The human circulatory system is closed - that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the hearts four chambers comes briefly to rest before it is ejected by contraction of the heart muscle. The contraction of the left ventricle lasts 250 ms and the speed of blood flow in the aorta (the large artery leaving the heart) is 0.80 m/s at the end of the contraction.

What is the average acceleration of a red blood cell as it leaves the heart?

Answer:

7.15 m/s

Explanation:

We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.

The truck moves at constant speed, we can use the equation for position under constant speed:

Xt = X0 + v*t

The car is accelerating with constant acceleration, we can use this equation

Xc = X0 + V0*t + 1/2*a*t^2

We know that both vehicles will meet again at x = 578

Replacing this in the equation of the truck:

578 = 24 * t

We get the time when the car passes the truck

t = 578 / 24 = 24.08 s

Before replacing the values on the car equation, we rearrange it:

Xc = X0 + V0*t + 1/2*a*t^2

V0*t = Xc - 1/2*a*t^2

V0 = (Xc - 1/2*a*t^2)/t

Now we replace

V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s

Answer:

The height of the cliff is 39.655 m

Given:

Height at which the arrow was shot, h = 1.3 m

Velocity of the arrow, u = 34 m/s

Angle, [tex]\theta' = 58.1^{\circ}[/tex]

Time of the fight, t = 4.1 s

Solution:

Let the Height of the cliff be H

Since, the motion of the object is projectile motion and the direction of motion  is vertical at some angle

Therefore, we consider the vertical component of velocity, [tex]u_{y} = usin\theta[/tex].

Now,

The Height of the cliff is given by applying the second equation of motion in the projectile:

Thus

[tex]s = u_{y}t - \frac{1}{2}gt^{2}[/tex]

[tex]s = 34sin60^{\circ}\times 4.1 - \frac{1}{2}\times 9.8\times 4.1^{2}[/tex]

s = 38.355 m

Now, the height of the cliff, H:

H = s + h = 38.355 + 1.3 = 39.655 m

The magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]

To find the tension on the ends of the clothesline, we can use the principle of equilibrium. When the mass is at rest in the middle of the clothesline, the tension in the clothesline must balance the gravitational force acting on the mass.

Let's denote:

- [tex]\( T \)[/tex] as the tension in the clothesline.

- [tex]\( m \)[/tex] as the mass tied to the clothesline.

- [tex]\( g \)[/tex] as the gravitational acceleration.

The gravitational force acting on the mass is given by [tex]\( F_{\text{gravity}} = m \cdot g \).[/tex]

When the mass is at rest in the middle of the clothesline, the horizontal components of the tensions on either side cancel each other out, leaving only the vertical components to support the weight.

Since the clothesline sags a distance of 3 meters in the middle, each side of the clothesline forms a right triangle with the sagging distance as one leg and half of the distance between the poles as the other leg. Therefore, each side of the clothesline forms a right triangle with legs of 8 meters and 3 meters.

Using the Pythagorean theorem, we can find the length of the clothesline (the hypotenuse of each triangle), which is the same as the length of each segment between the midpoint and the poles:

[tex]\[ L = \sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} \]\[ L = \sqrt{64 + 9} \]\[ L = \sqrt{73} \][/tex]

Now, we can set up an equation for the vertical equilibrium:

[tex]\[ 2T \sin(\theta) = F_{\text{gravity}} \][/tex]

Where [tex]\( \theta \)[/tex] is the angle the clothesline makes with the horizontal.

We can find [tex]\( \sin(\theta) \)[/tex] using trigonometry:

[tex]\[ \sin(\theta) = \frac{3}{L} \][/tex]

Now, substitute this expression for [tex]\( \sin(\theta) \)[/tex] into the equation for vertical equilibrium:

[tex]\[ 2T \left(\frac{3}{L}\right) = m \cdot g \]\[ 2T \left(\frac{3}{\sqrt{73}}\right) = 3 \times 9.8 \]\[ T = \frac{3 \times 9.8 \times \sqrt{73}}{2 \times 3} \]\[ T = \frac{9.8 \times \sqrt{73}}{2} \][/tex]

Now, we can calculate the tension:

[tex]\[ T \approx \frac{9.8 \times \sqrt{73}}{2} \]\[ T \approx \frac{9.8 \times 8.544}{2} \]\[ T \approx \frac{83.9712}{2} \]\[ T \approx 41.9856 \][/tex]

So, the magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]

The magnitude of the tension at the ends of the clothesline is approximately [tex]\[T = 41.87 \, \text{N}\][/tex]

To solve for the tension in the clothesline, we first consider the forces and the geometry involved.

