A roller coaster car is released from rest as shown in the image below. If
friction is neglected, the car will oscillate back and forth across the "dip" in
the roller coaster. What is the approximate velocity of the roller coaster car
each time it reaches the bottom of the roller coaster in the image? (Recall
that g = 9.8 m/s2.)
TAS
81 m
O A. 40 m/s
B. 25 m/s
C. 30 m/s
D. 15 m/s

Answer:

In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

Explanation:

It is given that a student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.

In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass.

In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.

In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

As per the given problem, the student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.

Thus, we can conclude that the in the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

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Answer:Power is a measure of the amount of work that can be done in a given amount of time. Power equals work (J) divided by time (s). The SI unit for power is the watt (W), which equals 1 joule of work per second (J/s). Power may be measured in a unit called the horsepower

Explanation:

Answer:

Explanation:Power equals work (J) divided by time (s).

Answer:

A. True

B. False

C. False

D. False

Answer:

A

Explanation:

protons and electrons are far apart with lots of empty space in between.

Answer:

Explanation:

Writing out the Newton's Law pf Cooling:

dT/dt = -k * (T - A),

where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.

If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C,

T = 100

A = 25

dT = 100 - 90 = 10

dt = 1

Putting the figures into the equation:

10/1 = -k * (100 - 25)

k = -10/75°C

After 4 minutes, dT/4 = 10/75 (100 - 25) = 10

dT = 40

Temperature after 4 minutes = 100 - 40 = 60°C

The temperature of a cup of coffee varies according to Newton's Law of Cooling, the temperature of the coffee after 4 minutes is approximately 67°C.

To tackle this problem, we can apply Newton's Law of Cooling's differential equation and solve it using variable separation.

dT/dt = -k(T - A)

At t = 0 (initial condition): T = 100°C

At t = 1 minute: T = 90°C

dT/dt = -k(T - A)

At t = 0: dT/dt = -k(100 - 25)

So,

-10 = -k(75)

k = 10/75

Separating variables and integrating, we have:

1/(T - A) dT = -k dt

∫(1/(T - A)) dT = ∫(-k) dt

ln|T - A| = -kt + C

ln|100 - 25| = 0 + C

ln|75| = C

So, the equation will be:

ln|T - A| = -kt + ln|75|

ln|(T - 25)/(75)| = -kt

Now,

ln|((T - 25)/(75))| = -(10/75)(4)

|((T - 25)/(75))| = \(e^{(-40/75)\)

T - 25 = ± 75 *  \(e^{(-40/75)\)

T = 25 ± 75 *  \(e^{(-40/75)\)

T ≈ 25 ± 42.42

Therefore, the temperature of the coffee after 4 minutes is approximately:

T ≈ 25 + 42.42 = 67.42°C

Thus, the temperature of the coffee after 4 minutes is approximately 67°C.

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The pressure at the upper level of a water flow into the house by means of pipe is 1081 kPa.

Calculate the cross-sectional area of the lower pipe:

A₁ = πr₁²

where:

A₁ = cross-sectional area of the lower pipe (m²)

π = mathematical constant (3.14)

r₁ = radius of the lower pipe (m)

A₁ = π(0.12 m)² = 0.0452 m²

Calculate the cross-sectional area of the upper pipe:

A₂ = πr₂²

where:

A₂ = cross-sectional area of the upper pipe (m²)

π = mathematical constant (3.14)

r₂ = radius of the upper pipe (m)

A₂ = π(0.06 m)² = 0.0113 m²

Calculate the flow rate per unit area:

q = Q/A

where:

q = flow rate per unit area (m³/s)

Q = flow rate (m³/s)

A = cross-sectional area (m²)

q = 6 m³/s / 0.0452 m² = 13.28 m²/s

Calculate the velocity of the water in the lower pipe:

v₁ = q/A₁

where:

v₁ = velocity of the water in the lower pipe (m/s)

q = flow rate per unit area (m³/s)

A₁ = cross-sectional area of the lower pipe (m²)

v₁ = 13.28 m²/s / 0.0452 m² = 29.3 m/s

Calculate the velocity of the water in the upper pipe:

v₂ = q/A₂

where:

v₂ = velocity of the water in the upper pipe (m/s)

q = flow rate per unit area (m³/s)

