In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses m1 and m2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of v1, and car 2 was traveling northward at a speed of v2. After the collision, the two cars stick together and travel off in the direction.

Required:
a. Write the momentum conservation equation for the east-west components.
b. Write the momentum conservation equation for the north-south components.
c. Find the tangent of the angle.

Answers

Answer 1

Answer:

a)     vfₓ = m₁ / (m₁ + m₂) v₁,  b)    tan θ  = m₂ / m₁ v₂ / v₁, c)

Explanation:

Momentum is a vector quantity, so the consideration must be fulfilled in all axes

a) conservation of the moment east-west direction

the system is formed by the two cases, so that the forces during the sackcloth have been internal and therefore the mummer remains

before the crash

                 p₀ = m₁ v₁

after the crash

                 [tex]p_{f}[/tex]= (m1 + m2) vfₓ

                p₀ = pf

                m₁ v₁ = (m₁ + m₂) vfₓ

              vfₓ = m₁ / (m₁ + m₂) v₁

b) conservation of the North-South axis moment

before the shock

                p₀ = m₂ v₂

after the crash

              p_{f} = ( m₁ +m₂) [tex]vf_{y}[/tex]  

             p₀ = p_{f}

            me 2 v₂ = (m₁ + m₂) vfy

       

            [tex]vf_{y}[/tex] = m₂ / (m₁ + m₂) v₂

c) the angle with which the car moves is

             tan θ = Vfy / Vfₓ

             tan θ = [m₂ / (m₁ + m₂) v] / [m₁ / (m₁ + m₂) v₁]

             tan θ  = m₂ / m₁ v₂ / v₁

Answer 2

The momentum conservation equation for the north-south components is [tex]m_1u_1 = v(m_1 + m_2)[/tex]

The momentum conservation equation for the north-south components is [tex]m_2u_2 = v(m_1 + m_2)[/tex]

The tangent of the angle is 1.

The given parameters;

angle between the initial velocity of the cars, θ = 90

Apply the principle of conservation of linear momentum of inelastic collision as shown below;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

The momentum conservation equation for the east-west components is written as follows;

[tex]m_1(u_1cos \ 0) + m_2(u_2 cos 90)= v(m_1 + m_2)\\\\m_1u_1 = v(m_1 + m_2)[/tex]

The momentum conservation equation for the north-south components is written as follows;

[tex]m_1(u_1sin 0) + m_2(u_2sin90) = v(m_1 + m_2)\\\\m_2u_2 = v(m_1 + m_2)[/tex]

The tangent of the angle is calculated as follows;

[tex]tan \ \theta = \frac{p_y}{p_x} = \frac{v(m_1 + m_2)}{v(m_1 + m_2)} \\\\tan \ \theta = 1\\\\\theta = tan^{-1} (1) \\\\\theta = 45\ ^0[/tex]

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Related Questions

At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?

Answers

Answer:

v = 8.5 m/s

Explanation:

In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,

Weight = Centripetal Force

but,

Weight = mg

Centripetal Force = mv²/r

Therefore,

mg = mv²/r

g = v²/r

v² = gr

v = √gr

where,

v = minimum speed required = ?

g = 9.8 m/s²

r = radius = 7.4 m

Therefore,

v = √(9.8 m/s²)(7.4 m)

v = 8.5 m/s

Minimum speed for a roller coaster while travelling upside down  so that the person will not fall out = 8.5 m/s

For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.

In the given condition the weight of the person must be balanced by the centrifugal force.

and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person

According to the Newton's Second Law of motion we can write force balance

[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]

Given Radius of loop = r = 7.4 m

Putting the value  of r = 7.4 m  in equation (1) we get

[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]

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An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?

Answers

Answer:

R = 1138.9 Ω

Explanation:

Hello,

In this case, for the given power (P) and current (I), we can compute the resistance (R) via:

R = P / I²

Thus, we obtain:

R = 1.64x10⁵ W / (12.0 A)²

R = 1138.9 Ω

Best regards.

In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
ОА.
in the direction of the applied force
OB.
in the direction opposite of the direction of the applied force
OC. perpendicular to the direction of the applied force
OD
in a circular motion

Answers

Answer:

in the direction of the applied force

Explanation:

7. Which statement is true about teens that are in Marcia’s final state of identity formation?

Answers

Answer:

D. All of the above

Explanation:

The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.

