The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.0 m/s. Initially, the car is also traveling at a speed 19.0 m/s and its front bumper is a distance 23.1 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.570 m/s^2, then pulls back into the truck's lane when the rear of the car is a distance 25.0 m ahead of the front of the truck. The car is of length 5.00 m and the truck is of length 21.3 m.
1. How much time is required for the car to pass the truck?
2. What distance does the car travel during this time?
3. What is the final speed of the car?

Answer:

0.01

Explanation:

Given the data:

10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90

True value = 9.81

Mean value :

Σx / n

Sample size, n = 9

(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9

= 88.69 / 9

= 9.854

Standard deviation (σ) :

Sqrt (Σ(X - m)² / n)

[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9

Sqrt(0.113824 / 9)

Sqrt(0.0126471)

σ = 0.1124593

Standard Error = σ / sqrt(n)

Standard Error = 0.1124593 / 9

Standard Error = 0.0124954

Standard Error = 0.01 ( 1 significant digit)

Answer:

1.704 × 10¹⁹ electrons

Explanation:

The gravitational force due to the identical masses with mass, m = 3.17 kg at a distance r = 2.96 m from each other is F = Gm²/r².

Since equal number of electrons have to leave both masses, we have a charge q on each mass acting to oppose each other at distance r.

So, the electrical force of repulsion is thus

F'= kq²/r²

For the net force to be zero, the gravitational force of attraction must balance the electrical force of repulsion.

So, F = F'

So,  Gm²/r² = kq²/r²

Gm² = kq²

q² = Gm²/k

taking square root of both sides, we have

q = m√(G/k)

So, substituting the values of the variables into the equation, we have

q = m√(G/k)

= 3.17 kg√(6.67 × 10⁻¹¹ Nm²/kg²/9 × 10⁹ Nm²/C²)

= 3.17 kg√(0.741  × 10⁻²⁰ C²/kg²)

= 3.17 kg × 0.861 C/kg

= 2.73 C

Now, q = ne where n = number of electrons and e = electron charge = 1.602 × 10⁻¹⁹ C

n = q/e

= 2.73 C ÷ 1.602 × 10⁻¹⁹ C

= 1.704 × 10¹⁹ electrons

There are three methods of heat travel:

CONDUCTION -- The transfer of heat through a medium. This is how we cook food on top of a stove. The heat from the stove burner is conducted through a medium (a metal pot) to the food.

CONVECTION -- The transfer of heat due to the physical movement of an object. We can observe convection by looking at a pot of boiling water. Have you ever noticed that when a pot of water is boiling, the water seems to follow a vertical circular motion? This is convection. The parcel of heated water at the bottom of the pot rises, as it rises it gives off some of its heat. Because it loses some heat, the parcel is cooler than the surrounding water. It then sinks to the bottom of the pot and the process is started again. The path of the rising water followed by the sinking water traces out a circle.

RADIATION -- The transfer of heat by means of waves. This is the most difficult method of heat transfer to understand. Yet, we experience it every day. We feel the effects of radiation whenever we stand near a stove or oven which is being used. We feel the heat radiating from the stove or oven to our skin. Similarly, we have all been outside on a sunny, hot Summer's day. If we look up to the sky we can feel the rays of the Sun hitting our faces. The Sun is radiating its heat to the Earth.

It is through one of the above processes of heat transfer that causes the air temperature at deep-ocean station 41001 to be warmer than that of land station CLKN7 during the winter months. Which process do you believe to be the cause of the air temperature differences between these two stations? I'll give you a hint, it has something to do with the temperature of the ocean water. Lets look at a graph of both the average air and water temperatures from Station 41001.

As you can see from the graph, the January (month 1) and February (month 2) water temperatures are about 20 degrees while the respective air temperatures are about 15 degrees. This is a 5 degree difference in temperature between the air and the water at the same geographical location!!

