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AleksandrR [38]
2 years ago
6

Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

her nose. The glasses hit the ground in 2.2 seconds. Neglecting the effects of air resistance, answer the questions below. A. What is the height of the window from the ground? B. What was the impact velocity of the glasses?
Physics
1 answer:
DiKsa [7]2 years ago
6 0

Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

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A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi
Sholpan [36]

Answer:

i = 7.83 \mu A

Explanation:

Induced EMF in the coil is given by the equation

EMF = M\frac{di}{dt}

so we have

M = 31 \mu H

also we know that rate of change in current in solenoid is given as

\frac{di}{dt} = 2.5 A/s

so induced EMF of coil is given as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

now induced current in the coil will be given as

i = \frac{EMF}{R}

i = \frac{77.5 \times 10^{-6}}{9.9}

i = 7.83 \mu A

4 0
1 year ago
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b
Degger [83]

Answer:

The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

Explanation:

Given that,

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We need to find the correct relationships

Using balance equation

In horizontally,

The acceleration is a

T-Fd=ma...(I)

In vertically,

No acceleration

w=L

mg-L=0

Put the value of mg

L-fg=0....(II)

Hence,  The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

3 0
2 years ago
What best describes myotibrils
krok68 [10]
Myofibrils are composed of long proteins such as actin, myosin, and titin, and other proteins that hold them together. These proteins are organized into thin filaments and thick filaments, which repeat along the length of the myofibril in sections called sarcomeres. Muscles contract by sliding the thin (actin) and thick (myosin) filaments along each other.
8 0
1 year ago
Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache
Nina [5.8K]

Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

       R = v₀² sin 2θ / g

How we know the maximum height

      v_{f}² =v_{oy}² - 2 g y

      v_{f}= 0

      v_{oy} = √ 2 g y

      v_{oy} = √ 2 9.8 / 15

      v_{oy} = 1.14 m / s

Let's use trigonometry to find the speed

    sin θ = v_{oy} / vo

    vo = v_{oy} / sin θ

    vo = 1.14 / sin 60

    vo = 1.32 m / s

We calculate the range with the first equation

     R = 1.32² sin(2 60) / 30

    R = 0.0503 m

3 0
1 year ago
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