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Olegator [25]
3 years ago
6

A skydiver reaches terminal velocity. Then he opens his parachute.

Physics
2 answers:
schepotkina [342]3 years ago
4 0

Answer:

it is 0.9999.10896

Explanation:

MatroZZZ [7]3 years ago
4 0

Answer:

Explanation:

Remark

He's pulled upwards by the parachute. The formula is given below. Details are found   searching for terminal velocity.

V_t=\sqrt\frac{2mg}{\rho A C_d}

Whatever that calculation turns out to be, a person has to be going a whole lot slower before he can safely hit the ground. The parachute is designed so that the person will be slowed very quickly to a safe velocity so when he hits the ground the impulse will not be as great as it could be. The force that the parachute delivers acts upward and opposes the force of gravity which is a downward force.

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Charges q, q, and – q are placed on the x-axis at x = 0, x = 4 m, and x = 6 m, respectively. At which of the following points do
MrRissso [65]

Answer:

can you show a graph but if not i believe the answer is x=6m

Explanation:

7 0
3 years ago
A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant vel
omeli [17]

Answer:

Explanation:

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate .

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration . It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

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3 0
3 years ago
Suppose you want to make a nested function call (i.e. a call to a function from inside of another function) using a jal rather t
WARRIOR [948]

Answer:

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8 0
4 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

6 0
3 years ago
Photoelectrons with a maximum speed of 8.00 • 106 m/sec are ejected froma surface in the presence of light with a frequency of 6
Andru [333]

The kinetic energy is given by:

K=\frac{1}{2}mv^2

We know the mass and the maximum speed, plugging their values in the expression above we have:

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Therefore, the answer is d.

5 0
1 year ago
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