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lana66690 [7]
3 years ago
13

A skydiver jumps from an airplane traveling 120. km/hr at a height of 1000. m. If his parachute must open 500. m above the groun

d, how long does he have for the chute to open?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

The parachute to open in 15 sec.

Explanation:

Given that,

Velocity = 120 km/hr

Height = 1000 m

If his parachute must open 500. m above the ground.

We need to calculate the time for parachute to open

Using formula of time

v=\dfrac{d}{t}

t=\dfrac{d}{v}

Where, d = distance

t = time

v = velocity

Put the value into the formula

t=\dfrac{500}{120\times\dfrac{5}{18}}

t=15\ sec

Hence, The parachute will be open in 15 sec.

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Grace pushes her foot against the wall with a force of 35 N while she stands
Airida [17]

Answer:

The correct answer is B. 0.64 m/s²

Explanation:

According to the Newton's Second law of motion acceleration of an object by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force.

Mathematically,

                           F ∝ a

                           F = ma

Given data:

Force = F = 35 N

Mass = m = 55 kg

acceleration = a = ?

                           F = ma

                           a = F/m

                           a = 35/55

                          a = 0.64 m/s²

5 0
3 years ago
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 2
Aleks04 [339]

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

6 0
3 years ago
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