James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure,  moratorium, and achievement.

Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?

Answers

Answer:

a) E = 2.7x10⁶ N/C

b) F = 54 N

Explanation:

a) The electric field can be calculated as follows:

[tex] E = \frac{Kq}{d^{2}} [/tex]

Where:

K: is the Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

d: is the distance

Now, we need to find the electric field due to charge 1:

[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]

The electric field due to charge 2 is:

[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]

The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]                                                                                                

Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.  

b) The force on a charge q₃ situated there is given by:

[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]

[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]

Therefore, the force on a charge q₃ situated there is 54 N.  

I hope it helps you!

(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].

(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.

The answer can be explained as follows.

Electric Field

Given that the two charges are;

[tex]q_1 = 50\times 10^{-6}\,C[/tex] and [tex]q_2 = -25\times 10^{-6}\,C[/tex]

(a) At the midpoint; [tex]r = 0.5\,m[/tex].

We know that the electric field due to charge [tex]q_1[/tex].

[tex]E_1 = k\,\frac{q_1}{r^2}[/tex]

Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]

[tex]E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C[/tex]

The electric field due to charge [tex]q_2[/tex] is given by;

[tex]E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C[/tex]

Therefore, the net electric field in the midpoint is given by;

[tex]E_{net} =E_2+E_1[/tex][tex]\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C[/tex]

The direction is towards the right side.

Electrostatic Force

(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.

So the force on the charge is ;

[tex]F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N[/tex]

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What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube

Answers

Answer:

f = 357.29Hz

Explanation:

In order to calculate the fundamental frequency in the closed tube, you use the following formula:

[tex]f_n=\frac{nv}{4L}[/tex]       (1)

n: order of the mode = 1

v: speed of sound = 343m/s

L: length of the tube = 24cm = 0.24m

You replace the values of the parameters in the equation (1):

[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]

The fundamental frequency of in the tube is 357.29Hz

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:

Answers

Answer:

The ratio  is  [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

Explanation:

Generally  the Moment of inertia of a spherical object (shell) is mathematically represented as

              [tex]I = \frac{2}{3} * m r^2[/tex]

Where m is  the mass of the spherical object

       and   r is the radius  

Now the the rotational kinetic energy can be mathematically represented as

       [tex]RE = \frac{1}{2}* I * w^2[/tex]

Where  [tex]w[/tex] is the angular velocity which is mathematically represented as

             [tex]w = \frac{v}{r}[/tex]

=>           [tex]w^2 = [\frac{v}{r}] ^2[/tex]

So

             [tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]

            [tex]RE = \frac{1}{3} * mv^2[/tex]

Generally the transnational  kinetic energy of this motion is  mathematically represented as

                [tex]TE = \frac{1}{2} mv^2[/tex]

So  

      [tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]

       [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

What must the charge (sign and magnitude) of a 1.60 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 680 N/C

Answers

Answer:

Explanation:

The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .

Let the charge be - q .

force on charge

= q x E where E is electric field

= q x 680

weight = 1.6 x 10⁻³ x 9.8

so

q x 680 = 1.6 x 10⁻³ x 9.8

q = 1.6 x 10⁻³ x 9.8 / 680

= 23 x 10⁻⁶ C

- 23 μ C .

A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the electric field between the plates

Answers

Answer:

Majorly the electric field is reduced among other effect listed in the explanation

Explanation:

In capacitors the presence of di-electric materials

1. decreases the electric fields

2. increases the capacitance of the capacitors.

3. decreases the voltage hence limiting the flow of electric current.

 The di-electric material serves as an insulator between the metal plates of the capacitors

6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.

Answers

Answer:

The correct options are:

B) They have the same wave speed as visible light

D) They have a lower frequency than gamma rays.

Explanation:

B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s

D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).

We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.

in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
​plz help me out
I will mark u as brainliest if u answer correct

Answers

Answer:

The correct answer is option 3 .

Please check the answer once :)

Classify the bending of light as exhibited by the ray diagrams. According to your data, is light refracted away from or toward the normal as it passes at an angle into a medium with a higher index of refraction?