We can figure out what heat transfer process is influencing the air temperature at station 41001 by applying the three methods to our situation and then we can choose the one that seems most logical.

First, lets look at conduction. This process involves the transfer of heat through a conductive medium. Well, nothing exists between the air and the water surface. In our situation, the heat is going directly from the water to the air without passing through a conductive medium. Therefore, this is not the applicable process that is causing the warm winter-time air temperatures at station 41001.

Convection involves the movement of heated objects. The physical movement must be a result of the heating, such as with the pot of boiling water where the vertical movement is caused by the intense heat applied to the bottom of the pot. Because the ocean water isn't moving into or through the atmosphere as a result of the sun's heating of the water, convection isn't the process influencing air and water temperature difference. Ocean water is moving through the lower few feet of the air as ocean surface waves, but this doesn't occur because of the sun's heat.

The final process, radiation, is causing the winter-time air temperatures over water to be warmer than the winter-time air temperatures over land. The heat of the ocean is being given off (radiated) into the air, thus making the air substantially warmer.

Answer:

[tex] \underline{ \boxed{ yes}}\\[/tex]

Explanation:

[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]

Answer:                  

32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice

Answer:

the answer will be 24.40 ohms law

Explanation:

Answer:

10N/m

Explanation:

Calculating workdone=Force/Distance

Therefore=15N/1.5m

=10N/m

Answer:

α = 36.21 °

β = 143.79°

Explanation:

To do this, we need to know the expression to calculate the angle.

In this case:

α₁ = tan⁻¹ (Fy₁/Fx₁)   (1)

Now, let's analize the given data.

We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.

Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.

The expression to calculate the force would be Coulomb's law so:

F = K q₁q₂ / r²    (2)

The value of K is 9x10⁹ N m² / C². Let's calculate the forces:

F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²

Fy = 1.04x10⁷ N

F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²

Fx = 1.42x10⁷ N

Now that we have both forces, we can calculate the magnitude of the force:

F = √(Fx)² + (Fy)²

F = √(1.04x10⁷)² + (1.42x10⁷)²

F = 1.76x10⁷ N

Finally, the direction would be applying (1):

α = tan⁻¹ (1.04x10⁷/1.42x10⁷)

α = 36.21 °

And counter clockwise it would be:

β = 180 - 36.21 = 143.79°

Hope this helps

Answer:

[tex]\mathrm{d.\:Static,\: Sliding,\:Rolling}[/tex]

Explanation:

Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

The absorb energy: Q = 600 Joule

Work done: W = 650 Joule.

Let, the change in internal energy of the engine= dU.

According to conservation of energy:

The absorb energy =  change in internal energy  + Work done

Q = dU + W

dU = Q - W

= 600 joule - 650 joule

= - 50 joule.

Hence, the change in internal energy of the engine is -50 joule.

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Answer:

The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation.The water cycle involves the exchange of energy, which leads to temperature changes.

Answer:

so be more specific please

Answer:

Graphing the momentum against the change in moment yields a linear relationship.

Explanation:

This is an impulse experiment,

          I = ∫ F .dt

where the force and time of the collision are measured, therefore if we assume an average force the integral reduces to

           I = F t

Furthermore, the momentum is equal to the change in moment of the ball, this change in moment can be found using the energy relations measuring the height of the ball and calculating its speed, in the two intervals for the descent and for the exit, possibly the heights are different so the moment change is different from zero.

Starting point. Higher

          Em₀ = U = mgh

Lower end point, just before hitting the scale

          [tex]Em_{f}[/tex] = K = ½ m v²

in the path in the air there is no friction

          Em₀ = Em_{f}

          m g h = ½ m v²

          v = [tex]\sqrt{2gh}[/tex]

this height is different for the descent and ascent of the ball, so we have two moments

         Δp = [tex]p_{f}[/tex] - p₀

         Δp = m (v_{f} -v₀)

therefore we have the relationship

         I = Δp

Graphing the momentum against the change in moment yields a linear relationship.

