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ElenaW
14 days ago
5

While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

, the hairdryer moves to the right and comes to rest when the cordmakes an angle of 5​ degrees​ with the vertical, as shown below. In a different experiment, you determine that the same hairdryer is pushing 0.06 m​3​ of air through itself every two seconds. The mass ofthe hairdryer is 420 g. Determine the speed of the air leaving the hairdryer, v ​air​. Assume that themass of 1 m​3​ air is 1.2 kg and that the hairdryer is blowing air perpendicular to the wire.

Physics
1 answer:
Yuliya22 [1.1K]14 days ago
6 0

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

(1).\: tan (5^o) = \dfrac{F_t}{Mg}

by substituting M =0.420kg, g = 9.8m/s^2 into the equation and resolving for F_t we find:

F_t = Mg\:tan(5^o)

F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)

\boxed{F_t = 0.3601N.}

This thrust is linked to the speed of air ejection v through the equation

(2).\: F_t = v\dfrac{dM}{dt}

where dM/dt signifies the rate of air ejection, which is known to be

0.06m^3/2s  = 0.03m^3/s

and since 1m^3 = 1.2kg,

0.03m^3/s \rightarrow 0.036kg/s

\dfrac{dM}{dt}  = 0.036kg/s,

by inserting these values into equation (2), we obtain the value of F_t as:

0.3601N = 0.036v

resulting in

v= \dfrac{0.3601}{0.036}

\boxed{v =10m/s.}

which indicates the air velocity discharged from the hairdryer.

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Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

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Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
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Answer:

2.32\times 10^{-11}

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The second number is 5.2\times 10^{-4}.

We must multiply these two numbers together.

4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}

In scientific notation: 2.32\times 10^{-11}

Therefore, this is the solution you are looking for.

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Answer:

C) True. The distance S increases over time, with v₁ = gt and v₂ = g (t-t₀), illustrating that v₁> v₂ for the same t.

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Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

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Stone 2

Released shortly thereafter, let's assume a delay of one second, we can utilize the same timing mechanism

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