Given:

Distance between poles, [tex]\( L = 16 \) meters[/tex]

Sagging distance,[tex]\( h = 3 \) meters[/tex]

Mass, [tex]\( m = 3 \) kilograms[/tex]

Gravitational acceleration, [tex]\( g = 9.8 \) m/s\(^2\)[/tex]

Determine the Weight of the Mass

The weight [tex]\( W \)[/tex] of the mass is given by:

[tex]\[W = mg = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N}\][/tex]

Analyzing the Geometry

[tex]\[\text{Half the distance between poles} = \frac{L}{2} = \frac{16}{2} = 8 \, \text{meters}\][/tex]

[tex]\[\text{Sagging distance} = h = 3 \, \text{meters}\][/tex]

Calculate the Length of the Hypotenuse

[tex]\[\sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} = \sqrt{64 + 9} = \sqrt{73} \, \text{meters}\][/tex]

Analyze Forces in Vertical and Horizontal Directions

[tex]\[2T_v = W \quad = \quad T_v = \frac{W}{2} = \frac{29.4 \, \text{N}}{2} = 14.7 \, \text{N}\][/tex]

The vertical component of the tension [tex]\( T \)[/tex] in each half of the clothesline is:

[tex]\[T_v = T \sin \theta\][/tex]

where \( \theta \) is the angle the clothesline makes with the horizontal. From the triangle, we know:

[tex]\[\sin \theta = \frac{h}{\sqrt{(L/2)^2 + h^2}} = \frac{3}{\sqrt{73}}\][/tex]

[tex]\[T \sin \theta = 14.7 \, \text{N}\][/tex]

Solve for the Tension[tex]\( T \)[/tex]

[tex]\[T \sin \theta = 14.7 \, \text{N} \quad = \quad T \cdot \frac{3}{\sqrt{73}} = 14.7 \, \text{N}\][/tex]

[tex]\[T = \frac{14.7 \times \sqrt{73}}{3}\][/tex]

[tex]\[T = 14.7 \times \frac{\sqrt{73}}{3}\][/tex]

[tex]\[T = 14.7 \times \frac{8.544}{3}\][/tex]

[tex]\[T = 14.7 \times 2.848 = 41.87 \, \text{N}\][/tex]

Answer:

a) 919 mts

b) 392 mts

Explanation:

In order to solve this, we will use the formulas of acceletared motion problems:

[tex](1)Y=Yo+Vo*t+\frac{1}{2}*a*t^2\\(2)Vf^2=Vo^2+2*a*y[/tex]

We are looking to obtain the initial velocity of the canister right after it was relased, we will use formula (2):

[tex]Vf^2=(0)^2+2*(3.10)*(240)\\Vf=38.6 m/s[/tex]

we need to calculate the time the canister takes to reach the ground, we will use formula (1):

[tex]0-240m=38.6*t+\frac{1}{2}(-9.8m/s^2)*t^2\\\\-4.9*t^2+38.6*t+240=0\\t=11.9seconds[/tex]

in order to know the new height of the rocket we have to use the formula (1) again:

[tex]Y=240+38.6*(11.9)+\frac{1}{2}*(3.10)*(11.9)^2\\Y=919mts[/tex]

We can calulate the total distance the canister traveled before reach the ground by (2):

[tex](0)^2=(38.6)^2+2*(-9.8)*y\\Y=76m[/tex]

So the canister will go up another 76m, so the total distance will be:

[tex]Yc=Yup+Ydown+240m\\Yc=76+76+240\\Yc=392 mts[/tex]

Answer:

Their kinetic energy will increase but potential energy will decrease.

Explanation:

Given that

Initial velocities of electron and proton is zero.

We know that ,electron have negative charge and proton have positive charge it means that they will attract to each other.We know that opposite charge attract to each other and same charge repels to each other.

It means that the velocities of proton and electron will increase and that leads to increase in the kinetic energy of proton and electron.

