. All synthetic materials ultimately come from

Calculate the potential energy of a body of a mass 2kg held 4 meters above the ground if g=10m/s?

Drupady [299]

Answer:

16000

Explanation:

Mass(m)=2Kg (1kg= 1ooo g then 2 kg=2000 g)

Velocity(v)= 4 meter

Acceleration due to gravity (g)=10m/s

We know that,

P.E= 1/2 mv^2

or, 1/2 × 2ooo × 4^2

or, 1/2×2000 ×16

or, 2000×8

Therefore= 16000

7 0

2 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago

The next four questions refer to the situation below.

Anna11 [10]

Answer:

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}, t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

$v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

$v_{sg1}$ = D / $t_{out}$

D = v_{sg1} t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

$v_{sg 2} = v_{sr} - v_{rg}$

$v_{sg 2}$ = D / $t_{in}$

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

$v_{sg1} t_{out} = v_{sg2} t_{in}$

t_{out} = t_{in}

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

7 0

3 years ago

The six statements below represent Newton's three laws of motion and Kepler's three laws of planetary motion. Match each stateme

mote1985 [20]

Answer:

1. Force = mass x acceleration - Newton

2. A planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times - Kepler

3. For any force, there is an equal and opposite reaction force - Newton .

4. An object moves at constant velocity if there is no net force acting upon it - Newton

5. The orbit of each planet about the Sun is an ellipse with the Sun at one focus - Kepler.

6. More distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2-a3 - Kepler.

Explanation:

The three laws of planetary motion formulated by Johannes Kepler or Kepler's laws of planetary motion:

The first law states that the planets move around the Sun in an elliptical orbit with the Sun at one of the foci.
The second law states that the line segment joining a planet to the Sun sweeps out equal areas in equal time.
The third law states that the square of the orbital period (p) of a planet is directly proportional to the cube of the mean distance (a) from the Sun (or semi-major axis of its orbit) i.e., p² is proportional to a³.

The three laws of motion formulated by Sir Isaac Newton or Newton's laws of motion:

The first law, also known as the law of inertia states that an object at rest or moves at a constant velocity will remain at rest or keep moving at a constant velocity unless it is acted upon by a force.
The second law states that the total force (F) applied on an object is directly related to the acceleration (a) of that object produced by the applied force and the mass (m) of the object, i.e., F = ma (assuming the mass m is constant).
The third law, also known as the law of action and reaction states that when an object exerts a force on another object, then the latter exerts a force equal in magnitude and opposite in direction on the former object i.e., for every action, there is an equal and opposite reaction. The example includes the recoiling of a gun when it fires a bullet forward.

5 0

4 years ago

At high elevations or high latitudes, some of the water that falls on land does not immediately soak in , run off, evaporate or

LUCKY_DIMON [66]

Answer:

The water is stored in ice sheets and as snow

Explanation:

Temperature reduces with an increase in altitudes. The standard laps rate is 6.5°C per 1,000 m gained in elevation

At very high elevations, therefore, the air is usually very cold such that when an elevation of 4,500 meters is reached at the equator, it is possible to observe snowfall and the water remain temporarily stored on the surface of the mountain as ice and snow

7 0

3 years ago