Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

Answer:

in CGS system G is denoted as gram

Answer:

The hydrostatic force on one end of the trough is 54994.464 N

Explanation:

Given;

liquid density, ρ = 810 kg/m³

side of the equilateral triangle, L = 8m

acceleration due to gravity, g =  9.8 m/s²

Hydrostatic force is given as;

H = ρgh

where;

h is the vertical height of the equilateral triangle

Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.

let the half side of the triangle = x

x = ⁸/₂ = 4m

The half section of the triangle forms a right angled triangle

h² = 8² - 4²

h² = 48

h = √48

h = 6.928m

F = ρgh

F = 810 x 9.8 x 6.928

F = 54994.464 N

Therefore, the hydrostatic force on one end of the trough is 54994.464 N

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

Answer:

depending on how high it goes at 100m it has taken 2 secondes

Explanation:

At the highest point, the arrow is changing from moving up to moving down. At that exact point, its speed AND its velocity are both ZERO.

And air resistance actually makes no difference.

0.17T

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

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]

Answer:

the size are components relative to the whole.

Explanation:

they are particularly good at showing percentage or proportional data

Answer:

C) It is either ferromagnetic or paramagnetic

Explanation:

The complete question is given below

We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?

A) It must be ferromagnetic.

B) It must be paramagnetic.

C) It is either ferromagnetic or paramagnetic.

D) It must be diamagnetic.

A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.

Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.

The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.

In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.

Answer:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution

Answer:

2P

Explanation:

See attached file

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.

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

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

At constant speed, we will have;

[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

Answer:

The answer is A

Explanation:

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

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]

For more information please refer to the link below

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Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

Substitute these values into equation (i) as follows;

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² +  [tex]\frac{1}{2}[/tex]m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) +  ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

Answer:

ML²/6

Explanation:

Pls see attached file

The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,

The length of the rod is L. It has a non-uniform distribution of mass given by:

dm/dx = Cx

where C has units kg/m²

dm = Cxdx

the total mass M of the rod can be calculated by integrating the above relation over the length:

[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]

Thus,

C = 2M/L²

Now, the moment of inertia of the small element dx of the rod is given by:

dI = dm.x²

dI = Cx.x²dx

[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]

I = ML²/2

Learn more about moment of inertia:

https://brainly.com/question/6953943?referrer=searchResults

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 .

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  [tex]P_{avg} = F * v_{avg}[/tex]

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

            [tex]P_{avg} = \frac{W }{\Delta t }[/tex]

Where W is  is the Workdone which is  mathematically represented as

         [tex]W = F * s[/tex]

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                [tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]

Now this  displacement can be represented mathematically as  

            [tex]s = v_{avg} * \Delta t[/tex]

Where [tex]v_{avg }[/tex] is the average  velocity and [tex]\Delta t[/tex] is the time  taken  

So  

            [tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]

=>         [tex]P_{avg} = F * v_{avg}[/tex]

Answer:

Pavg =  Fvavg

Explanation:

Since the P (power) done by the F (force) is:

P = Fs/t

and we are looking for the velocity, so then it would be:

P = Fv

with the average velocity the answer is:

Pavg =  Favg

If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v  is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).

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.

Answer:

  m = 4

Explanation:

The expression that explains the constructive interference of a diffraction pattern is

         a sin θ = m λ

where a  is the width of the slit and λ the wavelength

         sin θ = m λ / a

The maximum value is for when the sine is 1, let's substitute

         1 = m λ/a  

         m = a /λ

let's reduce the magnitudes to the SI system

        a = 2.1 um = 2.1 10⁻⁶

        lam = 475 nm = 475 10⁻⁹ m

let's calculate

        m = 2.1  10⁻⁶ / 475 10⁻⁹

        m = 4.42

with m must be an integer the highest value is

         m = 4

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm

(a) 7.392cm

(b) -15.32 cm/s

(c) -184cm/s²

(d) 0.4πs and 8.00cm

The general equation of a simple harmonic motion (SHM) is given by;

x(t) = A cos (wt + Φ)        --------------(i)

Where;

x(t) = position of the body at a given time t

A =  amplitude or maximum displacement during oscillation

w = angular velocity

t = time

Φ = phase constant.

Given from question:

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

(a) At time t = 0;

The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;

x(0) = 8.00 cos (5(0) + π / 8)

x(0) = 8.00 cos (π /8)

x(0) = 8.00 x 0.924

x(0) = 7.392 cm

Therefore, the position of the piston at time t = 0 is 7.392cm

(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;

v(t) = [tex]\frac{dx(t)}{dt}[/tex]

v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]

v(t) = 8 (-5 sin (5t + π / 8))

v(t) = -40sin(5t + π / 8)     --------------------(iii)

Now, substitute t=0 into the equation as follows;

v(0) = -40 sin(5(0) + π / 8)

v(0) = -40 sin(π / 8)

v(0) = -40 x 0.383

v(0) = -15.32 cm/s

Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s

(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;

a(t) = [tex]\frac{dv(t)}{dt}[/tex]

a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]

a(t) = -200 cos (5t + π / 8)

Now, substitute t = 0 into the equation as follows;

a(0) = -200 cos (5(0) + π / 8)

a(0) = -200 cos (π / 8)

a(0) = -200 x 0.924

a(0) = -184.8 cm/s²

Therefore, the acceleration of the piston at time t = 0 is -184cm/s²

(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;

x(t) = A cos (wt + Φ)                   --------------(i)

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

From these equations it can be deduced that;

Amplitude, A = 8.00cm

Angular velocity, w = 5 rads/s

But;

w = [tex]\frac{2\pi }{T}[/tex]           [Where T = period of oscillation]

=> T = [tex]\frac{2\pi }{w}[/tex]

=> T = [tex]\frac{2\pi }{5}[/tex]

=> T = 0.4π s

Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm

Answer:

in the direction of the applied force

Explanation:

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

Answer:

The total work done by the person is given as = m g h

= 4.5kg x 9.8m/s²x1.2m

= 52.92J

This is the work done in moving the block in a vertical distance

However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.

Explanation:

Answer:

Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.

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]

Answer:

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

Answer:

2.95m

Explanation:

Using h= 2.5+ v²/2g

Where v= 3m/s

g= 9.8m/s²

h= 2.95m

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.

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