We know that potential energy U

[tex]U\alpha -\dfrac{1}{r}[/tex]

So when r will decrease then U will increase but in negative direction it means that U will decrease.

So we can say that their kinetic energy will increase but potential energy will decrease.

Answer:

Maximum current in the inductor will be 3.824 A

Explanation:

In first case inductive reactance = 59.2 ohm

Frequency = 60 Hz

We know that inductive reactance is given by [tex]X_L=\omega L[/tex]

[tex]59.2=2\pi f\times L[/tex]

[tex]59.2=2\times 3.14\times  60\times L[/tex]

[tex]L=0.157H[/tex]

In second case frequency f = 45 Hz

Now inductive reactance [tex]X_L=\omega L =2\times 3.14\times 45\times .157=44.368ohm[/tex]

Now current [tex]i=\frac{V}{X_L}=\frac{120}{44.368}=2.70A[/tex]

Maximum current [tex]i_{max}=\sqrt{2}i=1.414\times 2.70=3.824A[/tex]

To find the maximum current when the inductor is connected to a 45Hz source, one needs to first calculate the inductance of the inductor using the provided inductive reactance at 60Hz. Then, with the inductance determined, the inductive reactance at 45Hz can be computed. Finally, Arnold's law (I = V/X₁) is used to determine the current.

In this problem, the concept of inductive reactance and Ohm's law are involved. First, we know that the inductive reactance (X₁) is directly proportional to the frequency, and it is given by the formula X₁ = 2πfL. From the information provided, we can find the inductance (L) of the inductor by using this formula at 60Hz, then looking at how X₁ changes when we use a 45Hz source. Once we determine L, we can find the new inductive reactance at 45Hz.

Next, we use Ohm's Law (I = V/X₁) to find the current at 45Hz, where V is the rms voltage across the inductor and X₁ is the inductive reactance. This would give us the maximum current in the inductor.

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Answer:[tex]\Delta E=2.0715\times 10^{-21} J[/tex]

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

[tex]E=\frac{3}{2}\cdot \frac{R}{N_A}T[/tex]

Where

R=Gas constant

[tex]N_A[/tex]=Avogadro's number

T=Temperature in kelvin

[tex]\frac{R}{N_A}=1.381\times 10^{-23}[/tex]

So kinetic energy increases by

[tex]\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )[/tex]

[tex]\Delta E=2.0715\times 10^{-21} J[/tex]

Answer:

Explanation:

Conduction of heat per second Q  through a metal sheet having thermal conductivity K , thickness d , area A and temperature difference between cold and hot surface ( T₂ -T₁ ) is given by the following equation

Q = [tex]\frac{KA(T_2-T_1)}{d}[/tex]

Area A = π R² = 3.14 X 10000 X 10⁻⁶

= 3.14 X 10⁻2 m²

For aluminium plate :-

T₂ - T₁ = [tex]\frac{Q\times d}{KA}[/tex]

= [tex]\frac{600\times5\times10^{-3}}{240\times3.14\times10^{-2}}[/tex]

T₂ - T₁ = .4

T₂ = 110.4 ° C

For copper plate :

[tex][tex]\frac{600\times5\times10^{-3}}{390\times3.14\times10^{-2}}[/tex][/tex]

T₂ - T₁ = .25

T₂ = 110.25 ° C

The temperature of the surface in contact with the stove is approximately 113.0 °C for aluminum and 111.3 °C for copper, considering the thermal conductivity and dimensions of the pan.

This is a heat transfer problem involving conduction through different materials. The heat conduction through a flat wall, like the bottom of a pan, can be calculated using Fourier's law of heat conduction: q = -kA(T₂ - T₁)/d, where q is the heat transfer rate (600 W in this case), 'k' is the thermal conductivity, 'A' is the area across which heat is transferring, 'T₁' and 'T₂' represent the temperatures on either side of the wall, and 'd' represents the wall thickness.

First, convert the diameter of the pan to the radius in meters, which is (200mm/2)/1000 = 0.1 m. The area, A, is then π*(0.1 m)² = 0.0314 m². The thickness d given is already in meters: 5 mm = 0.005 m.

For aluminum, rearrange Fourier's law to find T₂, the temperature on the stove side of the pan: T₂ = T₁ + qd/(kA), where T₁ = 110°C and k for aluminum = 240 W/m K. Plugging in all values we get: T₂ = 110 + (600*0.005)/(240*0.0314) ≈ 113.0 °C.

