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Blizzard [7]
3 years ago
10

1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?

Physics
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

The answer is 49m/s.

Explanation:

Given,

acceleration (a) =9.8m/s^2

initial velocity (u)=0 (as it was in rest condition in helicopter)

time (t) =5 s

now, final velocity (v) after 5s =?

we have,

a =  \frac{v - u}{t}

or \: 9.8 =  \frac{v - 0}{5}

or, v = 49m/s

Therefore, the velocity after 5s is 49m/s.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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At the bottom of a large cylindrical tank filled with fresh water the gauge pressure is 11.6 psi. What is the height (in feet) o
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Answer:

8.19m

Explanation:

Parameters given:

Pressure, P= 11.6 psi = 79979.185 Pa

Gauge pressure is given as:

P = h*d*g

=> h = P/(d*g)

Where

h = height of tank

d = density

g = acceleration due to gravity

Density of water = 997 kg/m³

Therefore, the height of the tank is:

h = 79979.185/(997 * 9.8)

h = 8.19m

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3 years ago
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
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Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
A NATO base in northern Norway is warmed with a heat pump that uses 7.0 °C ocean water as the cold reservoir. Heat extracted fro
tatiyna

Answer:

3.33, 4.84

Explanation:

A) Actual coefficient of performance can be calculated by the formula stated below

Actual COP = heat delivered/ work required

Actual COP = 2,000/600

Actual COP = 3.33

B) Th = High temperature = 80°C + 273 = 353K

Tl = Low temperature = 7°C + 273 = 280K

The theoretical maximum coefficient of performance can ve calculated by the formula

Theoretical COP = Th/(Th-Tl)

Theoretical COP = 353/(353-280)

Theoretical COP = 353/73

Theoretical COP = 4.84

3 0
4 years ago
What is a good definition of acceleration
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Rate of change of velocity is acceleration
8 0
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In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi
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components of the speed of the coin is given as

v_x = v cos60

v_x = 6.4 cos60 = 3.2 m/s

v_y = vsin60

v_y = 6.4 sin60 = 5.54 m/s

now the time taken by the coin to reach the plate is given by

t = \frac{\delta x}{v_x}

t = \frac{2.1}{3.2}

t = 0.656 s

now in order to find the height

h = vy * t + \frac{1}{2} at^2

h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2

h = 1.52 m

so it is placed at 1.52 m height

3 0
3 years ago
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