A₂ = cross-sectional area of the upper pipe (m²)

v₂ = 13.28 m²/s / 0.0113 m² = 117.0 m/s

Calculate the head loss:

hL = (v₁² - v2₂²) / 2g

where:

hL = head loss (m)

v₁ = velocity of the water in the lower pipe (m/s)

v₂ = velocity of the water in the upper pipe (m/s)

g = acceleration due to gravity (9.8 m/s²)

hL = (29.3 m/s)² - (117.0 m/s)² / 2(9.8 m/s²) = 23.2 m

Calculate the pressure at the upper level:

p₂ = p₁ + ρghL

where:

p₂ = pressure at the upper level (Pa)

p₁ = pressure at the lower level (Pa)

ρ = density of water (1000 kg/m³)

g = acceleration due to gravity (9.8 m/s²)

hL = head loss (m)

p₂ = 400 kPa + 1000 kg/m³(9.8 m/s²)(23.2 m) = 1081 kPa

Therefore, the pressure at the upper level is 1081 kPa.

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i) The initial deceleration of the car is -1.6 m/s² and  the time taken is 5 seconds

ii) The final deceleration is -1 m/s²

iii) The total dispalcement = 1016 m

The initial deceleration of the car is given by the formula below:

v² = u² + 2as

where;

Solving for a;

12² = 20² + 2 * a * 80

a = -1.6 m/s²

Time taken, t = v - u / a

t = 12 - 20 / (-1.6)

t = 5 seconds

Final deceleration:

a = v - u / t

a = 0 - 12 / 12

a = -1 m/s²

iii) Displacement at constant velocity = 12 * 1 * 60

Displacement at constant velocity = 720 m

Final displacement, s = ut + 0.5at²

s = 12 * 12 + 0.5 * 1 * 12²

s = 216 m

Total dispalcement = 80 + 720 + 216

Total dispalcement = 1016 m

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Answer:

To calculate the resistance of each circuit, we can use Ohm's law:

Resistance = Voltage / Current

For Circuit A: Resistance = 6 V / 2.3 A = 2.61 Ω

For Circuit B: Resistance = 6 V / 0.6 A = 10 Ω

For Circuit C: Resistance = 6 V / 0.2 A = 30 Ω

For Circuit D: Resistance = 6 V / 1.8 A = 3.33 Ω

To graph the current versus the resistance of each circuit, we can plot the resistance on the x-axis and the current on the y-axis. The resulting graph will show a decreasing linear relationship between current and resistance, with higher resistance resulting in lower current flow. This is in accordance with Ohm's law, which states that the current in a circuit is directly proportional to the voltage and inversely proportional to the resistance.

Explanation:

The current and voltage are directly proportional to each other and the resistance is inversely proportional to the current. The relation of three quantities is given by Ohm's law.

Ohm's law is defined as the applied voltage is directly proportional to the current flow through the circuit. V = IR where R is the resistance of the circuit and it obstructs the current flow. The unit of resistance is the ohm.

From the given,

Resistance, R₁  = V/I₁ = 6 /2.3 = 2.60Ω

Resistance, R₂  = V/I₂ = 6/0.6 = 10 Ω

Resistance, R₃  = V/I₃ = 6/0.2 = 30Ω

Resistance, R₄  = V/I₄ = 6/1.8 = 3.33Ω

In current versus resistance graph shows the linearly decreasing function, as the current increases resistance decreases and vice versa. Hence, the resistance and current are inversely proportional to each other.

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The slope of a linear position graph tells the velocity of the object, for an object which is moving at a constant speed the position graph with linear shape, and the velocity graph with horizontal shape, and for an object which is moving at a constant acceleration the position graph will be the in parabolic shape and velocity graph is of linear shape.

The rate of change in an object's velocity with respect to time is known as acceleration in mechanics. The vector quantity of accelerations. The direction of the net force that is acting on an object determines its acceleration.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector over a time interval divided by the time interval.

There are several types of acceleration :

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Answer:

Agricultural Engineer.

Agricultural Food Scientist.

Agricultural Inspector.

