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Maslowich
3 years ago
9

Kyle is flying a helicopter and is rising at 5.0m/s when he releases the bag. After 2.0s

Physics
2 answers:
ELEN [110]3 years ago
8 0
A. What is the bag’s velocity?
Answer: 25 m/s 

b. How far has the bag fallen?
Answer: -30 m

c. How far below the helicopter is the bag?
Answer: 20 m below the helicopter

ch4aika [34]3 years ago
5 0
1) v=u+gt = -5 +10*2=15 m/s
2) r=ut+gt^2/2=-5*2+10*2^2/2=-10+20=10m
3) r of heli= vt =5*2=10m
So 10m by bag from answer number 2 plus additional 10m by helicopter equals 20m
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Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
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As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

Q = heat given to the system

\Delta U = U_f - U_i

W = work done by the system

now here we can say

\Delta U = 180 - 78 = 102 J

W = 64 J

now we can say that heat will be given as

Q = 64 + 102 = 166 J

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

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Which rain forest is located mostly on islands
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The tropical rain forest is located mostly on Islands. Mainly found in Africa.
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Why does current flow through a circuit?
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Answer:

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A stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s. What is the frequency perceived by
inn [45]

For a stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s, the frequency perceived  is mathematically given as'

F=81.721Hz

<h3>What is the frequency perceived by a firefighter racing toward the station at 11km/h?</h3>

Generally, the equation for the doppler effect  is mathematically given as

F'=\frac{vs+v}{vs}*f

Therefore

F=81(343+3.05556)/343

F=81.721Hz

In conclusion, the frequency is

F=81.721Hz

Read more about frequency

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4 0
2 years ago
A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

3 0
3 years ago
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