a pool and stops at the
Which best describes his
motion?
A. Linear
B. Projectile
C. Vibrational

Answer:

5

Explanation:

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

Proton number = 2

Neutron number = 3

Mass number =.?

The mass number of a given atom is simply the sum of the protons and neutrons in the nucleus of the atom. Mathematically, it is expressed as:

Mass number = proton + neutron

With the above formula, we can obtain the mass number of the atom as follow:

Proton number = 2

Neutron number = 3

Mass number =.?

Mass number = proton + neutron

Mass number = 2 + 3

Mass number = 5

Thus, the mass number the atom is 5.

Answer:

The speed of the knife after passing through the target is 9.33 m/s.

Explanation:

We can find the speed of the knife after the impact by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{k}v_{i_{k}} + m_{t}v_{i_{t}} = m_{k}v_{f_{k}} + m_{t}v_{f_{t}} [/tex]

Where:

[tex] m_{k}[/tex]: is the mass of the knife = 22.5 g = 0.0225 kg

[tex] m_{t}[/tex]: is the mass of the target = 300 g = 0.300 kg

[tex] v_{i_{k}}[/tex]: is the initial speed of the knife = 40.0 m/s

[tex] v_{i_{t}} [/tex]: is the initial speed of the target = 2.30 m/s

[tex]v_{f_{k}}[/tex]: is the final speed of the knife =?

[tex] v_{f_{t}} [/tex]: is the final speed of the target = 0 (it is stopped)

Taking as a positive direction the direction of the knife movement, we have:

[tex] m_{k}v_{i_{k}} - m_{t}v_{i_{t}} = m_{k}v_{f_{k}} [/tex]  

[tex] v_{f_{k}} = \frac{m_{k}v_{i_{k}} - m_{t}v_{i_{t}}}{m_{k}} = \frac{0.0225 kg*40.0 m/s - 0.300 kg*2.30 m/s}{0.0225 kg} = 9.33 m/s [/tex]

Therefore, the speed of the knife after passing through the target is 9.33 m/s.

I hope it helps you!              

Answer:

Block of aluminum

Explanation:

The block of aluminum because of the conductivity, and is a metal, so it gets hotter faster

The temperature of wood and the aluminium will depend upon thier thermal coductivity.

The thermal conductivity of any material is defined as the amount of heat from a material can flow in a unit area per unit degree centigrade for a unit distance.

The thermal conductivity of wood is around 0.90 to 0.197 w/mk and that of aluminium is around 88 to 251 w/mk.

We can see that the aluminium block will absorb much amount of heat as compared to the wood block. Because the thermal conductivity of aluminium is much higher then the wood.

Hence temperature of wood and the aluminium will depend upon thier thermal coductivity.

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Answer:

C

Explanation:

if you measure your body's flexibility then you can keep track of how flexible you have gotten over time

Answer:

true..

Explanation:

Newton's 2nd law of motion is greater the mass of an object the more force it will take to accelerate the object

According to Newton’s second law of motion, weight depends on an object’s mass and the net force acting on the object. The given statement is true.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

The following equation help to understand the statement;

Force is defined as the product of mass and acceleration. Its unit is Newton.

F=ma

F=mg

W=mg

The weight of an item is determined by its mass and the net force exerted on it, according to Newton's second rule of motion. The provided assertion is correct.

Hence,the given statement is true.

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We have that for the Question " constant applied force Fp of 11.0 N pushes a box with a mass=7 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is .110." it can be said that

the final velocity vf is

[tex]vf=3.83m/s[/tex]

and the applied force Fnew is

[tex]F_{new}=3.43N[/tex]

From the question we are told

A constant applied force Fp of 11.0 N pushes a box with a mass=7 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is .110. Assuming the box starts from the rest, what is the final velocity vf of the box at 15.0 m point? If there were no friction between the box and the floor, what applied force F new would give the box the same final velocity?

