A garden hose shoots water horizontally from the top of a tall building toward the wall of a second building 20 meters away. If the speed with which the water leaves the hose is 5 m/sec, how long does it take the water to reach the second building, and what distance does the water fall in this time?​

Hi there!

To solve, we must begin by calculating the total WORK done on the object.

W = F · d (Force · displacement)

Plug in the given values. Remember to convert km to m:

1 km = 1000 m

0.6 km = 600 m

W = 60 · 600 = 36000 J

Now, we can solve for power:

P = W/t

Convert minutes to seconds:

1 min = 60 sec

20 min = 1200 sec

P = 36000/1200 = 30 W ⇒ Choice D.

His hair got charged due to continuous friction causing them to get attracted to other objects and thus standing up.

Answer:The comb and hair have become charged with opposite charges.

Explanation:When a comb is run through your hair charges pass between your hair and the comb, so the comb becomes charged either positively or negatively, and the hair oppositely charged. When the comb is brought close to paper an opposite charge is induced in the paper, and the opposite charges attract.

All the three concurrent forces acting on a body will not produce equilibrium.

The given parameters:

1. 1 N, 3 N and 5 N

2. 2N, 2N and 2 N

3. 3N, 4N and 5 N

4. 4N, 4N and 5 N

Concurrent forces lie on the same plane and their line of action pass through a common point.

A body under concurrent forces is in equilibrium if the resultant of the forces on the body is zero.

[tex]\Sigma F = 0\\\\F_1 + F_2 + F_3 = 0\\\\F_1 + F_ 2 = - F_3[/tex]

where;

[tex]F_3[/tex] is the equilibrant force

First set of concurrent forces;

[tex]1 \ N \ + \ 3\ N = 4 \ N\\\\F_ 3 = 5 \ N\\\\5 \ N > 4 \ N[/tex]

Second set of concurrent forces;

[tex]2 \ N \ + \ 2 \ N = 4 \ N\\\\F_ 3 = 2 \ N\\\\4 \ N > 2 \ N[/tex]

Third set of concurrent forces;

[tex]3 \ N \ + \ 4 \ N = 7 \ N\\\\F_ 3 = 5 \ N\\\\7 \ N > 5 \ N[/tex]

Fourth set of concurrent forces;

[tex]4 \ N \ + \ 4 \ N = 8 \ N\\\\F_ 3 = 5 \ N\\\\8 \ N > 5 \ N[/tex]

Thus, we can conclude that all the three concurrent forces acting on a body will not produce equilibrium.

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Answer:

Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.

Explanation:

Answer:

There ya go

Explanation:

The correct words to fill the given blanks would be as follows:

- Peak

- Trough

- Equilibrium Line

- Oscillation

- Frequency, Hertz

- Period, seconds

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Answer:

Explanation:

Average velocity in the 24 m interval is 24 / 4 = 6 m/s

Average velocity in the 64 m interval is 64 / 4 = 16 m/s

There is a 4 second interval between the two points where average velocity equals actual velocity

a = Δv/t = (vf - vi) / t = (16 - 6) / 4 = 2.5 m/s²

s = v₀t + ½at²

24 = v₀(4) + ½(2.5)4²

4v₀ = 24 - 20

v₀ = 1 m/s

Not asked for but the velocity at the end of the first segment and beginning of the second segment is 11 m/s and final velocity is 21 m/s

[tex]\\ \sf\Rrightarrow T=F[/tex]

[tex]\\ \sf\Rrightarrow T=ma[/tex]

[tex]\\ \sf\Rrightarrow T=100(2)[/tex]

[tex]\\ \sf\Rrightarrow T=200N[/tex]

Answer:

[tex]16m/s^{2}[/tex]

Explanation:

Please ask if you have more questions!

The net force on the rocket at the given applied force is 52 N.

The given parameters:

The net force on the rocket is calculated by applying Newton's second law of motion as follows;

F(net) = F(up) - F(down)

F(net) = 150 - mg

F(net)  = 150 - (10 x 9.8)

F(net) = 150 - 98

F(net) = 52 N

Thus, the net force on the rocket at the given applied force is 52 N.

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Explanation:

m = 70 kg

s = 3m

t =2s

g = 10 m/s²

(a)How much work is done by the man?

W = Fs

= mg X s

= 70 x 3 x 10

= 210 x 10

= 2100 Joule

(b)What is the power exerted by the man?

P = W/t

= 2100/2

P = 1050 Watt

Answer:

Explanation:

a) F = ma

a = F/m

a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s

b) t = v/a

t = 200 / 7.2 x 10⁻⁶

t = 2.8 x 10⁷ s       about 10½ months

c) v² = u² + 2as

s = (v² - u²) / 2a

s = (200² - 0²) / (2( 7.2 x 10⁻⁶))

s = 2.8 x 10⁹ m    nearly 7 times around the earth

And all this assumes NO FRICTION.

Answer:

3 v

Explanation:

Answer:

Change in velocity is the integral of all the differential accelerations acting over the period of acceleration.

