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matrenka [14]
2 years ago
14

6. A ball kicked from ground level at an initial velocity of 60 m/s and an angle with ground

Physics
1 answer:
Oksana_A [137]2 years ago
3 0

Answer:

66.5 is the answer

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Place the following list in order of occurrence from oldest to most recent:
Andre45 [30]

Answer:

Explanation:

2.Big bang, 3.contraction of the solar nebula, 5.stellar ignition in our sun, 4.outgassing of earths secondary atmosphere, 1.appearance of the first ocean on earth ,7.evolution of photosynthesis, 6.build- up of oxygen in earths atmosphere.

4 0
3 years ago
Using the equation for decay, calculate the amount left of a radioactive sample amount N0 if the decay constant is 0.00125 secon
makkiz [27]

The amount left of a radioactive sample amount N0 if the decay constant is 0.00125 seconds and the time is 180 seconds is 0.7999 N.

<h3>What is half-life?</h3>

The time it takes for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life T1/2 and the decay constant is given by T1/2 = 0.693/λ.

  1. N=N0e−λt
  2. given λ = 0.00125 seconds
  3. t = 180 seconds
  4. Now putting values.
  5. N=N0e−λt = 0.799
  6. N= 0.7999.

Read more about the radioactive :

brainly.com/question/2320811

#SPJ1

5 0
2 years ago
A mass of 2000 kg. is raised 5.0 m in 10 seconds. What is the potential energy of the mass at this height?
kolbaska11 [484]

Answer:

98,000J

Explanation:

Given parameters :

Mass = 2000kg

Height = 5m

Time = 10s

P. E = mgh

P. E = 2000x 5x 9.8

P.E = 98,000J

5 0
3 years ago
What energy transformation occurs during the combustion of coal in a power plant?
inn [45]
Hello there, the correct answer is:

B.
4 0
3 years ago
Read 2 more answers
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
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