For copper, using k for copper = 390 W/m K in the above formula will give: T₂ = 110 + (600*0.005)/(390*0.0314) ≈ 111.3 °C.

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Answer:

c.m=(1/∑mi)*∑mi*yi

[tex]c.m.=\frac{m*d+2m*3d}{m+2m}=7d/3[/tex]

Explanation:

We solve this problem, with the equation for the center of mass:

c.m=(1/∑mi)*∑mi*yi

[tex]m_i:[/tex]   represent the different mass of the system

[tex]y_i:[/tex]    represent the position of the different mass of the system

c.m=(1/∑mi)*∑mi*yi=(1/(m+2m))*(m*d+2m*3d)=7d/3

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

[tex]34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s[/tex]

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2[/tex]

Acceleration is 51.94 ft/s²

[tex]v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s[/tex]

Final velocity at this time is 257.63 ft/s

Answer:

Q = - 256 X 10⁻⁷ C .

Explanation:

Electric field due to a charge Q at a distance d from the center is given by the expression

E = k Q /d² Where k is a constant and it is equal to 9 x 10⁹

Put the given value in the equation

9 x 10² = [tex]\frac{9\times10^9\times Q}{(14+2)^2\times10^{-4}}[/tex]

Q = [tex]\frac{9\times16^2\times10^{-2}}{9\times10^9}[/tex]

Q = - 256 X 10⁻⁷ C .

It will be negative in nature as the field is directed towards the center.

Final answer:

To find the charge on the surface of the copper ball, we use Coulomb's law and the provided electric field magnitude, resulting in an approximate charge of 2.56 x 10⁻⁹ C or 2.56 nC, indicating a positive charge.

Explanation:

The question involves calculating the amount of charge on the surface of a copper ball using the given electric field intensity. The electric field due to a charged sphere at a distance from its surface can be calculated using Coulomb's law and the principle of superposition, which for a sphere of charge translates to E = kQ/r², where E is the electric field, k is Coulomb's constant (8.99 x 10⁹ Nm²/C²), Q is the charge, and r is the distance from the center of the sphere to the point of interest (in this case, 14 cm from the surface or 16 cm from the center considering the radius of the copper ball is 2 cm).

Substituting the given values into the equation, we have 9.0 x 10² = (8.99 x 10⁹)Q/(0.16)². Solving for Q gives us Q = 9.0 x 10² x (0.16)² / (8.99 x 10⁹) C, which yields approximately 2.56 x 10⁻⁹ C or 2.56 nC. The charge is positive, as the electric field is directed towards the center of the ball, indicating a positive source.

Answer:

a ) 2.13 X 10⁶ m/s .

b ) 1.28 X 10⁶ m/s

Explanation:

When electrons are repelled by negative plates and attracted by positive plates , it will increase their kinetic energy.

Increase in their energy = 4.15 eV

= 4.5 X 1.6 X 10⁻¹⁹ J

= 6.64 x 10⁻¹⁹ J

Initial kinetic energy

= 1/2 mv²

= 1/2 x 9.1 x 10⁻³¹ x ( 1.76 x 10⁶)²

= 14.09 X 10⁻¹⁹ J

Total energy

= 6.64 x 10⁻¹⁹+14.09 X 10⁻¹⁹

= 20.73 x 10⁻¹⁹

If V be the increased velocity

1/2 m V² = 20.73 X 10⁻¹⁹

.5 X 9.1 X 10⁻³¹ V² = 20.73 X 10⁻¹⁹

V = 2.13 X 10⁶ m/s .

b ) When electrons are released from positive plate , their speed are reduced because of attraction between electrons and positively charge plates.

Initial kinetic energy

= 14.09 x 10⁻¹⁹ J (see above )

reduction in kinetic energy  

= 6.64 x 10⁻¹⁹ J ( See above )

Total energy with electron -

= 14.09 x 10⁻¹⁹  - 6.64 x 10⁻¹⁹

= 7. 45 x 10⁻¹⁹ J .

If V be the energy of electrons reaching the negative plate,

1/2 m V² =7. 45 x 10⁻¹⁹

V = 1.28 X 10⁶ ms⁻¹

Answer:

The final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Explanation:

A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.

GIVEN:

The distance traveled by the ball is equal to h.

