Which term is defined as an investigative process that recognizes that scientific concepts and

While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000

Minchanka [31]

Answer:

a

The hiker (you ) is 200 m below his/her(your) starting point

b

The resultant displacement in the north east direction is

$a = 6562.0 \ m$

The resultant displacement in vertical direction (i.e the altitude change )

$b =6503.1 \ m$

Explanation:

From the question we are told that

The displacement in the morning is $S_{morning} = (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)$

The displacement in the afternoon is $S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)$

Generally the direction west is negative , the direction east is positive

the direction south is negative , the direction north is positive

resultant displacement is mathematically evaluated as

$(2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east$

$(4000\ m\ north) + (2500 \ m ,\ north) = 6500 \ m ,\ north$

$(100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m$

From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

$a = \sqrt{900^2 + 6500^2}$

=> $a = 6562.0 \ m$

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

$b = \sqrt{6500^2 +(-200)^2 }$

=> $b =6503.1 \ m$

5 0

3 years ago

You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe

NARA [144]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

$I_{dB}=10\log(\dfrac{I}{I_{0}})$

For one amplifier,

$I_{1}=10\log(\dfrac{130}{I_{0}})$ ...(I)

For other amplifier,

$I_{2}=10\log(\dfrac{200}{I_{0}})$ ...(II)

For difference in dB levels

$I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})$

$I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})$

$I_{2}-I_{1}=10\log(\dfrac{200}{130})$

$I_{2}-I_{1}=1.870\ dB$

Hence, The sound level will be 1.870 dB louder.

7 0

2 years ago

An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (a

polet [3.4K]

Answer:

the magnitude of Vpg = 493.711 km/h

Explanation:

given data

speed Vpg = 560 km/h

speed Vwg = 80 km/h

solution

we get here magnitude of the plane velocity w.r.t. ground is

we know that the Vpg = Vpw + Vwg .....................1

writing the component of the velocity that is

Vpw = (0 km/h î - 560 km/h j )

Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)

adding these

Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i

Vpg = (42.025 ) î (-491.92 km/h)j

now we take magnitude

the magnitude of Vpg = $\sqrt{(42.025^2+(-491.92)^2)} km/h$

the magnitude of Vpg = 493.711 km/h

5 0

2 years ago

Who do you think took Ms. C's lunch?

seropon [69]

Answer:

the main character

Explanation:

i think

4 0

2 years ago

A ball is projected into the air. Where is potential energy the greatest? A) A. B) B. C) C. D) A. and C.

Ksju [112]

The point with the greatest potential energy is B.

The potential energy of an object is depends on the relative distance between the object and the ground and its mass. The higher the object is from the ground the greater the potential energy posses.

Potential energy (P.E) = mgh

Where m is the mass, g is the gravity and h is the height from the ground to where the object is.

Since the mass and gravity is constant in this case, only the height will determine the point with the greatest P.E and that point is B.

5 0

2 years ago

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