Answer:

98%

Explanation:

Given parameters

Mass of motor = 10kg

Height = 2m

Time = 2s

Power input = 100w

Unknown

Efficiency = ?

Solution

Efficiency is the percentage of the power output to the power input.

Power is the rate at which work is done.

Power output = mass x g x height / time

g is the acceleration due to gravity

Power output = 10x 2 x 9.8 / 2 = 98W

Efficiency = power output/ power input x 100

Efficiency = 98/100 x 100 = 98%

Answer:

[tex]1140\:\mathrm{J}[/tex]

Explanation:

Work is given by [tex]W=F\Delta x[/tex], where [tex]F[/tex] is force and [tex]\Delta x[/tex] is displacement.

Plugging in given values, we get:

[tex]W=95\:\mathrm{N}\cdot12\:\mathrm{m}=\fbox{$1140\:\mathrm{J}$}[/tex].

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

or

p'=2p

So, the new momentum becomes twice the initial momentum.

Answer:

More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).

Answer:

The answer is below

Explanation:

Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.

Momentum = mass * velocity

The principle of conservation of momentum states that momentum cannot be created or destroyed but  can be transferred. Therefore the momentum before and after an action is equal.

Initial momentum = Final momentum

Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.

m = 71 kg, M = 500 kg, v = 6 m/s

Initial both the raft and diver are at rest, hence u and U is zero, hence:

mu + MU = mv + MV

71(0) + 500(0) = 71(6) + 500(V)

0 = 426 + 500(V)

500(V) = -426

V = -426/500

V = -0.852 m/s

Answer:

[tex]t = 1.82[/tex]

Explanation:

Given

[tex]u = 7.70m/s[/tex] -- initial velocity

[tex]s = 30.2m[/tex] --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

[tex]s = ut + \frac{1}{2}gt^2[/tex]

Where

[tex]g = 9,8m/s^2[/tex]

So, we have:

[tex]30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2[/tex]

[tex]30.2 = 7.70t + 4.9 * t^2[/tex]

Subtract 30.2 from both sides

[tex]30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2[/tex]

[tex]0 = 7.70t + 4.9 * t^2 - 30.2[/tex]

[tex]0 = 7.70t + 4.9t^2 - 30.2[/tex]

[tex]7.70t + 4.9t^2 - 30.2 = 0[/tex]

[tex]4.9t^2 + 7.70t - 30.2 = 0[/tex]

Solve using quadratic formula:

[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

Where

[tex]a = 4.9;\ b = 7.70;\ c = -30.2[/tex]

[tex]t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}[/tex]

[tex]t = \frac{-7.70\±\sqrt{651.21}}{9.8}[/tex]

[tex]t = \frac{-7.70\±25.52}{9.8}[/tex]

Split the expression

[tex]t = \frac{-7.70+25.52}{9.8}[/tex] or [tex]t = \frac{-7.70-25.52}{9.8}[/tex]

[tex]t = \frac{17.82}{9.8}[/tex] or [tex]t = -\frac{33.22}{9.8}[/tex]

Time can't be negative;  So, we have:

[tex]t = \frac{17.82}{9.8}[/tex]

[tex]t = 1.82[/tex]

Hence, the time to hit the ground is 1.82 seconds

Answer:

the car has greater momentum.

the car has greater kinetic energy.

Explanation:

FOR MOMENTUM:

Momentum is given as the product of mass and velocity of an object:

[tex]P = mv[/tex]

where,

P = momentum

m = mass

v = velocity

For Truck:

[tex]P_{truck} = (3530\ kg)(21\ m/s)\\P_{truck} = 74130 Ns[/tex]

For Car:

[tex]P_{car} = (1620\ kg)(54\ m/s)\\P_{car} = 87480\ Ns[/tex]

Therefore, car has greater momentum.