Initial velocity of the ball is zero.

Concept:

Two type of force comes into play in dropping of ball.

1. Gravitational force

Due to gravitational force ball forced to fall downward towards the surface.

2. Air resistance

Air develops drag force on the ball in opposite direction of the gravitational force.

Assumption:

There is no air resistance in falling of ball. So, there will be no drag force on the ball.

Calculation:

It is given that a ball is dropped from a building of height h.

The Newton’s equations of motion are as follows:

[tex]V^{2}-u^{2}=2as[/tex]

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

[tex]v=u+at[/tex]

STEP 1

Newton’s equation of motion is expressed as follows:

[tex]V^{2}-u^{2}=2as[/tex]

Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.

Here, initial velocity is zero. So, u = 0 m/s.

Vertical distance traveled by the ball is h. So, s= h.

Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.

Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.

Substitute 0 for m/s, g for a and h for s in above equation as follows:

[tex]V^{2}-u^{2}=2as[/tex]

[tex]V^{2}-(0)^{2}=2gh[/tex]

[tex]V^{2}=2gh[/tex]

[tex]v=\sqrt{2gh}[/tex]

Thus, the final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

[tex]R = u² sin2θ/g[/tex]

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

[tex]u² = \frac{Rg}{sin2\theta}[/tex]

[tex]u^2 = \frac{9.1 x 9.80}{sin26}[/tex]

[tex] u^2 = 113.17[/tex]

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

[tex]v^2 = (v_osin\theta)^2 -2gh[/tex]

[tex]h =\frac{(v_o^2sin\theta)^2}{2g}[/tex]

[tex]h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}[/tex]

h = 1.10 m

We have that for the takeoff speed and maximum height above ground is

From the question we are told

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo leaves the ground at a 26° angle. If this is so:

1) What is its takeoff speed

2) What is its maximum height above the ground?

Generally the equation for the Projectile motion   is mathematically given as

[tex]R=\frac{u^2sin 2\theta}{g}\\\\Therefore\\\\9.1=\frac{u^2sin 52}{9.8}[/tex]

u=10.69m/s

And

Generally the equation for the Projectile Height   is mathematically given as

[tex]Hmax=\frac{u^2sin^2 2\theta}{2g}\\\\Therefore\\Hmax=\frac{114.27*0.184}{19.6}[/tex]

Hmax=1.072m

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Answer:

40m (40m east)

Explanation:

The fist move is 70m east, and east is the positive direction so the truck initially is at +70m.

The second move is 120m to the west, since east is the positive direction, west must be the negative direction, this means the truck now is at:

[tex]+70m -120m = -50m[/tex]

The third move is 90m to the east, again, this is the positive direction, so the new position is:

[tex]-50m + 90m = +40m.[/tex]

The truck is at + 40m, so it ended up 40m away from its initial position and this is the resultant dispacement.

Answer: 950 Kg/m^3

Explanation: We can deduce from the Archimedes principle that there is a relation between the density and the volumes displaced, as follows:

Density*Volume= Mass

So for equilibrium Density of body= Density of water *Vw/Vb

Being Vw/Vb the relation between  the displaced water and the body volume, and given the water density as 1000 Kg/m^3 we got:

Density(B)= 0.95 * 1000 Kg/m^3.

Answer:

I think D is the answer.

D. The development of reading and writing in young children

Final answer:

The glass globe with a net electric charge of 5.31 × 10^-6 C has 3.32 × 10^13 fewer electrons than it did in its neutral state, because positive net charge indicates electron removal.

Explanation:

When a glittering glass globe is given a net electric charge of 5.31 × 10^-6 C, it means that electrons have been added or removed to achieve this net charge. The charge on an electron is known to be approximately -1.6 × 10^-19 C. To determine if the globe has more or fewer electrons than in its neutral state, we compare the sign of the given charge. Here, the positive charge indicates that electrons have been removed. To calculate the quantity of electrons removed, we use the formula:

Number of electrons (n) = Net charge (Q) / Elementary charge (e).

Plugging in the values:

n = 5.31 × 10^-6 C / 1.6 × 10^-19 C/e-

n = 3.32 × 10^13 electrons.

Therefore, the globe now has 3.32 × 10^13 fewer electrons than it did in its neutral state.

