What happens to the rate of deposition as a river gains velocity?
O The rate of deposition decreases
O The rate of deposition increases
O The rate of deposition stays the same
O Deposition stops completely

Answers

Answer 1

Answer:  A streain with a velocity ... The rate at which particles are deposited by a stream is least affected ... (4) Particles lose potential energy during erosion and gain potential .

Explanation:


Related Questions

Two identical pendulums have the same period when measured in the factory. While one pendulum swings on earth, the other is taken on a spaceship traveling at 95%% the speed of light. Assume that both pendulums operate under the influence of the same net force and swing through the same angle.
When observed from earth, how many oscillations does the pendulum on the spaceship undergo compared to the pendulum on earth in a given time interval?
a. more oscillations
b. fewer oscillations
c. the same number of oscillations

Answers

Answer:

Explanation:

As a result of impact of time widening, a clock moving as for an observer seems to run all the more gradually than a clock that is very still in the observer's casing.  

At the point when observed from earth, the pendulum on the spaceship takes more time to finish one oscillation.  

Hence, the clock related with that pendulum will run more slow (gives fewer oscillations as observed from the earth)  than the clock related with the pendulum on earth.

Ans => B fewer oscillations

An arrow is launched vertically upward at a speed of 50 m/s. What is the arrow’s speed at the highest point? Ignore air resistance

Answers

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.

A proton with an initial speed of 400000 m/s is brought to rest by an electric field.
Part A- Did the proton move into a region of higher potential or lower potential?
Part B - What was the potential difference that stopped the proton?
?U = ________V
Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________eV

Answers

Answer:

moves into a region of higher potential

Potential difference = 835   V

Ki = 835 eV

Explanation:

given data

initial speed = 400000 m/s

solution

when proton moves against a electric field  so that it will move into higher potential  region

and

we know Work done by electricfield  W is express as

W = KE of proton   K

so

q × V   =  0.5 × m × v²     ......................1

put here va lue

1.6 × [tex]10^{-19}[/tex] × V   =   0.5 × 1.67 × [tex]10^{-27}[/tex] × 400000²

Potential difference V = 1.336 × 10-16 / 1.6  × 10-19      

Potential difference = 835   V

and

KE of proton in eV is express as

Ki  =   V numerical

Ki = 835 eV

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you at 13.0 m/s. The maximum force you apply in catching the water balloon is twice the average force. How long must the interaction time of your catch be to make sure the water balloon doesn't soak you

Answers

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

[tex]|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s[/tex]

The interaction time to avoid that the water balloon breaks is 0.029s

How can global warming lead to changes to the Earth’s surface? a. Global warming can lead to an increased number of earthquakes, which change the Earth’s surface. b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses. c. Global warming leads to a decrease in water levels of coastal wetlands. d. Global warming cannot lead to changes to the Earth’s surface.

Answers

Answer:

Option:  b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.

Explanation:

Global warming is the reason for the changes in environment and climate on earth. Melting of glaciers leads to an increase in water level and a decrease in landmass. One of the most climactic consequences is the decrease in Arctic sea ice. Melting polar ice along with ice sheets and glaciers across Greenland, North America, Europe, Asia, and South America suspected to increase sea levels slowly. There is an increase in the glacial retreat due to global warming, which leaves rock piles that covered with ice.  

Answer:

B: Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.

Explanation:

Global warming is primarily caused by the increase in greenhouse gases, such as carbon dioxide, in the Earth's atmosphere. This leads to a rise in global temperatures, which has various impacts on the Earth's surface. One significant effect is the melting of glaciers and ice caps in polar regions and mountainous areas.

As temperatures increase, glaciers and ice sheets start to melt at a faster rate. This melting results in the release of massive amounts of water into rivers, lakes, and oceans. Consequently, there can be an increase in the frequency and intensity of flooding events in regions downstream from these melting glaciers.

Moreover, the melting of glaciers and ice caps contributes to a rise in sea levels. As the melted ice enters the oceans, it adds to the overall volume of water, leading to a gradual increase in sea levels worldwide. This rise in sea levels poses a threat to coastal areas, as they become more vulnerable to coastal erosion, storm surges, and saltwater intrusion into freshwater sources.

Additionally, the loss of glacial land masses due to melting can have long-term effects on ecosystems. Glaciers act as freshwater reservoirs, releasing water gradually throughout the year. With their decline, the availability of freshwater for agriculture, drinking water, and other human needs can be significantly affected.

Therefore, global warming can indeed lead to changes in the Earth's surface, particularly through the melting of glaciers and subsequent impacts on sea levels, flooding, and glacial land masses.

E23 verified.

A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.

