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Julli [10]
3 years ago
10

A ballon is losing altitude at a constant rate of 25 kilometers per day. if a landing is defined as an impact with a velocity of

less than 1 meter per second and a crash is greater than 1 meter per second, determine whether or not the ballon will land or crash based on how fast it will land
Physics
1 answer:
zubka84 [21]3 years ago
5 0

(25 km/day) x (1000 m/km) x (day/86,400sec) = 0.29 meter/sec. That's a gentle, graceful landing.

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if a 10kg object is lifted a distance of 10m from the rest what was the final velocity of the object?
Anastasy [175]
I think it will be 100 kg
5 0
3 years ago
Assuming things about someone based on your experiences with similar people you have encountered is called
AysviL [449]

Answer:

presumptuous

Explanation:

it's what you call someone who assumes something

3 0
3 years ago
problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi
Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
3 years ago
An applied force varies with position according to F = k1 x n − k2, where n = 3, k1 = 3.6 N/m3 , and k2 = 76 N. How much work is
aniked [119]

Answer:

The work done is 205 kJ.

Explanation:

Hi there!

Work can be calculated using the following equation:

W = F · Δx

Where:

W = work

F = applied force

Δx = displacement

In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:

W = ∫ F · dx  

F = 3.6 N/m³ · x³ - 76 N

W = ∫ (3.6 x³ - 76)dx

W = 0.9 x⁴ - 76x

Evaluating from xi to xf:

W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m

W = 205 kJ

8 0
3 years ago
A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling
Setler [38]
  • m1=1500kg
  • m_2=3000kg
  • v_1=5m/s
  • v_2=7m/s

Using law of conservation of momentum

\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3

\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3

\\ \sf\Rrightarrow 7500-21000=4500v_3

\\ \sf\Rrightarrow -13500=4500v_3

\\ \sf\Rrightarrow v_3=-3m/s

6 0
2 years ago
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