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car.  It can be calculated using third equation of motion as :

[tex]v^2-u^2=2ad[/tex]

[tex]a=\dfrac{v^2-u^2}{2d}[/tex]

[tex]a=\dfrac{0-(22.22)^2}{2\times 80000}[/tex]

[tex]a=-0.00308\ m/s^2[/tex]

Value of g, [tex]g=9.8\ m/s^2[/tex]

[tex]a=\dfrac{-0.00308}{9.8}\ m/s^2[/tex]

[tex]a=(-0.000314)\ g\ m/s^2[/tex]

Hence, this is required solution.

Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.

Here's how to solve the problems about the falling stone:

1. Time to reach the bottom:

First half of the journey (upward):

Use the equation v = u - gt, where v is the final velocity (0 m/s at the top), u is the initial velocity (15.0 m/s), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.

Solving for t, we get t = u/g = 15.0 m/s / 9.81 m/s^2 = 1.53 seconds.

Second half of the journey (downward):

The stone falls freely again with an initial velocity of 0 m/s. The time taken will be the same as the upward journey, 1.53 seconds.

Total time to reach the bottom:

Add the times for both halves: 1.53 seconds + 1.53 seconds = 3.06 seconds.

2. Speed just before hitting:

Use the same equation v = u - gt, but this time u is 0 m/s (at the top) and t is the total time (3.06 seconds).

Solving for v, we get v = 0 m/s - 9.81 m/s^2 * 3.06 seconds = -29.8 m/s (negative sign indicates downward direction).

Therefore, the stone's speed just before hitting is 29.8 m/s downwards.

3. Total distance traveled:

The stone travels twice the height of the cliff: once going up and once going down.

Total distance = 2 * cliff height = 2 * 75.0 m = 150.0 m.

Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.

An equipotential line is like the contour line of a map that had lines of equal altitude. In this case the "altitude"is the electrical potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions these lines form equipotential surfaces. The movement along an equipotential surface does not perform work, because that movement is always perpendicular to the electric field.

Answer:

The momentum is not conserved.

Explanation:

When the superhero throws the piano, he exerts a force on the piano. Newton's third law says that each action has an opposite and equal reaction. In other words, this means that when the superhero throws the piano, the piano also exerts a force on the superhero, therefore, the superhero should recoil. You can visualize a gun being shot to visualize this phenomenon, when the gun is shot, it will tend to go backwards, trying to conserve momentum.

The scenario in the movie is not physically possible and violates the principles of Newton's laws of motion and conservation of momentum.

The scenario in the movie where the superhero hovers in the air and throws a piano while remaining stationary is not physically possible. It violates the principles of Newton's laws of motion and the conservation of momentum. In order for the superhero to hover and throw a piano, they would need some external force or propulsion, like jetpack or levitation powers, which are not commonly associated with superheroes.

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Answer:

An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.

Explanation:

An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.

if the +ve charge moves in the very same direction as the electrical field vector then the velocity of the particle will increase. It will slow down if it moves in the reverse direction.

if the adverse charge moves in the very same direction as the electrical field vector, the particle velocity will decelerate. If it moves in the reverse direction, it will speed up.

Answer and Explanation:

When a particle is placed in an Electric field having a charge Q and mass M, then it experiences an electric force on it. and it is given by:

[tex]F = QE[/tex]

This force is the overall force which is responsible for the acceleration of the particle.

We know, from Newton's second law:

F = Ma

and the force on a particle in electric field:

[tex]F_{E} = QE[/tex]

Therefore, from these two forces:

Ma = QE

a = [tex]\frac{QE}{M}[/tex]

In case of uniform Electric field, the particle moves with constant acceleration.

This acceleration is also capable of bending the object and to move it in a circle.

Answer:

1) Distance traveled equals 23.1 meters.

2) Final velocity equals 21.658 m/s.

Explanation:

The problem can be solved using second equation of kinematics as

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

where

s is the distance covered

u is the initial speed of the ball

a is the acceleration the ball is under

t is time of travel

Applying the given values in the above equation we get

[tex]s=4.0\times 1.8+\frac{1}{2}\times 9.81\times 1.8^{2}\\\\s=23.1meters[/tex]

Part 2)

The velocity after 't' time can be obtained using first equation of kinematics.