Generally the equation for the Force  is mathematically given as

F_p-\mu mg=ma

Therefore

11-0.110*7*9.8=7*a

a=0.49m/s^2

Using Newtons equation

v^2=u^2+2as

v^2=0+2*0.49*15

v=3.83m/s

Therefore

F=ma

F=0.49*7

F_{new}=3.43N

Therefore

the final velocity vf is

[tex]vf=3.83m/s[/tex]

and the applied force Fnew is

[tex]F_{new}=3.43N[/tex]

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Answer:

[tex]6.28\times 10^{-5}\ \text{T}[/tex]

[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]

[tex]I_l[/tex] = Current in circular loop = 13 A

[tex]r_l[/tex] = Radius of circular loop = 13 cm

[tex]N[/tex] = Number of turns = 58

[tex]r_c[/tex] = Radius of coil = 0.94 cm

[tex]I_c[/tex] = Current in coil = 1.9 A

[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]

Magnitude of magnetic field in circular loop

[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]

The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].

Torque is given by

[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]

The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].

cold, dry, and stable air masses .

Answer:

2000 N.

Explanation:

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

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

The mechanical advantage of a machine is simply defined as:

Mechanical advantage = Load / Effort

MA = L / E

With the above formula, we can obtain the effort used to raise the load of 8000 N as follow:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

MA = L / E

4 = 8000 / E

Cross multiply

4 × E = 8000

Divide both side by 4

E = 8000 / 4

E = 2000 N

Thus, the effort used to raised the load is 2000 N.

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have ―

We've been asked to calculate acceleration.

[tex]\qquad \implies\boxed{\red{\sf{ a = \dfrac{v-u}{t} }}}\\[/tex]

[tex] \twoheadrightarrow \quad \sf {a = \dfrac{30-18}{3} \; ms^{-2} } \\ [/tex]

[tex] \twoheadrightarrow \quad \sf {a =\cancel{ \dfrac{12}{3} \; ms^{-2} }} \\ [/tex]

[tex]\twoheadrightarrow \quad \boxed{\red{\sf{ a = 4 \; ms^{-2} }}}\\[/tex]

Therefore, acceleration of the car is 4 m/s².

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

        a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength  λ = 630 nm = 630 10⁻⁹ m and the width of the slit  a = 5 10⁻⁶ m

         sin  θ = m λ / a

         sin  θ = 3 630 10⁻⁹ / 5 10⁻⁶

         sin  θ = 3.78 10⁻¹ = 0.378

          θ = sin⁻¹  0.378

to better see the result let's find the angle in radians

          θ = 0.3876 rad

let's reduce to degrees

         θ = 0.3876 rad (180º /π rad)

         θ = 22.2º

Answer:

-40 degrees

Explanation:

I just took the Temperature Quiz and got it right. :)

Answer:

Instantaneous velocity  [tex]= 20[/tex] meter per second

Instantaneous acceleration  [tex]= 6[/tex] meter per second square

Explanation:

Given equation of distance X = [tex]2t+3t^2[/tex]

Instantaneous velocity [tex]= \frac{dX}{dt}[/tex]  [tex]= 2 + 6 t[/tex]

Substituting the value of t = 3 seconds, we get -

[tex]\frac{dX}{dt} = 2 + 6*3 = 20[/tex] meter per second

Instantaneous acceleration  [tex]= \frac{d^2X}{dt^2}[/tex]  [tex]= 6[/tex] meter per second square

Answer:

1. Rₑq = 4 Ω

2. R₂ = 6 Ω

3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V

4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A

Explanation:

1. Determination of the equivalent resistance

Voltage (V) = 12 V

Current (I) = 3 A

Resistance (Rₑq) =?

V= IRₑq

12 = 3 × Rₑq

Divide both side by 3

Rₑq = 12 / 3

Rₑq = 4 Ω

Thus, the equivalent resistance (Rₑq) = 4 Ω

2. Determination of R₂.

Equivalent resistance (Rₑq) = 4 Ω

Resistance 1 (R₁) = 12 Ω

Resistance 2 (R₂)

Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:

Rₑq = R₁ × R₂ / R₁ + R₂

4 = 12 × R₂ / 12 + R₂

Cross multiply

4(12 + R₂) = 12R₂

48 + 4R₂ = 12R₂

Collect like terms

48 = 12R₂ – 4R₂

48 = 8R₂

Divide both side by 8

R₂ = 48 / 8

R₂ = 6 Ω

3. Determination of the total voltage (Vₜ), V₁ and V₂.

From the question given above, the total voltage is 12 V

Since the resistors are arranged in parallel connection, the same voltage will go through them.

Thus,

Vₜ = V₁ = V₂ = 12 V

4. Determination of the total current (Iₜ), I₁ and I₂

From the question given above, the total current (Iₜ) is 3 A

Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.

Vₜ = V₁ = 12 V

R₁ = 12 Ω

I₁ =?

V₁ = I₁R₁

12 = I₁ ×12

Divide both side by 12

I₁ = 12 / 12

I₁ = 1 A

Next, we shall determine I₂. This can be obtained as follow:

Iₜ = 3 A

I₁ = 1 A

I₂ =?

Iₜ = I₁ + I₂

3 = 1 + I₂

Collect like terms

I₂ = 3 – 1

I₂ = 2 A

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

[tex]\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big)[/tex])

[tex]\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)[/tex]

Similarly;

[tex]\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)[/tex]

However; the relative difference is: [tex]\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}[/tex]

[tex]\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}[/tex]

[tex]= 0.2828 \\ \\ \mathbf{\simeq 29\%}[/tex]

Answer:

image distance = -8.47 cm

diameter of the image = 0.769 cm

Explanation:

Given that,

Mass of the particle, m = 4 kg

Speed of the particle, v = 2.5 m/s

The radius of the circle, r = 2 m

We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :

[tex]L=mvr[/tex]

Substitute all the values,

[tex]L=4\times 2.5\times 2\\\\L=20\ kg{\cdot}m^2s[/tex]

So, the angular momentum of the particle is 20 kg-m² s.

Answer:

A.

Explanation:

Some of these photons are absorbed by the atoms in the ground state and the photons are lost to the laser process. However, some photons cause stimulated emission in excited-state atoms, releasing another coherent photon. In effect, this results in optical amplification

Answer:

D

Explanation:

Given that all of the following are characteristics of a laser diode except

Solution.

A laser diode emit light of a single colour. That is, it emits a monochromatic light in which the light emitted monochromatically will definitely be coherent. That is, a light with a single wavelength emitted by a wide wavelength.

Also, all diode have positive and negative junction which is known as pn junction.

The incorrect among all the options is option D which is the answer.

Answer:

 v = 4.8 10⁻⁴ m / s

Explanation:

To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy

          U = K

          q V = ½ m v²

          v = [tex]\sqrt { \frac{2qV}{m} }[/tex]

in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C

let's calculate

          v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]

          v = 4.8 10⁻⁴ m / s

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

Answer:

100 milliliters

Explanation:

Answer:

two least important criteria are A and B.

Explanation:

In an industrial production process there are criteria on the location of the production plants. Among the most important criteria we have that a good quantity is produced from the producer and that the process is cost-beneficial - cats to increase the profit of the company.

There are other desirable but minor criteria

A) The size of the city is of minor importance

b) Finding raw material locally is interesting, but its importance is less, the raw material can be sent from other locations.

Consequently the two least important criteria are A and B.

Answer:

A. The process should be located in a major city.

B. The process should use local, organic reactants.

Explanation:

ap.ex

Answer:

a) the x-component of the net momentum is 44200 kgm/s [tex](-x)[/tex]

b) the y-component of the net momentum is 31200 kgm/s [tex](y)[/tex]

c) the magnitude of the net momentum is 54102.5 kgm/s

Explanation:

Given the data in the question;

a) x-component of the net momentum of this system.

the second vehicle ( sedan ) doesn't have momentum along x-axis, the momentum along x-axis is strictly contributed by the pick up

so;

Px = 2600 kg × 17.0 m/s [tex](-x)[/tex]

Px = 44200 kgm/s [tex](-x)[/tex]

Therefore, the x-component of the net momentum is 44200 kgm/s [tex](-x)[/tex]

b) y-component of the net momentum of this system

Also, momentum along y-axis is entirely provided by the sedan

Py = 1300 kg × 24.0 m/s [tex](y)[/tex]

Py = 31200 kgm/s [tex](y)[/tex]

Therefore, the y-component of the net momentum is 31200 kgm/s [tex](y)[/tex]

c) magnitude of the net momentum?

magnitude of the net momentum P = √( Px² + Py² )

so we substitute  

P = √( (44200)² + (31200)² )

P = √( 2927080000 )

P = 54102.5 kgm/s

Therefore, the magnitude of the net momentum is 54102.5 kgm/s

It's the one that has a concave line running upward.

Answer:

its the middle one

Explanation:

i did it on edge 2021

Explanation:

When there is a high frequency sound the speed of the vibrations is faster and makes a high pitch sound. When there is a low frequency sound the speed of the vibrations is slower and makes a lower pitch sound.