Answers

Answer:

the ray of light should approach normal

Explanation:

When light passes through two means of different refractive index, it fulfills the equation

              n₁ sin  θ₁ = n₂ sin θ₂

where index 1 and 2 refer to each medium

In this problem, they tell us that light passes to a medium with a higher index, which is why

               n₁ <n₂

let's look for the angle in the second half

            sinθ₂ = n₁ /n₂  sin θ₁

            θ₂ = sin⁻¹ (n₁ /n₂  sin θ₁)

let's examine the angle argument the quantity n₁ /n₂ <1   therefore the argument decreases, therefore the sine and the angle decreases

Consequently the ray of light should approach normal

Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish really is? Does the fish see the person's face closer or farther than it really is? Explain your answer.

Answers

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

        θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Answer:

The fish appears closer than it really is because light from the fish is refracted away from the normal as it enters the air pocket in the goggles. This is because air has a smaller index of refraction than water. The person will trace rays back to an image point in front of the actual fish. The fish will see the person's face exactly where it actually is because the light from the face is not refracted as it travels through water only, and does not change from one medium to another.

Explanation:

A proton moves at a speed 1.4 × 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.85 m. What is the field strength?

Answers

Answer:

0.17T

Explanation:

When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e

[tex]F_{C}[/tex] = [tex]F_{M}[/tex]     --------------(i)

But;

[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex]   [m = mass of the particle, r = radius of the path, v = velocity of the charge]

[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]

Substitute these into equation (i) as follows;

[tex]\frac{mv^2}{r}[/tex] = qvB

Make B subject of the formula;

B = [tex]\frac{mV}{qr}[/tex]            ---------------(ii)

Known constants

m = 1.67 x 10⁻²⁷kg

q = 1.6 x 10⁻¹⁹C

From the question;

v = 1.4 x 10⁷m/s

r = 0.85m

Substitute these values into equation(ii) as follows;

B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]

B = 0.17T

Therefore, the magnetic field strength is 0.17T

key points that can be found in the realist philosophical position​

Answers

Answer:

Key points that can be found in the realist philosophical position​ are as follows:

The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices,  conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.

I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5

Answers

Answer:

8.9 seconds

Explanation:

The height of the object at time t is:

y = h + vt − 4.9t²

where h is the initial height, and v is the initial velocity.

Given h = 30 and v = 40:

y = 30 + 40t − 4.9t²

When y = 0:

0 = 30 + 40t − 4.9t²

4.9t² − 40t − 30 = 0

Solving with quadratic formula:

t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)

t = [ 40 ± √(1600 + 588) ] / 9.8

t = 8.9

It takes 8.9 seconds for the object to land.

How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under clear skies is 1000 W/m2

Answers

Answer:

344.8 x10^-8J/m³

Explanation:

Using=> energy intensity/ speed oflight

= 1000/2.9x10^8

= 344.8 x10^-8J/m³

The electromagnetic energy is 344.8 x10⁻⁸J/m³

We have to use the formula which says

Electromagnetic energy = energy intensity/ speed of light

We are given intensity as 1000 W/m²

Electromagnetic energy    = 1000/2.9 x 10⁸

                                             = 344.8 x 10⁻⁸J/m³

Therefore the electromagnetic energy is contained in each cubic meter will be  344.8 x 10⁻⁸J/m³

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Calculate the current through a 15.0-m long 20 gauge (having radius 0.405 mm) nichrome wire if it is connected to a 12.0-V battery. The resistivity of nichrome is 100 × 10-8 Ω ∙ m.

Answers

Given Information:  

Radius of wire = r = 0.405 mm = 0.405×10⁻³ m

Length of wire = L = 15 m

Voltage = V = 12 V

Resistivity =  ρ = 100×10⁻⁸ Ωm

Required Information:  

Current = I = ?

Answer:  

Current = I = 0.412 A

Explanation:  

The current flowing through the wire can be found using Ohm's law that is

V = IR

I = V/R

Where V is the voltage across the wire and R is the resistance of the wire.

The resistance of the wire is given by

R = ρL/A

Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(0.405×10⁻³)²

A = 0.515×10⁻⁶ m²

So the resistivity of the wire is

R = ρL/A

R = (100×10⁻⁸×15)/0.515×10⁻⁶

R = 29.126 Ω

Finally, the current flowing through the wire is

I = V/R

I = 12/29.126

I = 0.412 A

Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.

An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...

p.

2p.

4p.

8p.

12p.