Answers

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

2. A 2.0-kg block slides down an incline surface from point A to point B. Points A and B are 2.0 m apart. If the coefficient of kinetic friction is 0.26 and the block is starting at rest from point A. What is the work done by friction force

Answers

Answer:a

Explanation:

What direct current will produce the same amount of thermal energy, in a particular resistor, as an alternating current that has a maximum value of 2.59 A?

Answers

Answer:

The direct current that will produce the same amount of thermal energy is 1.83 A

Explanation:

Given;

maximum current, I₀ = 2.59 A

The average power dissipated in a resistor connected in an AC source is given as;

[tex]P_{avg} = I_{rms} ^2R[/tex]

Where;

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }[/tex]

[tex]P_{avg} = (\frac{I_o}{\sqrt{2} } )^2R\\\\P_{avg} = \frac{I_o^2R}{2} ----equation(1)[/tex]

The average power dissipated in a resistor connected in a DC source is given as;

[tex]P_{avg} = I_d^2R --------equation(2)[/tex]

where;

[tex]I_d[/tex] is direct current

Solve equation (1) and (2) together;

[tex]I_d^2R = \frac{I_o^2R}{2} \\\\I_d^2 = \frac{I_o^2}{2} \\\\I_d=\sqrt{\frac{I_o^2}{2} } \\\\I_d = \frac{I_o}{\sqrt{2}} \\\\I_d = \frac{2.59}{\sqrt{2} } \\\\I_d = 1.83 \ A[/tex]

Therefore, the direct current that will produce the same amount of thermal energy is 1.83 A

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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A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40 (kinetic) assuming the car is not skidding while traveling along the curve, what is the magnitude of the centripetal acceleration of the car

Answers

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

ac = 3.92 m/s²

Based on the data provided, the centripetal acceleration is 3.92 m/s²

What is centripetal acceleration?

Centripetal acceleration is the acceleration of a body moving in a circular path which is directed toward the center of the circle.

In the given question, the frictional force must balance the centripetal force for the car not to skid.

Frictional Force = Centripetal Force

where,

Frictional Force = μR

R = mg

F = μmg

Centripetal Force = m

Then

μmg = ma

a = μg

ac = 0.4 * 9.8 m/s²

ac = 3.92 m/s²

Therefore, the centripetal acceleration is 3.92 m/s².

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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?

Answers

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.

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?

Answers

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

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.

A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2

Answers

Answer:

ML²/6

Explanation:

Pls see attached file

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

Moment of inertia:

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

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The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 88,000 m/s. What are the masses of the two stars

Answers

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/([tex]\frac{1}{4}[/tex]year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

What happens when you increase the number of slits per millimeter (decrease the spacing between slits)?

Answers

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.

Blue light (λ = 475 nm) is sent through a single slit with a width of 2.1 µm. What is the maximum possible number of bright fringes, including the central maximum, produced on the screen? (Hint: What is the largest angle that can be used?)

Answers

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 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:

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

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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A loop of wire with cross-sectional area 1 m2 is inserted into a uniform magnetic field with initial strength 1 T. The field is parallel to the axis of the loop. The field begins to grow with time at a rate of 2 Teslas per hour. What is the magnitude of the induced EMF in the loop of wire

Answers

Answer:

The magnitude of the EMF is 0.00055  volts

Explanation:

The induced EMF is proportional to the change in magnetic flux based on Faraday's law:

[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]

Since in our case there is only one loop of wire, then N=1 and we get:

[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]

We need to express the magnetic flux given the geometry of the problem;

[tex]\Phi=B\,\,A[/tex]where A is the area of the coil that remains unchanged with time, and B is the magnetic field that does change with time. Therefore the equation for the EMF becomes:

[tex]emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts[/tex]

When a hydrometer (see Fig. 2) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface

Answers

Answer:

5.79 in

Explanation:

We are given that

Diameter,d=0.30 in

Radius,r=[tex]\frac{d}{2}=\frac{0.30}{2}=0.15 in[/tex]

Weight of hydrometer,W=0.042 lb

Specific gravity(SG)=1.10

Height of stem from the water surface=3.15 in

Density of water=[tex]62.4lb/ft^3[/tex]

In water

Volume  of water displaced [tex]V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3[/tex]

Volume of another liquid displaced=[tex]V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3[/tex]

Change in volume=V-V'

[tex]V-V'=\pi r^2 l[/tex]

Substitute the values

[tex]6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l[/tex]

By using

1 ft=12 in

[tex]\pi=3.14[/tex]

[tex]l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}[/tex]

l=2.64 in

Total height=h+l=3.15+2.64= 5.79 in

Hence, the height of the stem protrude above the liquid surface=5.79 in

A 40.0 kg ballet dancer stands on her toes during a performance with 25.0 cm2 in contact with the floor. What is the pressure exerted by the floor over the area of contact if the dancer is stationary

Answers

Explanation:

40×10ms^-2

400N.

