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3 years ago

13

A skydiver falls out of a plane from rest, and experiences no air resistance. Eventually, this skydiver reaches a velocity of 33

m/s. How much time did this take? *

1 answer:

shusha [124]3 years ago

3 0

Answer:

the time taken for the motion is 3.37 s

Explanation:

Given;

initial velocity of the skydiver, u = 0

final velocity of the skydiver, v = 33 m/s

The time taken for the motion is calculated as;

v = u + gt

33 = 0 + 9.8t

33 = 9.8t

t = 33 / 9.8

t = 3.37 s

Therefore, the time taken for the motion is 3.37 s

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According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something

Tasya [4]

Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

The gauge pressure at the mouth is $p_1$

The radius of the column is $r_2 = 4 \ mm = 0.004 \ m$

The speed of the liquid outside the body is $v_2 = 3.1 \ m/s$

The area of the column is $A_2$

The area inside the mouth $A_1 = 10 A_2$

Generally according to continuity equation

$v_1 A_1 = v_2 A_2$

=> $v_ 1 = v_2 * \frac{A_2}{A_1}$

=> $v_ 1 = 3.1 * \frac{1}{10}$

=> $v_ 1 = 0.31 \ m/s$

So

$A_1 = 10A_2$

=> $\pi * r_1^2 = 10(\pi * r_2^2)$

=> $r_1 = 10 * r_2$

substituting values

$r_1 = 10 * 0.004$

$r_1 =0.04 \ m$

Now the height of inside the mouth is $h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m$

Now the height of the column is $h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m$

Generally according to Bernoulli's equation

$p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now $\rho = 1000 \ kg m^{-3}$ which is the density of water

$p_2$ is the gauge pressure of the atmosphere which is zero

So

$p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-$

$[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$

$p_1 = 4051.4 \ Pa$

8 0

3 years ago

Which statement correctly describes these electric field lines?

Otrada [13]

I don’t know, which statement ahh I see white screen lol

4 0

3 years ago

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Lynna [10]

Explanation:

Given Data

Total mass=93.5 kg

Rock mass=0.310 kg

Initially wagon speed=0.540 m/s

rock speed=16.5 m/s

To Find

The speed of the wagon

Solution

As the wagon rolls, momentum is given as

P=mv

where

m is mass

v is speed

put the values

P=93.5kg × 0.540 m/s

P =50.49 kg×m/s

Now we have to find the momentum of rock

momentum of rock = mv

momentum of rock = (0.310kg)×(16.5 m/s)

momentum of rock =5.115 kg×m/s

From the conservation of momentum we can find the wagons momentum So

wagon momentum=50.49 -5.115 = 45.375 kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

Speed of wagon=45.375/(93.5-0.310)

Speed of wagon= 0.487 m/s

Throwing rock backward,

momentum of wagon = 50.49+5.115 = 55.605 kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)

speed of wagon= 0.5967 m/s

7 0

3 years ago

Which of the following is not a synthesis reaction?

Nastasia [14]

C is the awnser because c is single replacement

6 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago