Which factor affects the amount of runoff that occurs in an area?

land use
the water table
the saturation zone
amount of nutrients in soil

Answer:

2156 N

Explanation:

Data obtained from the question include:

Mass of satellite (m) = 220 Kg

Force (F) of gravity =?

The force of gravity exerted on the satellite on the surface of the earth can be obtained by using the following formula:

Force (F) of gravity = mass (m) × acceleration due to gravity (g)

F = mg

Mass of satellite (m) = 220 Kg

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

Force (F) of gravity =?

F = mg

F = 220 × 9.8

F = 2156 N

Thus, the force of gravity exerted on the satellite on the surface of the earth is 2156 N

Answer:

(a) 3.43 m/s

(b) 3.43 m/s

(c) 95.8 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

where P is pressure (either absolute or gauge), ρ is density, v is velocity, g is acceleration due to gravity, and h is elevation.

(a) Let's choose point 1 at the surface of the fluid in the container, and point 2 at point Z at the exit of the tube.  I'll say 0 elevation is at point Z, and I'll use gauge pressure.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 Pa + ½ ρ (0 m/s)² + ρ (9.8 m/s²) (0.60 m) = 0 Pa + ½ ρ v² + 0

ρ (9.8 m/s²) (0.60 m) = ½ ρ v²

5.88 m²/s² = ½ v²

v = 3.43 m/s

(b) The tube's cross section is constant, so the fluid's speed is the same at all points in the tube. v = 3.43 m/s.

(c) Use Bernoulli equation again, choosing point 2 to be at Y.  I'll say 0 elevation is at the surface of the fluid, and again use gauge pressure.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + 0 + 0 = P + ½ (700 kg/m³) (3.43 m/s)² + (700 kg/m³) (9.8 m/s²) (0.20 m)

0 = P + 4116 Pa + 1372 Pa

P = -5488 Pa

The gauge pressure is -5488 Pa, so the absolute pressure is 101,300 Pa + -5488 Pa = 95812 Pa, or approximately 95.8 kPa.

Answer:

Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.

Answer:

hhhhhhhh

Explanation:

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(c) left will decrease, right will increase

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support.  (Remember that the magnitude of torque is force times the perpendicular distance.  Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned  as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)  

                                  0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

c)

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support,  Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)  

                                  0.6

Right support = 43 kN

NEIL ARMSTRONG WAS THE FIRST MAN WHO TRAVELLED TO THE MOON.

Answer:

On July 20, 1969, Neil Armstrong became the first human to step on the moon. 

Answer:

Explanation:

Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is as seen below

HF ⇄ H⁺ + F⁻

There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ coloumbs.

The required value of effective charge on HF molecule, due to H and F is  1.602 × 10⁻¹⁹ Coulombs.

The given problem is based on the concept of effective charges. The net positive charge carried out by the electrons of atomic species, after forming a polyelectronic atom  is known as Effective charge.

As per the given problem, the Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is given as,

HF ⇄ H⁺ + F⁻

There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ Coulombs.

Thus, we can conclude that the required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.

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Answer:

1- t^3

2- t^2

3- t1

Explanation:

The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:

ac = v²/r

where,

ac = centripetal acceleration

v = speed

r = radius

for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:

1- t^3

2- t^2

3- t1

Answer:

A

Explanation:

hope this helps

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]acceleration = \frac{force}{mass} \\[/tex]

From the question

force = 20 N

mass = 4 kg

We have

[tex]a = \frac{20}{4} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

(a) 0.829 m/s

(b) 3.27 m/s

(c) 0.000153 m²

55.8%

Explanation:

(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)

Q = vA

(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)

v = 0.829 m/s

(b) Use Bernoulli equation.  Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head.  Choose 0 elevation to be at point 1.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)

1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa

v = 3.27 m/s

(c) Flow rate is constant.

Q = vA

(0.001 m³ / 2.00 s) = (3.27 m/s) A

A = 0.000153 m²

Flow rate is proportional to the pressure difference and the radius raised to the fourth power.

Q ∝ ΔP r⁴

Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴

Q₂/Q₁ = (1.120) (0.840)⁴

Q₂/Q₁ = 0.558

The flow decreases to 55.8% of the original value.

Answer:

Explanation:

Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed.  As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.

In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.

The new flow rate = (1.12)*(0.84)^4

=0.5576 or 55.76% of the original flow rate

Answer:

Δt = 1.1 s

Explanation:

Given information:

H= 50.0 m

g= 9.8 m/s²

   [tex]H = \frac{1}{2} * g* t^{2}[/tex]

Solving for t, we get:

t₁= 3.2 s

We can find the final velocity for the object when it hits the ground, using the following expression:

[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]

Solving for vf, we get:

vf = 33.9 m/s

Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:

[tex]v_{f} = v_{o} + g*t[/tex]

(Assuming the downward direction to be positive).

Solving for t, we get:

t₂ = 2.1 s

So the difference in time when both objects hit the ground, it's simply

Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s

Answer:

kenietic and potential i guess

Explanation:

Answer:

The current in the loop is 10.5 A.