[tex]v=u+at[/tex]

Applying the given values we get

[tex]v=4+9.81\times 1.80\\\\\therefore v=21.658m/s[/tex]

Final answer:

The ball travels 23.07 meters in 1.80 seconds, and the velocity of the ball at that time is 21.66 m/s downward.

Explanation:

The student wants to know how far a ball projected vertically downward at a speed of 4.00 m/s will travel in 1.80 s and the ball's velocity at that time. To solve these questions, we use the kinematic equations that govern linear motion with constant acceleration due to gravity, ignoring air resistance.

The distance traveled by the ball can be found using the equation:

s = ut +  [tex]rac{1}{2}at^2[/tex]

Where:

s is the distance traveled,

u is the initial velocity (4.00 m/s),

t is the time (1.80 s),

a is the acceleration due to gravity (9.81 m/s^2).

Plugging in the values, we have:

s = 4.00 m/s  imes 1.80 s + 0.5  imes 9.81 [tex]m/s^2[/tex] imes [tex](1.80 s)^2[/tex]

s = 7.20 m + 15.87 m

s = 23.07 m

The ball travels 23.07 meters in 1.80 seconds.

The velocity of the ball after 1.80 seconds can be calculated using the equation:

v = u + at

Where:

v is the final velocity,

u is the initial velocity,

a is the acceleration,

t is the time.

So, v = 4.00 m/s + 9.81 [tex]m/s^2[/tex] imes 1.80 s

v = 4.00 m/s + 17.66 m/s

v = 21.66 m/s

The velocity of the ball after 1.80 seconds is 21.66 m/s downward.

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

Answer:

The equivalent capacitance will be [tex]15\mu F[/tex]  

Explanation:

We have given two capacitance [tex]C_=10\mu F\ and\ C_2=20\mu F[/tex]

They are connected in parallel

So equivalent capacitance [tex]C=C_1+C_2=10+20=30\mu F[/tex]

This equivalent capacitance is now connected in series with [tex]30\mu F[/tex]

In series combination of capacitors the equivalent capacitance is given by [tex]\frac{1}{C}=\frac{1}{30}+\frac{1}{30}[/tex]

[tex]C=\frac{30}{2}=15\mu F[/tex]

So the equivalent capacitance will be [tex]15\mu F[/tex]  

The equivalent capacitance of the arrangement where capacitors of 10 uF and 20 uF are connected in parallel, and in series with a 30 uF capacitor, is 15 uF.

To determine the equivalent capacitance of a combination in which capacitors are connected both in parallel and in series, one must consider the rules for each type of connection. For capacitors connected in parallel, the equivalent capacitance is simply the sum of their individual capacitances. When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. In your case, the 10 uF and 20 uF capacitors are connected in parallel, so their equivalent capacitance is 10 uF + 20 uF = 30 uF. This combined capacitance is then in series with the 30 uF capacitor.

The next step is to calculate the equivalent capacitance for the series connection using the formula:

1/Cseries = 1/C1 + 1/C2

Where C1 is the equivalent parallel capacitance (30 uF) and C2 is the capacitance of the third capacitor (30 uF). Substituting the values:

1/Cseries = 1/30 uF + 1/30 uF = 1/15 uF

So, Cseries = 15 uF. This is the total equivalent capacitance of the combination of capacitors.

The coefficient of friction in this scenario is calculated as the ratio of the frictional force to the normal force. After calculating the frictional force to be 50N and the normal force to be 300N, the coefficient of friction is 0.167.

Dana is pulling a box on a level surface with a rope that makes an angle of 30 degrees. The box has a weight of 300 N and the tension in the rope is 100 N. We first must understand that the force of friction is balancing the force that Dana is applying to move the box.

The vertical component of the tension (Tsin30) should be equal to the force of friction since they are balancing each other. Therefore, Tsin30 = Friction. Substituting the given values (100N*sin30 = Friction), we can calculate friction as 50 N.

The coefficient of friction can be calculated as the ratio of the frictional force to the normal force. As there is no vertical acceleration, the normal force is equal to the weight of the box which is 300N. Therefore, the coefficient of friction (mu) equals friction/normal force is equal to 50N/300N which is 0.167.

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The coefficient of friction is approximately 0.247.

The coefficient of friction between the box and the surface is given by the ratio of the frictional force to the normal force. The frictional force is the horizontal component of the tension in the rope, and the normal force is the vertical component of the tension in the rope plus the weight of the box acting downward.