Answers

Answer:

2P

Explanation:

See attached file

An ac series circuit contains a resistor of 20 ohms, a capacitor of 0.75 microfarads of 120 x 10-3 H. If an effective (rms) voltage of 120 V is applied, what is the effective (rms) current when the circuit is in resonance

Answers

Answer:

The effective (rms) current when the circuit is in resonance is 6 A

Explanation:

Given;

resistance of the resistor, R = 20 ohms

capacitance of the capacitor, C = 0.75 microfarads

inductance of the inductor, L =  0.12 H

effective rms voltage, [tex]V_{rms}[/tex] = 120

At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).

The effective (rms) current, = [tex]V_{rms}[/tex] / R

                                              = 120 / 20

                                              = 6 A

Therefore, the effective (rms) current when the circuit is in resonance is 6 A

f the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in the string at the top of the circle

Answers

Answer:

the size are components relative to the whole.

Explanation:

they are particularly good at showing percentage or proportional data

4. Chloe has a vertical velocity of 3 m/s when she leaves the 1 m diving board. At this instant, her center of gravity is 2.5 m above the water. How high above the water will Chloe go

Answers

Answer:

2.95m

Explanation:

Using h= 2.5+ v²/2g

Where v= 3m/s

g= 9.8m/s²

h= 2.95m

Value of g in CGS system

Answers

Answer:

in CGS system G is denoted as gram

In cgs, G can be written as G ≈ 6.674×10−8 cm3⋅g−1⋅s−2.

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t = 0 and subsequently experiences a constant angular acceleration α = 1.3 rad/s2. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 1.7 s.

Answers

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

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

Put the value in the equation

[tex]\omega=0.17+1.3\times1.7[/tex]

[tex]\omega=2.38(k)\ m/s[/tex]

We need to calculate the angular displacement

Using angular equation of motion

[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]

Put the value in the equation

[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]

[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]

[tex]\theta= 124.18^{\circ}[/tex]

We need to calculate the velocity at point A

Using equation of motion

[tex]v_{A}=v_{0}+\omega\times r[/tex]

Put the value into the formula

[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]

[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]

[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]

We need to calculate the acceleration at point A

Using equation of motion

[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]

Put the value in the equation

[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]

[tex]a_{A}=0.791j-0.851i[/tex]

[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]

Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]

(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]

A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.

Answers

Answer:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution

⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?

Answers

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

According to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of

Answers

Answer:

Order of 10^(-35) m.

Explanation:

The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.

Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.

This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 971 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
You find that the forces are attractive and the magnitude of each force is:______

Answers

Answer:

The magnitude of each force is 2.45 x 10⁻¹⁶ N

Explanation:

The charge of proton, +q = 1.603 x 10⁻¹⁹ C

The charge of electron, -q = 1.603 x 10⁻¹⁹ C

Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m

Apply Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

q₁ and q₂ are the charges of proton and electron respectively

F is the magnitude of force between them

Substitute in the given values and solve for F

[tex]F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N[/tex]

Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N

A 2kg block is sitting on a hinged ramp such that you can increase the angle of the incline. The coefficient of static friction between the block and the ramp is 0.67 and the coefficient of kinetic friction is 0.25.
a. What angle do you have to tilt the ramp to get the block to slide?
b. What acceleration does the block experience at this angle when kinetic friction takes over?

Answers

Answer:

θ = 33.8

a = 3.42 m/s²

Explanation:

given data

mass m = 2 kg  

coefficient of static friction μs = 0.67

coefficient of kinetic friction μk = 0.25

solution

when block start slide

N = mg cosθ    .............1

fs = mg sinθ   ...............2

now we divide equation 2 by equation 1 we get

[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]

[tex]\frac{\mu s N }{N}[/tex]  = tanθ

put here value we get

tan θ = 0.67

θ = 33.8

and

when block will slide  then we apply newton 2nd law

mg sinθ - fk = ma    ...............3

here fk = μk N = μk mg cosθ

so from equation 3 we get

mg sinθ -  μk mg cosθ = ma

so a will be

a = (sinθ - μk cosθ)g

put here value and we get

a = (sin33.8 - 0.25 cos33.8) 9.8

a = 3.42 m/s²

Which of the following technologies is based on the work of Ibn al-Haytham?
A. Telescopes to observe the visible light of distant stars
B. Radiation treatments for breast cancer
C. Radar to detect the movement of storms
O D. An orbiting observatory to detect X-rays from space objects

Answers

Answer:

The answer is A

Explanation:

Its A because he created a telescope to be able to observe stars.

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