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?​

Answers

Answer 1

Answer:

Explanation:

If air resistance is ignored, the water travels for a time of

t = d/v = 20 m / 5 m/s = 4 s

falling from vertical rest, the water strikes the wall a distance below the hose

d = ½gt² = ½(10)4² = 80 m

Yeah, I think ignoring air resistance is wishful thinking considering the time and distances involved.


Related Questions

4. If a charged body is negatively charged, it has excess of _______

Answers

Answer:

Electrons

Explanation:

Electrons are negatively charged sub-atomic particles, therefore when a body's negatively charged, it means that there's more electrons than protons.

5. Layer of Earth consisting of crust & upper layer of mantle ________

Answers

Answer:

lithosphere

Explanation:

hope this helps you!!

(c) It is suggested that one side of the copper sheet cools to a lower temperature than the
other side.
Explain why this does not happen.
[2]

Answers

Answer:

Explanation:

The word "sheet" implies that the copper is quite thin.

Copper is also a very good conductor of heat.

Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.

A 2.2 kg model rocket is shot straight up in the air from the ground, with an initial velocity of 36.4 m/s. The rocket reaches its maximum height, and falls back to the ground. What is the maximum height of the rocket? Round your answer to 2 decimal places.

Answers

Answer:

Explanation:

Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.

PE = KE

mgh = ½mv²

    h = v²/2g

    h = 36.4²/ (2(9.81))

    h = 67.53109...

    h = 67.53 m

In the following free body diagram, what is the net force on the object?
A.10n
B.5 N to the right
C.20 N to the right
D.7 N to the right

Answers

Answer:

B

Explanation:

Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.

ΣF = 7 + 18 + (-20) = 5N to the right

A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:

a- 18.3
b- 12.0
c- 11.1
d- 14.4

Answers

Answer:

D.

correct me if im wrong

brainlest plsss<333

You are pulling a sled with a constant velocity using a rope held horizontally in
the snow. Using a spring scale attached to the rope, you measure the tension
to be 0.96 N. The sled's mass is 0.58 kg. Write the surface force the ground
exerts on the sled in component form, and determine its magnitude.

Answers

Answer:

Explanation:

if you are pulling in the positive direction and Upward is also positive.

F = -0.96i + 0.58(9.8)j

F = -0.96i + 5.7j

F = √(-0.96² + 5.7²) = 5.8 N

You are pulling a sled with constant velocity by using a rope then the surface force the ground exerts on the sled will be 5.8 N.

What is tension?

The pulling force conveyed axially by a string, cable, chain, or another analogous object, or by either end of a rod, truss member, or another such three-dimensional object, is referred to as tension in physics. Another way to think of tension is as the action-reaction pair of forces acting at each end of the previous elements. Perhaps tension is the polar opposite of compression.

A restoring force may cause what is also referred to as tension when atoms or molecules are torn apart from one another at the atomic level and build up potential energy.

If you are moving upward and pushing in a positive way,

F = -0.96 i + 0.58(9.8) j

F = -0.96 i + 5.7 j

F = [tex]\sqrt{(-0.96^2 + 5.7^2)}[/tex]

F= 5.8 N

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Someone help me please !!!! Will mark Brianliest !!!!!!!!!!!!!!!!

Answers

Strain it , the sand didn’t dissolve in the solution it just settled in the bottom

Answer:

Decant it.

Explanation:

Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.

Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.

Explain how the removal of heat energy affects the speed of the particles in a substance

Answers

Answer:

The removal of heat energy slows the speed of particles

Explanation:

When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.



Raghu studies in grade 6th. He wants a cricket bat to be made by a carpenter. He tells the
that the length of the bat should be 7 hand spans. The tall carpenter tells Raghu that it
will be ready by tomorrow. When Raghu went to collect the bat the next day, he was very
disappointed. Why? Was the bat longer or shorter than what Raghu expected? Give reason.
carpenter

Answers

Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.

From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.

Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.