First, let's calculate the horizontal and vertical components of the tension in the rope. The horizontal component can be found using the cosine of the angle, and the vertical component can be found using the sine of the angle:

[tex]\[ T_{\text{horizontal}} = T \cdot \cos(\theta) \] \[ T_{\text{vertical}} = T \cdot \sin(\theta) \][/tex]

Given that the tension in the rope (T) is 100 N and the angle [tex](\(\theta\))[/tex] is 30 degrees, we have:

[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \cos(30^\circ) \] \\T_{\text{vertical}} = 100 \text{ N} \cdot \sin(30^\circ) \][/tex]

Using the values of cosine and sine for 30 degrees:

[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \frac{\sqrt{3}}{2} \approx 86.6 \text{ N} \]\\T_{\text{vertical}} = 100 \text{ N} \cdot \frac{1}{2} = 50 \text{ N} \][/tex]

The normal force (N) is the sum of the weight of the box (W) and the vertical component of the tension:

[tex]\[ N = W + T_{\text{vertical}} \] \[ N = 300 \text{ N} + 50 \text{ N} \] \[ N = 350 \text{ N} \][/tex]

The frictional force (f) is equal to the horizontal component of the tension since the box is moving at constant velocity (implying that the net force in the horizontal direction is zero):

[tex]\[ f = T_{\text{horizontal}} \] \[ f = 86.6 \text{ N} \][/tex]

Now, the coefficient of friction is the ratio of the frictional force to the normal force:

[tex]\[ \mu = \frac{f}{N} \] \[ \mu = \frac{86.6 \text{ N}}{350 \text{ N}} \] \[ \mu \approx 0.247 \][/tex]

Answer:

speed when the block had slid 3.40 m is 2.68 m/s

Explanation:

given data

distance = 6.80 m

speed = 3.80 m/s

to find out

speed when the block had slid 3.40 m

solution

we will apply here equation of motion that is

v²-u² = 2×a×s   ..............1

here s is distance, a is acceleration and v is speed and u is initial speed that is 0

so put here all value in equation 1 to get a

v²-u² = 2×a×s

3.80²-0 = 2×a×6.80

a = 1.06 m/s²

so

speed when distance 3.40 m

from equation 1 put value

v²-u² = 2×a×s

v²-0 = 2×1.06×3.40

v² = 7.208

v = 2.68

so speed when the block had slid 3.40 m is 2.68 m/s

Answer: Ok, so we know that them start on the same place.

First, the first one rides 1290 to the east, and 1410 to the north.

So if we put the 0 at the place where they start, north is the Y axis and east the X axis, we could describe the campground is in (1290,1410)

a) when the second cyclist turns, he traveled 1890m to the north, se the point where he is, is described by (0, 1890). We want to know how far is from the campground which we already know that is located in (1290,1410), so if we subtract we get distance = (0,1890) - (1290,1410) = (-1290, 1890 - 1410) = (1290, 480). The total distance will be D = [tex]\sqrt{1290^{2} + 480^{2}  }[/tex] = 1376.4 meters.

b) You want to know in what angle should he turn now.

So again, the cyclist is on the point (0,1890) and the campground is on the point (1290,1410) so he want to go 1290 meters to east, and -480 meters to north. If we define this two amounts as the cathetus of triangle rectangle, the route that he must follow is the hypotenuse of said triangle.

So the angle that he must turn (counting from the east, or the +x axis, counterclockwise) is defined by Tg(A) = -480/1290 = -0.372.

A = aTg(-0.372) = -20° aprox.

Answer:

Time taken by the coin to reach the ground is 1.69 s

Given:

Initial speed, v = 11.8 m/s

Height of the building, h = 34.0 m

Solution:

Now, from the third eqn of motion:

[tex]v'^{2} = v^{2} + 2gh[/tex]

[tex]v'^{2} = 11.8^{2} + 2\times 9.8\times 34.0 = 805.64[/tex]

[tex]v' = \sqrt{805.64} = 28.38 m/s[/tex]

Now, time taken by the coin to reach the ground is given by eqn (1):

v' = v + gt

[tex]t = \frac{v' - v}{g} = \frac{28.38 - 11.8}{9.8} = 1.69 s[/tex]