Agricultural Manager.

Agricultural Specialist.

They are;

Answer:

Folding

Explanation:

Answer:

SI unit is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion.

Explanation:

hope it helps

Si unit is used almost all around the world, so our use of it allows

Scientist from disparate regions to use a single standered in communicating scientific data without vocabulary confusion

Answer:

Explanation:

airspeed is speed with respect to air .

Let speed of airplane with respect to ground be V .

wind is blowing east at 55 km/h

airplane is heading north with speed V

Resultant of V and 55 km /h is 230 km/h

V² + 55² = 230²

V² = 52900 - 3025

V = 223.32 km/h

Let θ be the required angle

Tanθ = 55 / 223.32 = .246

θ = 14⁰ .

Answer:

\(\boxed {\boxed {\sf 5 \ meters/second}}\)

Explanation:

Speed is equal to distance over time.

\(s=\frac{d}{t}\)

The distance is 100 meters and the time is 20 seconds.

\(d= 100 \ m \\t= 20 \ s\)

Substitute the values into the formula.

\(s=\frac{100 \ m }{20 \ s}\)

Divide.

\(s= 5 \ m/s\)

Aman's speed is 5 meters per second.

Answer:

19.6 Newtons

Explanation:

\(\displaystyle \textsf{The weight of the block is W= mg where m is the mass and g the acceleration due to gravity}\)

\(\sf W = 5kg, \; and\; g = 9.80 m/s^2\)

\(\sf W = 5 x 9.8 = 49\; Newtons\)

\(\textsf {If the coefficient of static friction is } $\mu_s$, the force to be applied for the block to slip is given by the equation}:\\\mathsf {f_s = \mu_s \times W}\\\\\textsf {So force required, given $\mu_s$ = 0.4 is: }\\\\\\\mathsf {f_s = 0.4 \times 49 = 19.6 } \textsf{ Newtons }\)

The initial velocity of the cart.

Newton, There can be a mass is four.080, So acceleration might be equal to 2.50 m in step with cent within the rectangular. Initial is that amount that relies upon total mass. The greater mass the more inertia. So Mass is 2000 kg and acceleration is 3 ms square. So this offers us an internet pressure identical to 6000 newtons or 6.0 and 210 to the power

In case you roll a ball, it initially will keep rolling except friction or something else stops it by means of pressure. you could also think about the way that your body maintains transferring ahead when you hit the brake on your bike. Translational Inertia = ma, in which m is the mass, and a is the acceleration of the object. Calculate the rotational inertia or the instant of inertia velocity by way of multiplying the mass of the object with a square of the gap between the item and the axis, the radius of rotation.

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Answer:

Gravity between you and the sun

Answer:

Explanation:

To find the focal length, we will use the following equation

Where do is the distance of the object and di is the distance of the image.

So, replacing do = 8.0 cm and di = 16.0 cm, we get

Therefore, the focal length is 5.3 cm

Answer:

the answer is c

Explanation:

see the light appears from there and with the dotted lines you can clearly see the green line touches the dot okay, then the light appears smaller because of water and light source

Hi there!

a)

We can begin by using the equation for energy density.

\(U = \frac{1}{2}\epsilon_0 E^2\)

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
\(\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}\)

Creating a Gaussian surface being the lateral surface area of a cylinder:
\(A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}\)

Now, we can calculate the energy density using the equation:
\(U = \frac{1}{2} \epsilon_0 E^2\)

Plug in the expression for the electric field and solve.

\(U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}\)

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
\(V = \pi r^2L \\\\dV = 2\pi rLdr\)

Now, we can take the integral of the above expression. Let:
\(r_i\) = inner cylinder radius

\(r_o\) = outer cylindrical shell inner radius

Total energy-field energy:

\(U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr\)

Plug in the equation for the electric field energy density and solve.

\(U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\\)

Bring constants in front and integrate. Recall the following integration rule:
\(\int {\frac{1}{x}} \, dx = ln(x) + C\)

Now, we can solve!

\(U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\)

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

\(\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}\)

And here's our equation!