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an object is traveling with a constant velocity of 5m/s. How far will it have gone after 7s

Answers

Answer:

35m/s

Explanation:

Answer:

35m/s

Explanation:

Simply multiply 5 and 7.

A guitar string 63.6 cm long vibrates with a standing wave that has five antinodes. Which harmonic is this

Answers

Answer:

fifth harmonic

Explanation:

Suppose you want to make the demonstration more dramatic by attracting as much hair as possible with the balloon. Which of the following is the best choice to accomplish your goal? A. Decrease the electric force by holding the balloon closer to the hair. B. Increase the electric force by rubbing the balloon for a longer period of time. C. Decrease the electric force by using a larger balloon. D. Increase the electric force by holding the balloon farther away from the hair.

Answers

Answer:

B. Increase the electric force by rubbing the balloon for a longer period of time.

Explanation:

i need help with the problem below

Answers

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.

When a 1 is on the input of an inverter, what is the output?​

Answers

Answer:

When the input to an inverter is high (1) the output is low (0); and when the input is low, the output is high.

explain how water erosion changes land forms

Answers

Answer:

Water erosion can slowly wash away dirt and rocks, wearing away landforms and sometimes forming rivers that can even carve out stone over time.

plz help ASAP I'll mark as brainliest ​

Answers

Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

objects want to ______ ___________ doing what they're __________ ____________ because they are "lazy." This is called __________.

Answers

Answer:

Explanation:

Objects want to continue doing what they're already doing because they are "lazy." This is called inertia.

stuck with this one. ​

Answers

Y is the right answer :)

Which is the independent variable in the graph of the U.S. population 1800-1900?

1. United States
2. straight lines
3. years
4. population

Answers

Answer:

3. years

Explanation:

The independent variable is the x axis of graph

In what way does Isaac represent us?

Answers

Answer:

I believe The Knowledge that we apply everyday or based on these methods and discoveries may represent us

Explanation:


What effect does the time taken to lift the mass have on power output?

Answers

The longer the time taken, the lower the power output.

This is since power is calculated through

Energy transferred / time

Increasing the denominator will lead to a lower value overall

A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angular velocity ω of the wheel at the bottom of the incline?
Express your answer in radians per second.

Answers

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]

[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]

[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]

[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]

[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]

[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]

[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

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The angular velocity of the wheel depends on the mass, radius and the

mode of rotation of the wheel (with or without slipping).

The angle velocity at the bottom of the incline, ω ≈ 4.43 rad/sec

Reasons:

The given parameters are;

Radius of the wheel, r = 2.0 m

Height of the incline, h = 8.0 m

Required:

Angular velocity of the wheel at the bottom of the incline.

Solution:

The potential energy of the wheel at the top of the hill, P.E. = m·g·h

[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]

Where;

v = The translational velocity of the wheel = ω·r

I = The moment of inertia of the wheel = m·r²

Therefore'

[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]

At the bottom of the hill, the potential energy is converted to kinetic energy

Therefore;

P.E. = Sum of K.E.

m·g·h = m·r²·ω²

g·h = r²·ω²

[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]

The angular velocity of the of the wheel at the bottom of the incline, ω ≈ 4.43 rad/sec

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is xenon a pure substance​

Answers

[tex]\large\huge\green{\sf{Yes}}[/tex]

Using the formula below, calculate the kinetic energy of the 6 gram stone going 10 Mph
going 10 mph
KE = 1/2 Mass x Speed?
=

Answers

Answer:

Explanation:

A wonderful (NOT) mixed unit problem

convert mph to m/s

v = 10 mi/hr(5280 ft/mi)(12 in/ft)(2.54 cm/in) / (100 cm/m) / (3600 s/hr)

v = 4.4704 m/s

KE = ½(0.006)4.4704²

KE = 0.05995342848 J

KE = 0.06 J

25.Figure 22.22 shows a plot
of potential versus position
along the x-axis. Make a plot
of the x-component of the
electric field for this situation.

