lower elastic properties

I hope this helps

Answer:

Upthrust = 19.6 N

Explanation:

When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.

Upthrust = density of liquid x gravitational force x volume of object

i.e U = ρ x g x vol

Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]

So that;

U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])

   = 19.6 Kg m/[tex]s^{2}[/tex]

U = 19.6 Newtons

The upthrust on the iron is 19.6 N.

Answer:

a circular motion a constant force can be measured

Explanation:

The force is expressed by the relation

         F = m a

The bold are vectros.

Therefore, when your partner moves away, he has a reading of a force, so that this force remains constant there must be an acceleration at all times, one way to achieve these is with a circular motion with constant speed, in this case the module of the velocity is constant, but the direction changes at each point and there is an acceleration at each point.

Consequently with a circular motion a constant force can be measured

The force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.

A force can be defined as an influence that can change the motion of an object. The force is expressed by the relation

[tex]F = m a[/tex]

The force is dependent on the mass and acceleration of the object. The acceleration is a vector quantity, so the force will be a vector quantity.

Given that, in a lab room, you are sitting on a wheelchair at one end and at the other end, lab partner then moves away from you, pulling on the string that is attached to the force sensor.

When the lab partner moves away and pulled the string, there must be an acceleration during the motion. If the lab partner moves in a circular motion, then the velocity is constant but the direction changes at each point. There is an acceleration at each point that will be constant.  

As the force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.

To know more about the force, follow the link given below.

https://brainly.com/question/26115859.

Answer:

the angle between the electron's velocity and the magnetic field is 19⁰

Explanation:

Given;

magnitude of the magnetic field, B = 83 x 10⁻⁵ T

acceleration of the electron, a = 34 x 10¹³ m/s²

speed of the electron, v = 72 x 10⁵ m/s

mass of electron, m = 9.11 x 10⁻³¹ kg

The magnetic force experienced by the electron is calculated as;

F = ma = qvB sinθ

where;

q is charge of electron = 1.602 x 10⁻¹⁹ C

θ is the angle between the electron's velocity and the magnetic field.

[tex]sin(\theta ) = \frac{ma}{qvB} \\\\sin(\theta ) = \frac{(9.11\times 10^{-31})(34\times 10^{13})}{(1.602\times 10^{-19})\times (72\times 10^5) \times (83 \times 10^{-5})} \\\\sin(\theta ) = 0.3235\\\\\theta =sin^{-1}(0.3235)\\\\\theta =18.9^0[/tex]

[tex]\theta \approx 19^ 0[/tex]

Therefore, the angle between the electron's velocity and the magnetic field is 19⁰

Answer:

v² = u² + 2as

v² = 3600 + 6400

v² = 10000

v = 100

Explanation:

final velocity is 100 m/s

According to third equation of kinematics

[tex]\boxed{\sf v^2-u^2=2as}[/tex]

[tex]\\ \sf\longmapsto v^2=u^2+2as[/tex]

[tex]\\ \sf\longmapsto v^2=(60)^2+2(10)(320)[/tex]

[tex]\\ \sf\longmapsto v^2=3600+3400[/tex]

[tex]\\ \sf\longmapsto v^2=10000[/tex]

[tex]\\ \sf\longmapsto v=\sqrt{10000}[/tex]

[tex]\\ \sf\longmapsto v=100m/s[/tex]

Answer: [tex]1361.11\ N/C,\text{upward}[/tex]

Explanation:

Given

Mass of particle is [tex]m=2.5\ gm[/tex]

Charge of particle is [tex]q=18\ \mu C[/tex]

Electrostatic force must balance the weight of the particle

[tex]\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C[/tex]

Direction of the electric field is in upward direction such that it opposes the gravity force.

Using the left-hand rule,

[tex](4\,\vec\imath+\vec k)\times(2\,\vec\imath+\vec\jmath-3\,\vec k) = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\4&0&1\\2&1&-3\end{vmatrix} = -\vec\imath+14\,\vec\jmath+4\,\vec k[/tex]

Then in the right-handed rectangular coordinates, the cross product is the negative of this,

[tex]\boxed{\vec\imath-14\,\vec\jmath-4\,\vec k}[/tex]

Answer:

The Michelson-Morley was designed to detect the motion of the earth through the ether.

No such relation was found and the speed of light is assumed to be the same in all reference frames.

The Michelson-Morley experiment was designed to measure: A. the velocity of the Earth relative to the ether.

Michelson-Morley experiment is an experiment which was first performed in Germany by the American physicist named, Albert Abraham Michelson between 1880 to 1881.

However, the experiment was later modified and refined by Michelson and Edward W. in 1887.

The main purpose of the Michelson-Morley experiment was to measure the velocity of planet Earth relative to the luminiferous ether, which is a medium in space that is hypothetically said to carry light waves.

In conclusion, the Michelson-Morley experiment was designed to measure the velocity of the Earth relative to the hypothetical luminiferous ether.

Read more: https://brainly.com/question/13187705

Answer:

100m

Explanation:

i think this is the answer because the formula for distance is

d=speed×time in this case the speed is 10m/s and the time is 10s therefore the distance will be

10m/s×10s

=100m

I hope this helps

Answer:

100 m

Explanation: this is when you need to find velocity and the formula for velocity is displacement by time taken.

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

[tex]A = 10 \ cm^2[/tex]

   [tex]$=0.0010 \ m^2$[/tex]

d = 10 mm

  = 0.010 m

Then, Capacitance,

[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]

[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]

[tex]$C=2.655 \times 10^{12} \ F$[/tex]

[tex]$U_1 = \frac{1}{2}CV^2$[/tex]

[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]

Now,

[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]

And

[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]

In parallel combination,

[tex]$C_{eq}= C_k + C_{air}$[/tex]

[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]

[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]

[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]

Then energy,

[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]

[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]

b). Now the charge on the [tex]\text{capacitor}[/tex] is :

[tex]$Q=C_{eq} V$[/tex]

[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]

[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]

Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :

[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]

[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]

[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]

[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]

Without the dielectric,

[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]

[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]

[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]