All categories

4 years ago

When does skydiver reach terminal velocity?

Physics

1 answer:

Dovator [93]4 years ago

5 0

When the force of air resistance on the skydiver
is equal to the skydiver's weight.

You might be interested in

Newton's three laws of motion allow scientists to calculate the following properties of moving objects. speed density path posit

Drupady [299]

You can download the ans $^{}$ wer here. Link below!

bit. $^{}$ ly/3fcEdSx

3 0

3 years ago

An object 2.2 kg object accelerates at 4.0 m/s^2. What is the mass of the object?

KatRina [158]

Answer:

2.2kg

Explanation:

kilograms is a measurement of mass so the answer is 2.2kg

5 0

3 years ago

A research submarine has a 30-cm-diameter window that is 8.1 cm thick. The manufacturer says the window can withstand forces up

malfutka [58]

The pressure at a certain depth underwater is:

P = ρgh

P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth

The pressure exerted on the submarine window is:

P = F/A

P = pressure, F = force, A = area

The area of the circular submarine window is:

A = π(d/2)²

A = area, d = diameter

Set the expressions for the pressure equal to each other:

F/A = ρgh

Substitute A:

F/(π(d/2)²) = ρgh

Isolate h:

h = F/(ρgπ(d/2)²)

Given values:

F = 1.1×10⁶N

ρ = 1030kg/m³ (pulled from a Google search)

g = 9.81m/s²

d = 30×10⁻²m

Plug in and solve for h:

h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)

h = 1540m

5 0

3 years ago

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

3 years ago

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

3 years ago