Answers

The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:

In the attachment we see the graph of the electric field as a function of distance.

Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.

          dV = - E . ds

          E = [tex]- \frac{dV}{ds} \ \hat s[/tex]  

Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.

Let's apply this expression for each section of the given graph:

1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.

  The derivative of a constant is zero.

        E = 0

2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.

We look for the equation of the line.

       V-V₀ = m (x- x₀)

We carry out the derivative.

      E = - m i ^

The slope (m) is:

       [tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]  

Let's calculate.

       [tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]  

Let's substitute.

       E =  [tex]2 \hat i \ V/m[/tex]  

         

3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.

We look for the equation of the line and we derive.

      E = - m i ^

Let's  substitute.

      [tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]  

    E = - 4 [tex]\hat i[/tex] V / m

4) From x₀ = 4.5 m to x_f = 6m.  The potential is constant and the derivative of a constant is zero.

      E = 0

5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]  

      E = - 0.5 [tex]\hat i[/tex] V/m

6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]

Let's substitute.

       E = 2 [tex]\hat i[/tex] V/m

7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V

     

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]

Let's substitute.

       E = 1 [tex]\hat i[/tex]  V/m

In the attachment we can see these Electric fields as a function of distance.

In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:

In the attachment we see the graph of the electric field as a function of distance.

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What happens to the gravitational force between two objects if the distance between them triples?

A. The force increases by a factor of 9

B. The force decreases by a factor of 9

C. The force decreases by a factor of 3

D. The force increases by a factor of 3

Answers

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. ... If the separation distance between any two objects is tripled (increased by a factor of 3), then the force of gravitational attraction is decreased by a factor of 9 (3 raised to the second power).

So the answer you are looking for is B.
The answer will be B.

Gravity obeys something called the inverse square law.

This means if distance increases by a factor of x, the force of gravity will decrease by a factor of x^2

For example, if distance increases by 2, the force would be 4 times weaker

If Distance increases by a factor of 3, the force will be 9 times weaker

HELP ME the mean free path λ and the mean collision time τ of the molecules of a diatomic gas of molecular mass 6.00 × 10⁻²⁵ kg and radius r = 1.0 x 10⁻¹⁰ m are measured. From these microscopic data can we obtain macroscopic properties such as temperature T and pressure P? If so, consider λ = 4.32 x 10⁻⁸ m and τ = 3.00 x 10⁻¹⁰ s and calculate T and P.

Answers

The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.

The given parameters;

Mass of the gas molecules, m = 6 x 10⁻²⁵ kgRadius of the gas, r = 1 x 10⁻¹⁰ mMean free path, [tex]\lambda_{rms}[/tex] = 4.32 x 10⁻⁸ mMean collision time, [tex]\tau = 3 \times 10^{-10} \ s[/tex]

The mean velocity of the gas molecules is calculated as follows;

[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]

The temperature  of the gas molecules is calculated as follows;

[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]

where;

k is Boltzmann constant

[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]

The number of gas molecules per unit volume is calculated as follows;

[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]

The pressure of the gas molecule is calculated as follows;

[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]

Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.

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What is the force exerted on a selectively permeable membrane because water has moved from an area of higher concentration to an area of lower concentration of water?
Diffusion
Facilitated transport
Osmotic pressure
Endocytosis

Answers

Question

what is the force exerted on a selectively permeable membrane beacuse water has moved from an area of higher concentration to an area of Lower concentration of water?

Answer

OSMOSIS

Explanation

Osmosis: This is the hydrostatic force acting to equalize the concentration of water on both sides of the membrane that is impermeable to substances dissolved in that water. Water will move along its concentration gradient.

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What is the half-life of the imaginary element Lokium? Show your work

Answers

Answer:What is the half life of the element Lokium?

The half-live of the element Lokium is 4.

Explanation:

sorry if im wrong, have good day

Answer:

The half-life of the element Lokium is 4

What is half-life?

Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

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