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Lemur [1.5K]
2 years ago
7

Find the scaler product of this two vector

Physics
1 answer:
OLga [1]2 years ago
5 0

Answer:

The scalar product of a and b is: a · b = |a||b| cosθ

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With what speed must you approach a source of sound to observe a 25% change in frequency?
insens350 [35]
Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
115 = 100 + 100v/343
15 = 100v/343
v = 15*343/100
<span> v = 51,45 m/s </span>
5 0
2 years ago
A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a
nevsk [136]

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

7 0
3 years ago
The convection currents in the asthenosphere cause the movement of the
MA_775_DIABLO [31]
Convection currents generated within the asthenosphere push magma upward through volcanic vents and spreading centres to create new crust. Convection currents also stress the lithosphere above, and the cracking that often results manifests as earthquakes.
7 0
3 years ago
Help please!!!
exis [7]

Answer:

5 years worth of work (aka all of the homework i currently have)

3 0
3 years ago
A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F
vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

t = \frac{d}{v_{net}}

here net speed of the plane will be given as

v_{net} = v + v_w

t = \frac{1000}{205 + v_w}

similarly when it moves under the condition of headwind its net speed is given as

v_{net} = v - v_w

now time taken to cover the distance is 2 hours more

t + 2 = \frac{1000}{205 - v_w}

now solving two equations

\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}

solving above for v_w we got

v_w = 40.4 mph

6 0
3 years ago
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