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2 years ago

how do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate

1 answer:

3 0

While skydiving, its not just freely falling under Earth's gravity. Additional force called drag acts against the gravity which slows down the rate of fall. Drag is caused by the air molecules which pushes against the body as it falls through them. This is actually a significant amount of force which slows down the rate of fall of the body. Drag depends on the contact surface area and weight. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position because of the less contact surface area of the body with the air molecules while in the former case. No two persons have identical body shape and weight. Hence, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

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The y-component of the force F which a person exerts on the handle of the box wrench is known to be 86 lb. Determine the x-compo

emmainna [20.7K]

Answer:

x-component of force is 38.18 lb where as magnitude of Force is 93.16

Explanation:

Fy of the force F exerted on the handle of the box wrench = 86 lb

Considering the triangle in Fig 1

magnitude of perpendicular = P = 12

magnitude of base = B = 5

using Pythagoras theorem

$H= \sqrt{P^{2} + B^{2}}$

$H= \sqrt{12^{2} + 5^{2}}$

$H=13\\\implies cos \theta = \frac{5}{13} \\\implies sin \theta = \frac{12}{13}\\$

y-component of force is given given as:

$86 = Fsin\theta\\F = 86 (\frac{13}{12})\\F = 93.16 lb\\F_{x} =F cos \theta\\F_{x} = 93.16 (\frac{5}{13})\\F_{x} =38.18 lb.$

5 0

3 years ago

Change in momentum is equal to force times the amount of time the force is applied? True or False?

harina [27]

The answer to this question is false

7 0

2 years ago

A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another

Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t² ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s² (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t² ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

t = 1.8 sec

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

s₂ = 28.8 m

using this value of of t in equation (1)

s₁ = (5)(1.8)²

s₁ = 16.2 m

Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

6 0

3 years ago

Name the quantity which is measured by the area occupied below the velocity time graph

kiruha [24]

If you mark off a beginning time and ending time on the graph,
then the area under the part of the graph between those limits
is the distance covered during that period of time.

3 0

3 years ago

Read 2 more answers

Use appropriate units and significant figures. USE THE LAW OF COSINES AND LAW OF SINES.

N76 [4]

Answer:

The resultant velocity is 86.1 mi/h.

Explanation:

The law of cosines is given by:

$c^{2} = a^{2} + b^{2} - 2abcos(\theta)$