25/100m^2

0.25m^2

1.P=F/A

=400N/0.25m^2

=100Nm^-2

=100Pa

A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of
8 m/s to the east. What is the recoil velocity of the launcher?

Answers

Answer:

1.6 m/s west

Explanation:

The recoil velocity of the launcher is 1.6 m/s west.

What is conservation of momentum principle?

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of 8 m/s to the east.

Final momentum will be zero, so

m₁u₁ +m₂u₂ =0

Substitute the values for m₁ = 5kg, m₂ =1kg and u₂ =8 m/s, then the recoil velocity will be

5 x v +1x8 = 0

v = - 1.6 m/s

Thus, the recoil velocity of the launcher is  1.6 m/s (West)

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a person lifts 4.5kg block up a vertical distance of 1.2meters and then carries it horizontally for 7.3meters. Calculate the Total Work done by the person and the block.

Answers

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:

Find the terminal velocity (in m/s) of a spherical bacterium (diameter 1.81 µm) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10 ✕ 103 kg/m3. (Assume the viscosity of water is 1.002 ✕ 10−3 kg/(m · s).)

Answers

Answer:

The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

Explanation:

The terminal velocity of the bacterium can be calculated using the following equation:

[tex] F = 6\pi*\eta*rv [/tex]    (1)

Where:

F: is drag force equal to the weight

η: is the viscosity = 1.002x10⁻³ kg/(m*s)

r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm

v: is the terminal velocity

Since that F = mg and by solving equation (1) for v we have:

[tex] v = \frac{mg}{6\pi*\eta*r} [/tex]  

We can find the mass as follows:

[tex] \rho = \frac{m}{V} \rightarrow m = \rho*V [/tex]

Where:

ρ: is the density of the bacterium = 1.10x10³ kg/m³

V: is the volume of the spherical bacterium

[tex] m = \rho*V = \rho*\frac{4}{3}\pi*r^{3} = 1.10 \cdot 10^{3} kg/m^{3}*\frac{4}{3}\pi*(0.905 \cdot 10^{-6} m)^{3} = 3.42 \cdot 10^{-15} kg [/tex]

Now, the terminal velocity of the bacterium is:

[tex] v = \frac{mg}{6\pi*\eta*r} = \frac{3.42 \cdot 10^{-15} kg*9.81 m/s^{2}}{6\pi*1.002 \cdot 10^{-3} kg/(m*s)*0.905 \cdot 10^{-6} m} = 1.96 \cdot 10^{-6} m/s [/tex]

Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

I hope it helps you!

Complete the following statement:

Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different:_____

a. volume.
b. specific heat capacities.
c. expansion coefficients.
d. mass.
e. length.

Answers

Answer: c. expansion coefficients.

Explanation: Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients.

I found the answer on Quizlet. :)

Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients. The correct option is c.

What is the expansion coefficient?

The coefficient of thermal expansion (CTE) is the rate at which a material expands as its temperature rises. This coefficient is determined at constant pressure and without a phase change, i.e. the material is expected to remain solid or fluid.

Bimetallic strips, which are utilized as adjustable switches in electric appliances, are made up of metallic strips with differing expansion coefficients. The coefficient of thermal expansion indicates how the size of an object varies as temperature changes.

Therefore, the correct option is c. expansion coefficients.

To learn more about the expansion coefficient, refer to the link:

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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

Answers

Question:

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

Answer:

(a) 7.392cm

(b) -15.32 cm/s

(c) -184cm/s²

(d) 0.4πs and 8.00cm

Explanation:

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

define limitations in the operation conditions of a pn junction

Answers

Answer:

Such limitations are given below.

Explanation:

Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance. In connector operation, the maximum inversion voltage seems to be of significant importance.

Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Answers

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).

A trough is filled with a liquid of density 810 kg/m3. The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s2 for the acceleration due to gravity.)

Answers

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

Two lenses of focal length 4.5cm and 1.5cm are placed at a certain distance apart, calculate the distance between the lenses if they form an achromatic combination

Answers

Answer:

3.0cm

Explanation:

For lenses in an achromatic combination, the following condition holds, assuming the two lenses are of the same materials;

d = [tex]\frac{f_1 + f_2}{2}[/tex]     ---------(i)

Where;

d= distance between lenses

f₁ = focal length of the first lens

f₂ = focal length of the second lens

From the question;

f₁ = 4.5cm

f₂ = 1.5cm

Substitute these values into equation (i) as follows;

d = [tex]\frac{4.5+1.5}{2}[/tex]

d = [tex]\frac{6.0}{2}[/tex]

d = 3.0cm

Therefore, the distance between the two lenses is 3.0cm

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