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3 years ago

How do the forces on a skydiver affect the velocity of the skydiver as they fall? Hellp

2 answers:

8 0

When a skydiver falls, he constantly accelerates. Therby, falling faster and faster. But also he is slowed down a bit due to the force of air resistance.

solmaris [256]3 years ago

3 0

Answer:

Explanation:

As a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity.

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HELP PLS MARKING BRANLIST 100 pts TAKING TEST RN

AlladinOne [14]

Answer:

15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2

Explanation:

8 0

3 years ago

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

Explanation:

(a) Expression for change in temperature is as follows.

$|\Delta T| = |x - y|K$

= 15.1 K

= $|(x - 273.15) - (y - 273.15)|^{o}C$

= $|x - y|^{o}C$

= $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) Change in temperature from Celsius to Fahrenheit is as follows.

F = 1.8C + 32

C = $\frac{F - 32}{1.8}$

Since, K = C + 273

or, $\Delta K = \Delta C = \frac{\Delta F}{1.8}$

$\Delta F = 1.8 \Delta K$

= 1.8 (15.1)

= $27.18^{o}F$

or, = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

7 0

3 years ago

Read 2 more answers

34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf

ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

$d_1 = \frac{1}{2}(0.6) t^2$

also the distance moved by passenger

$d_2 = \frac{1}{2}(1.2) t^2$

so we will have

$d_1 + 5 = d_2$

$0.3 t^2 + 5 = 0.6 t^2$

$0.3 t^2 = 5$

t = 4.08 s

3 0

3 years ago

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago

In this diagram, medium A represents water and medium B represents air. The phenomenon in this diagram, moving from A to B, is c

Readme [11.4K]

Its C) refraction
hope its may help you

6 0

3 years ago

Read 2 more answers