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

18.36 m/s

Explanation:

We can solve this using conservation of energy. The energy in the system will be conserved since there are no outside forces acting upon it so the potential energy and kinetic energy will be equal. Giving us this formula to start:

1/2mv^2=mgh

m=mass

g=gravity

h=height

v=velocity

We can start by figuring out the total height the rock travels which we can do by subtracting the height of the frisbee by the height the rock started at.

19m-1.8m=17.2m

Now we can plug in our variables to solve for velocity.

First we negate mass since its on both sides and cancels out leaving us with.

1/2v^2=gh

Plug in.

1/2v^2=(9.8)(17.2)

1/2v^2=168.56

v^2=337.12

v=18.36m/s

Answer:

the time for the car to reach the final velocity is 0.56 s.

Explanation:

Given;

acceleration of the car, a = 50 m/s²

final velocity of the car, v = 100 km/h = 27.778 m/s

the initial velocity of the car, u = 0

The time for the car to reach the final velocity is calculated as;

v = u + at

27.778 = 0 + 50t

27.778 =  50t

t = 27.778 / 50

t = 0.56 s

Therefore, the time for the car to reach the final velocity is 0.56 s.

Raven's percentage error of the measurement is 5.26%

Percentage error = (absolute error / standard measurement) × 100

The pecenrage error of the measurement can be obtained as illustrated below:

Percentage error = (absolute error / standard measurement) × 100

Percentage error = (0.006 / 0.114) × 100

Percentage error = 5.26%

Thus, the perecentage error of Raven's measurement is 5.26%

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Answer:

I think the answer is 5.26%

but please let me know if I was wrong.

5.26% is the answer I got

Explanation:

Answer:

69

Explanation:

Perform addition first as;

(0.210 + 0.41) = 0.62 -------{2 significant figures}

Perform product

(1.10 x 102) = 112 -------------- {3 significant figures}

The combined product

0.62 x 112 =69.44

                 = 69 ----------2 significant figures

Rule

The least number of significant figures in the numbers performing the operation determines the number of significant figures in the answer.

The two social, economic, or environmental issues are Safety and security concerns and the Cost of storage and disposal.

People all across the world are impacted by global environmental changes, such as pollution, climate change, biodiversity loss, and freshwater decrease, which have social and economic effects in addition to physical ones. The continuity of families and communities, social ties, health, and occasionally survival are all impacted by these changes.

There are many environmental problems, but there are three main ones that influence most of them in general: the loss of biodiversity, water pollution, ocean acidification, and climate change.

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Answer:

For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. If the velocity is non-uniform all you can say is what the average speed is.

Once the ball is thrown, the only force acting on it is gravity, which means that it's acceleration is -9.81 m/s² (negative means downward).

List the known and unknown quantities from the question.

u = initial velocity = 20 m/s

v = final velocity = ? m/s

g = acceleration due to gravity = -9.81 m/s²

t = time interval = ? s

s displacement = 11 m

Before calculating the time it takes for the ball to reach 11 m, the final velocity needs to be calculated using the following kinematic equation.

v² = u² + 2gs

v = √(u² + 2gs)

= √((20 m/s)² + (2x-9.81 m/s² x 11 m)) = 13.57 m/s V=

Calculate the time it takes the ball to reach 11 m using the following kinematic equation.

V = u + gt

Solve for t.

t = (v-u)/g

t (13 57 m/s - 20 m/s)/(-981 m/s²) = 0.655 s

The current through each lamp is  0.545 A and the power dissipated in each lamp is  0.323 W.

In the diagram the batteries are 1.5V each and each lamp has a resistance of 1.1Ω when the switch is closed.

When S₁ is closed. The current is passing through lamps. The lamps are connected in the series connection the current passing through each is same.

Applying KVL through lamps

1.1 I + 1.1 I + 1.1 I + 1.1 I + 1.1 I - 1.5 - 1.5  = 0

5( 1.1 I) = 3

I = 3/5.5 = 0.545 A

Current through each lamp is 0.545 A

The resistor value of each lamp is same and current passing through each lamp is also same.

Therefore, power dissipation through each lamp is also same.

P = I²R = (0.545)²× 1.1 = 0.323 W

Hence, the current through each lamp is  0.545 A and the power dissipated in each lamp is  0.323 W.

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