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KIM [24]
3 years ago
9

Objects the exhibit projectile motion follow a ____________ path.

Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0
B. Parabolic

This is because an object that is in projectile motion has gravity acting on it.
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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

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5 0
2 years ago
How does a crowbar work?
mylen [45]
It works by you putting leverage on one side makes more force go to the other side so if you put a crowbar in between a door and you push on one side the other will push the opposite side with more force<span />
4 0
3 years ago
4. Derive<br>the relation, P= hd g​
Sergeu [11.5K]

p=F/A

or,P=d×V×G/A (m=d×V)

or,p= d× A×h×g/A (A and A are cut)

or,P=d×H×G

4 0
3 years ago
What is the mass of a dog that weighs 382 N?(unit=kg)
Luba_88 [7]

Answer:

The answer to your question is: mass = 38.93 kg

Explanation:

Data

mass = ?

Weight = 382 N

gravity = 9.81 m/s2

Formula

Weight = mass x gravity

mass = weight / gravity

mass = 382 / 9.81         substitution

mass = 38.93 kg           result

6 0
3 years ago
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
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