4. Brian stands at the edge of a cliff and throws a stone horizontally over the edge with a
speed of 18 m/s. The cliff is 50 m above a flat, horizontal beach: (2 pts each)
a. How long does it take for the stone to reach the bottom of the cliff?
b. What is the range of the stone?
5. A student shot a basketball at an initial velocity of 25 m/s at an angle of 30 degrees with
respect to the horizontal: (2 pts eac)
a. What is the vertical component of the initial velocity?
b. What is the maximum height reached by the arrow?
6. A fireman stands 50 m away from a burning building and directs a stream of water from
ground level at an angle of 30° above the horizontal. If the speed of the stream as it
leaves the hose is 40 m/s: (2 pts each)
a. How long will it take for the stream to reach the building?
b. At what height will the water strike the building?

Answer:

F(friction) = μ M g         definition of frictional force

μ  = F / (M g) = 11 N / 50 N = .22

The co-efficient of kinetic friction between the block and the table is 0.22.

To find the co-efficient of friction, the given values are,

Friction force - 11 N

Normal force - 50 N

         μ = 11 N / 50 N

             = 0.22

Thus, the co-efficient of the kinetic friction is 0.22

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Answer:

We are given that the student has a weight of 400N

So, the force applied by the boy has to be added to 400 since he has to apply force to keep his body on the stairs

We are given:

u = 0 m/s

v = v m/s

t = 4 s

a = a m/s/s

s = 3 m

w = 400 N

Converting to Mass

w = m * g    (where m is the actual mass and g is the value of gravity)

letting g = 10m/s/s

400  = 10 * m

m = 40 kg

Finding Acceleration:

From the second equation of motion:

s = ut + (a*t*t) / 2

3 = (0)(4) + (a)(4)(4)/2

3 = 8a

a = 3/8 m/s/s

Force Applied by the Student:

Now that we have the acceleration, we can find the force applied by the student to go up the stairs

F = ma (Newton's second law of motion)

F = (40)(3/8)

F = 15 N

Hence, the student is applying 15N to climb up the stairs and another 400 N to support his mass

Total Force applied by the Student:

F(total) = F(climbing) + F(weight)

F(total) = 15 + 400

F(total) = 415 N

Work done by the Student:

Work done = F.S   (dot product of force applied and displacement)

Work done = (415)(3)

Work done = 1245 Joules

Power Output of the Student:

power output is the amount of work done per second

Power = Work Done / t

Power = 1245 / 4

Power = 311.25 Watt

**note that the this is just the power consumed by the student to climb the stairs**

Explanation:

The mechanical energy can be expressed as kinetic and potential energy, given by the equation;

ME=KE+PE

but since the car isn't elevated off the ground nor is it subjected to the restoring force of a spring, we can then say PE=0

ME=KE

The equation of KE=0.5mv²

ME=0.5mv²

32 000=0.5(1600)v²

[tex]v {}^{2} = \frac{32 \: 000}{1600 \times 0.5} [/tex]

[tex]v {}^{2} = \frac{32 \: 000}{800} [/tex]

[tex] {v}^{2} = 40[/tex]

[tex]v = \sqrt{40} = 6.324[/tex]

v=6.324 m/s :)

Answer:

D.

Explanation:

Budgeting time, avoiding stress, and prioritizing.

Answer:

a)  0.002 cm³

b) 0.005 cm³

Explanation:

density d = 19 kg/cm³

Given mass m, find volume V

The definition of density is

d = m/V

so V = m/d

a) m = 38 g = 0.038 kg

so V = 0.038/19 = 0.002 cm³

b) m = 0.095 kg

V = 0.095/19 = 0.005 cm³

Answer:

ززث ع ثذآرثفاقآثہ تجفہجاءآڑجءبیجدءچءاءاڑڑشءڑدءحابڑءحبزیاڑچءازءچزیبزچدزچ،چدز،ڈزیجپا،جڈزیجڈیبزءفحدءڈحضغ دح دفھچءچدجءیچزءچددفحءدفءاڑحعخ حدھغچغھافءاڑاڑہشھدفڈچقڈ،شڈیچڑ

Answer:

it's because of friction

Explanation:

In slanted one the friction will be less so it would be more difficult to climb

Answer: I'm not very good at all. I need help with determine the velocity of an object just before it hits the ground, given the object's mass and the distance it falls. For example, a .060 kg (or 60 g) ball falls from a 12 meter building. Also, assume it is free falling (no air resistance) and the average time it takes to hit the ground it 1.16 seconds.

Explanation:

Answer:

B

Explanation:

Answer:

i think it's either B or C

Explanation:

Answer:

The answer is non-mineral nutrients.

Answer:

non-mineral nutrients.

Explanation:

Answer:

use math.way

Explanation:

i solved it.

The deflection will be toward the backside for bad cost and closer to the pinnacle of the spread page for the high-quality charge.

Charged particles are simply defined as particles with an electric charge on them. This may sometimes result in being an ion, as a molecule with excess or a reduced amount of electrons relative to protons.