FOR KINETIC ENERGY:

Kinetic Energy is given as:

[tex]K.E = \frac{1}{2} mv^{2}[/tex]

where,

K.E = Kinetic Energy

m = mass

v = velocity

For Truck:

[tex]K.E_{truck} = \frac{1}{2} (3530\ kg)(21\ m/s)^{2}\\K.E_{truck} = 778365\ J = 778.36\ KJ[/tex]

For Car:

[tex]K.E_{truck} = \frac{1}{2} (1620\ kg)(54\ m/s)^{2}\\K.E_{truck} = 2361960\ J = 2361.96\ KJ[/tex]

Therefore, car has greater kinetic energy.

Answer:

[tex]8.1\:\mathrm{Ns}[/tex]

Explanation:

The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:

[tex]\Delta p = m\Delta v[/tex], where [tex]m[/tex] is the mass of the basketball and [tex]\Delta v[/tex] is the change in velocity.

Since the basketball is changing direction, its total change in velocity is:

[tex]\Delta v = 20-(-15)=35\:\mathrm{m/s}[/tex].

Therefore, the basketball's change in momentum is:

[tex]\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}[/tex].

Thus, the impulse on the basketball is [tex]\fbox{$8.1\:\mathrm{Ns}$}[/tex] (two significant figures).

Answer:

See explanation

Explanation:

Now we have;

vi = (-30, 0) m/s

vf = (0, 20) m/s

Δvav= vf - vi= (30, 20) m/s

magnitude of Δvav= √30^2 + 20^2 = 36.0 m/s

Direction = tan-1(20/30) = 33.69°

For aav

aav= Δvav/t = (30/5, 20/5) = (6,4) m/s^2

magnitude of  aav = √6^2 + 4^2 = 7.2 m/s^2

direction of aav = tan-1(4/6) = 33.69°

Answer:

the total distance is 5km and the displacement is 1km

Explanation:

The total distance would be the addition of John running both ways so 3 km, 2 km.

However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.

Think about 2 km as a positive value for the first part of the question and a negative value for the second part.

Answer:

F = 15.47 N

Explanation:

Given that,

Q = 52 µC

q = 10 µC

d = 55 cm = 0.55 m

We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N[/tex]

So, the magnitude of the electrostatic force is 15.47 N.

The magnitude of electrostatic force will be "15.47 N".

According to the question,

Charges, Q = 52 μC

                q = 10 μC

Distance, d = 55 cm, or

                   = 0.55 m  

Constant, k = 9 × 10⁹

We know the relation,

→ Electrostatic force, F = k [tex]\frac{q_1 q_2}{d^2}[/tex]

By substituting the values, we get

                                      = 9 × 10⁹ × [tex]\frac{10\times 10^{-6}}{(0.55)^2}[/tex]

                                      = 15.47 N

Thus the above answer is correct.

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Answer:

These findings led to the formation of the modern cell theory, which has three main additions: first, that DNA is passed between cells during cell division; second, that the cells of all organisms within a similar species are mostly the same, both structurally and chemically; and finally, that energy flow occurs within.

Explanation:

hope this helped! <3

can i have the crown ?

Answer:

2 m/s²

Explanation:

From the given information:

The first mass m_1 = 0.6 kg

The second mass m_2 = 0.3 kg

The magnitude for the acceleration of 0.3 kg is:

a = net force/ effective mass

Mathematically, it can be computed as follows:

[tex]a = \dfrac{F}{m}[/tex]

[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]

[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]

a ≅ 2 m/s²

Answer:

7.01yard/sec

Explanation:

Given parameters:

Initial position  = 50yard

Final position = 12yard

Time  = 5.42s

Unknown:

Average speed of runner  = ?

Solution:

To solve this problem;

        Speed  = [tex]\frac{distance}{time}[/tex]  

Distance covered  = Initial position  - final position  = 50 - 12  = 38yards

So;

       Speed  = [tex]\frac{38}{5.42}[/tex]   = 7.01yard/sec