[tex]20 = u + \int\limits^4_0 {2t} \, dt[/tex]

20 = u + t²[tex]\left \{ {{4} \atop {0}} \right.[/tex]

20 = u + 4² - 0²

20 = u + 16

u = 4 m/s

Answer:

Explanation:

Conservation of momentum

Initial momentum is zero

3.0(6.0) + 7.0v = 0

v = -18/7 = - 2.57142857...

|v| = 2.6 m/s

The muzzle speed of the bullet before the collision is 415.3 m/s.

The given parameters:

Apply the principle of work-energy theorem to determine the final velocity of the block-bullet system;

[tex]F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;

[tex]m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s[/tex]

Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.

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Answer:

Explanation:

Conservation of momentum

In the x direction

2(15) + 2(-10) = 2(-5) + 2(vBfx)

vBfx = 10 m/s

In the y direction

2(30) + 2(5) = 2(20) + 2(vBfy)

vBfy = 15 m/s

vBf = 10i + 15j

KEi = ½(2)(15² + 30²) + ½(2)(-10² + 5²) = 1250 J

KEf =  ½(2)(-5² + 20²) + ½(2)(10² + 15²) = 750 J

KEf - KEi = 750 - 1250 = -500 J

Answer:

Increasing the number of trials reduces the impact of any one imprecise measurement. … To increase the number of attempts, you can find an average result for the experiment, as well as find and discrepancies as human error if you perform an experiment several times.

Explanation:

hope it helps :)

Answer:

It is because the increase in the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value.

Answer:

Explanation:

Assuming the run out at the bottom of the slope is horizontal

μ = Ff/N = Ff/mg

Ff = μmg

a = Ff/m = μmg/m = μg

v = u + at

v = u + μgt

v = 6.55 - 0.088(9.81)(2)

v = 4.82 m/s

After reaching the bottom of the slope, it will take a speed of 4.8m/s to reach. So, the correct option is (C).

Friction is the force which opposes the relative motion of solid surfaces, fluid layers, and material elements that are sliding against each other. There are many types of friction like

Friction can be expressed as

Ff = μFN

where, Ff is the force of friction (N)

μ is the coefficient of friction.

FN is the normal force

As per the above equation, it is actually the normal force, not the weight, which is responsible for friction.

For given information,

Assuming the bottom of the slope is horizontal

So, μ = Ff/N = Ff/mg

Ff = μmg

a = Ff/m = μmg/m = μg

v = u + at (u- initial speed, v= final speed, a= acceleration and t= time)

v = u + μgt (g= acceleration due to gravity)

u = 6.55 - 0.088(9.81)(2)

u = 4.82 m/s

Thus, after reaching the bottom of the slope, it will take a speed of 4.8m/s to reach. So, the correct option is (C).

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Answer:

1.884 meters per second

Explanation:

s = d/t

d = C = 2πr = 2π(6) = 37.68 m

t = 20 s

s = (37.68 m)/(20 s)

s = 1.884 m/s

The speed of the boy is 1.88 m/s while he is moving around a circular park of radius 6m and taking time 20 seconds.

Speed can be defined as a measurement of how quickly the distance an object traveled changes with time. Speed is a scalar quantity as it has magnitude but no direction as a unit of measurement.

An object that moves quickly and with high speed, traveling a lot of ground in a lesser time.

Given, the radius of the circular park, r = 6m,

The time taken to cover one round of the park, t = 20 s

The total distance traveled by the boy is equal to the circumference of the circular park, so

Circumference of park, C = d = 2πr = 2× 3.14 × 6 =37.68 m

The speed of the boy, S = d/t

S = 37.68 /20

S = 1.88 m/s

Therefore, the speed of the boy is equal to 1.88 m/s.

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The change in potential energy is 6324 Joules while the power is 263.5 Watts.

Potential energy is the amount of energy possessed by a body due to its position relative to another object.

a) Potential energy = weight * height = 510 N * 4 flight * 3.1 m per flight

Potential energy = 6324 Joules

b) Power = Energy / time = 6324 Joules / 22 seconds = 263.5 Watts

The change in potential energy is 6324 Joules while the power is 263.5 Watts.

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Answer:

Agrophobia

Explanation:

an anxiety disorder in which someone feals anxious and scared in a public place

Yes, it's possible to predict the speeds that a coaster will reach before it’s

ever placed on the track.

This is usually calculated with the potential energy which is

Potential energy = m g h

where m is mass, g is acceleration due to gravity and h is height.

The given formula above is used in predicting the speeds that a coaster will

reach before it’s ever placed on the track.

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Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

The magnitudes of the forces that the ropes must exert on the knot connecting are :

Given data :

Mass ( M ) = 12 kg

∅₂ = 63°

∅₃ = 45°

a) Force exerted by the first rope = weight of rope

∴  F₁ = mg

     = 12 * 9.81 ≈  118 kg

b) Force exerted by the second rope

applying equilibrium condition of force in the vertical direction

F₂ sin∅₂ + F₃ sin∅₃ - mg = 0  ---- ( 1 )

where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 )  applying equilibrium condition of force in the horizontal direction

Back to equation ( 1 )

F₂ =  [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]

   = [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]

   = 89.21 N

C ) Force exerted by the third rope

Applying equation ( 2 )

F₃ = ( F₂ cos∅₂ / cos∅₃ )

    = ( 89.21 * cos 63 / cos 45 )

    = 57.28 N

Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N

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