Generally, the Magnetic area points internal the airplane so through fleming left hand rule is

F = q ( V x B). Ie

hence

qVB = qE

V= E/B then no deflection.

V > E/B then magnetic pressure is dominant over electric powered force.

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Answer:

19.5

Explanation:

16,17,18,18,21,24,24,29

(fourth (18) and fifth(21)) ÷2=

(18+21)÷2=

39÷2=

19.5 is median

Answer:

a.

Explanation:

i did a quizz for it im sorry if im wrong

Answer:

Graph #2

Explanation:

Notice, on the position time graph, there is a constant rate of change. This means that there is a constant velocity, meaning it's a straight line. The line is decreasing so the velocity will be negative.

Answer:

neutrons protons electrons

Explanation:

after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.

Answer:

a = 2.5 m/s^2

Explanation:

u = 0

v = 25

t = 10

(using first eq. of motion)

a = (v - u) /t

a = (25 - 0) /10

a = 25/10

a = 2.5 m/s^2

The force of attraction between the balloon and the rod is 9 x 10⁻³ N.

The force of attraction between the balloon and the rod is calculated by applying Coulomb's law as shown below;

[tex]F = \frac{kq_1 q_2}{r^2}[/tex]

where;

F = (9 x 10⁹) x (2.0 x 10⁻⁸)²/(0.02)²

F = 9 x 10⁻³ N

Thus, the force of attraction between the balloon and the rod is 9 x 10⁻³ N.

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[tex]\:\:\:\:\:\:\:\:\:\:\star [/tex]We can calculate the force between the balloon and the rod applying "Coulomb's Law". Which states-

[tex]\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{F_{(air)} = K\times \dfrac{ q_1 q_2}{r^2}} \\ [/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\star\:\sf\underline{F_{(air)} = \dfrac{1}{4\pi \epsilon_{0} } \dfrac{q_1q_2}{r^2} }\\ [/tex]

Where-

As per question, we are given that -

Now that required values are given, so we can substitute the values into the formula and solve for Force -

[tex] \:\:\:\:\:\:\:\:\:\:\star\:\sf\underline{F_{(air)} = \dfrac{1}{4\pi \epsilon_{0} } \dfrac{q_1q_2}{r^2} }\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)}= 9\times 10^9 \times \dfrac{-2\times 10^{-8}\times 2\times 10^{-8} }{(0.02)^2}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)}= -9\times 10^9 \times\dfrac{4\times10^{-16}}{4\times 10^{-4}}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)}= -9\times 10^9 \times\dfrac{\cancel{4}\times10^{-16}}{\cancel{4}\times 10^{-4}}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^9 \times 10^{-16} \times 10^{4}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^9 \times 10^{-16+4} \\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^9 \times 10^{-12} \\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^{9-12}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^{-3}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf \underline{Force_{(air)} = -9\times 10^{-3}\:N}\\ [/tex]

Open star clusters tend to orbit within the disk. Globular star clusters contain hundreds of thousands of stars, and some – like Omega Centauri, shown above – contain millions of stars. Open star clusters tend to contain only hundreds of stars.

Answer:

perpendicular

Explanation:

The projectile motion of an object can be described by two independent vector components that are horizontal and vertical motion.

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.

The projectile's motion is divided into two parts: horizontal and vertical motion.

Because perpendicular motion components are independent of one another, these two motion components may be treated individually.

Hence two independent vector components are horizontal and vertical motion.

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Answer: When there is no change in chemical composition of a substance then it is known as physical change. When we say that mass is conserved during a physical change then it means that total mass of the substance or object will remain the same from beginning till the end

Answer:

Sample Response: When a physical change occurs, the mass of the substance is conserved. This means that the total mass of the substance remains the same from beginning to end. The physical properties of the substance, such as size and shape, may change, but the amount of matter in the substance does not change.

Explanation:

Explanation:

Crossing over is the swapping of genetic material that occurs in the germ line. During the formation of egg and sperm cells, also known as meiosis, paired chromosomes from each parent align so that similar DNA sequences from the paired chromosomes cross over one another.

Answer:

Explanation:

After cross-over, the chromosomes are randomly shuffled to the daughter cells as the cell divides. This process results in in gametes that with unique combinations of chromosomes and is called independent assortment.

Answer:

explain the meaning of the term work done and give an example of where work is done

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))| = [tex]e^{(-40/75)[/tex]

T - 25 = ± 75 *  [tex]e^{(-40/75)[/tex]

T = 25 ± 75 *  [tex]e^{(-40/75)[/tex]

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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Given parameters:

Volume of metal block = 0.035mL

Mass of block  = 765mg

Unknown:

Volume of metal block = ?

Solution:

Density is the mass per unit volume of a substance.

  Density  = [tex]\frac{mass}{volume}[/tex]

Mass is the amount of matter in a substance

Volume is space occupied by a substance

     Mass of the block in g;

                         765mg =    765 x 10⁻³ g

                                     = 0.765g

 Density = [tex]\frac{0.765}{0.035}[/tex]   = 21.86g/mL

The density of the metal block is 21.86g/mL