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NISA [10]
2 years ago
15

4. List 3 detailed examples that illustrate Newton's 2nd Law.

Physics
1 answer:
NNADVOKAT [17]2 years ago
3 0

Examples of Newton's Second Law of Motion

Pushing a Car and a Truck. ...

Pushing a Shopping Cart. ...

Two People Walking Together. ...

Hitting a Ball. ...

Rocket Launch. ...

Car Crash. ...

Object thrown from a Height. ...

Karate Player Breaking Slab of Bricks.

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5 0
3 years ago
A. If a T-Rex runs 20 meters in 4 seconds, what is it’s average speed?
mariarad [96]

5m/s

100m

Explanation:

Average speed is sum of distance distance traveled in a given time by a body.

 Average speed= \frac{total distance covered}{time taken}

Distance = 20m

time = 4s

  Average speed = \frac{20}{4} = 5m/s

For the spaceship;

   Distance covered = speed x time

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learn more:

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3 0
3 years ago
Someone please help with these 2
Ann [662]

Answer:

Explanation:

The formula that you are working with is F = m*a

Since mass is one part of the formula if you increase the mass, you are going to increase the force.

The second one is much more difficult to answer because it is basically incomplete. This is one way to interpret it. If you start at a certain speed and increase during a known time period then effectively you are defining acceleration which is "a" in the formula.

Without those modifications, there is no answer.

6 0
2 years ago
After completing a career search, Patrick has decided that he wants to be a paralegal. Which two skills or abilities are importa
Aleks04 [339]
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7 0
3 years ago
Read 2 more answers
A type of light bulb is labeled having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of
SashulF [63]

Answer:

0.2592 \ or \ 25.92\%

Explanation:

The exponential density function is given as

f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t

\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}

To find probability that bulb fails with the first 300hrs, we integrate from o to 300:

P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}|  \ 0\leq t\leq 300

P(0\leq X\leq 300)=-0.7408+1\\=0.2592

Hence probability of bulb failing within 300hrs is 25.92% or 0.2592

3 0
3 years ago
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