Explanation:

Given that,

Radius = 9.4 cm

Magnetic field = 0.7 G

Angle = 70°

We know that,

The magnetic field due to the current in a loop is

[tex]B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]

The magnetic field due to the current is equal to the magnetic field of earth.

[tex]B_{E}=B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]

We need to calculate the current in the loop

Using formula of magnetic field

[tex]B=\dfrac{\mu_{0}NI}{2r}[/tex]

[tex]I=\dfrac{2rB}{\mu_{0}N}[/tex]

Put the value into the formula

[tex]I=\dfrac{2\times9.4\times10^{-2}\times0.7\times10^{-4}}{4\pi\times10^{-7}\times1}[/tex]

[tex]I=10.5\ A[/tex]

Hence, The current in the loop is 10.5 A.

Answer:

500

Explanation:

The net forces should be subtracted, so the number would be 500 N.

Answer:

the net force acting on the piano is -500N.

Explanation:

[tex]F=F_{g} -F_{t} \\F=5,500-6,000\\F=-500N[/tex]

An external agent that can change, shape, size, position, and direction is  called force.

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Answer:the answer is 3

Explanation:

Answer:

did you tried putting it in standard form

Answer:

It may help to round all of the given numbers up to at least 1 or 2 decimal points or you could round up the number you get to 1 or 2 decimal places. For example, for this question round up your answer to 1.9 or 1.94

Explanation:

hope this helps!!

Answer:

potential energy

Explanation:

Answer:

i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack

ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack

iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track

Explanation:

The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack

The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack

The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track

Answer: the wavelength of this transition is 1.2039 um

Explanation:

Given that;

the energy level between the transitioning energy gap Eg = 1.0 + 0.03 = 1.03 eV

we know that λ = 1.24 / Eg

so we substitute our Eg into the above equation

λ = 1.24 / 1.03

λ = 1.2039 um

therefore the wavelength of this transition is 1.2039 um  

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

Given;

The initial temperature of the coffee T₀ = 95 °C

The temperature of the room = 21°C

Let T be the temperature at time of cooling t in mins

According to Newton's law of cooling;

[tex]\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}} = \int\limits kdt\\\\Log(T-21) =kt + Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453[/tex]

When the temperature is 50 °C, the time t in min is calculated as;

[tex]T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins[/tex]

Therefore, when the temperature of the coffee is 50 °C, the time will be 20.68 mins

In 10 minutes the hot coffee will attain the temperature of 50 degrees Celsius.

Initially the hot cup of coffee at the temperature of 95 degrees Celsius but after 5 minutes its temperature decreases from 95 to 85 degrees Celsius which is 15 degrees Celsius decrease so in other 5 minutes, the temperature decreases to 65 degrees Celsius.

Again after 5 minutes the temperature will further decrease finally the cup of coffee attain the temperature of 50 degrees Celsius so we can conclude that in 10 minutes the hot coffee will gain the 50 degrees Celsius temperature.

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Answer:

The acceleration of the car is 10.16m/s²

Explanation:

Given parameters:

  Initial velocity = 8.77m/s

   Final velocity = 47.8m/s

   Time duration  = 3.84s

Unknown:

Acceleration of the car = ?

Solution:

To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;

     Acceleration  = [tex]\frac{V - U}{T}[/tex]

V is the final velocity

U is the initial velocity

T is the time taken

  Input the parameters and solve for acceleration;

      Acceleration  = [tex]\frac{47.8 - 8.77}{3.84}[/tex]   = 10.16m/s²

The acceleration of the car is 10.16m/s²

Answer:

Explanation:

Let the velocity be v .

vertical component of the velocity = v sin 60 = √3 v /2

it reaches maximum height of 10 m .

v² = 2 gh

( √3 v/2 )² = 2 x 9.8 x 10

3 v² = 196 x 4

v² = 65.33 x 4

v = 16.2  m /s

Let time taken to reach height of 10 m

v = u - gt

v sin 60 = 9.8 t

16.2 x √3  /2 = 9.8 t

t = 1.43 s

horizontal distance covered = v cos 60 x t

16.2 x .5 x 1.43 = 11 .5 m

Explanation:

Let east = E, and, west = opposite to east = - E.

Here, displacement:

=> 2m east + 4m west + 1m east

=> 2E + 4(-E) + 1E

=> 2E - 4E + 1E

=> - 1E

=> 1(-E)

=> 1m west

And, distance,

=> 2m + 4m + 1m = 7m

The distance of a person is 7 m and the displacement of the person is 1m west.

To find the distance and displacement, the given values are,

A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.

Displacement:

Distance:

Let us consider East = E and west = opposite to east = - E.

Calculating the displacement:

= 2m east + 4m west + 1m east

= 2E + 4(-E) + 1E

= 2E - 4E + 1E

= - 1E

= 1(-E)

= 1m west.

The displacement is 1m west.

Now calculating the distance,

= 2m + 4m + 1m

= 7m

The distance is 7m.

Thus, the displacement and the distance is found as 1 m west and 7m.

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Answer:

Please find the answer in the explanation

Explanation:

Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.

What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.

What happens above the coil?

the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines

Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.

Below the coil?

The needle will move in an opposite direction.