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Sav [38]
2 years ago
11

Newton began his academic career in 1667. For how long was he a working scientist? Was he a very productive scientist?

Physics
1 answer:
rjkz [21]2 years ago
7 0
Newton's 3 laws are...

inertia: things tend to continue to do what they are doing.

Change: to make something change you need a force to change it. the force needed = the mass times its acceleration

<span> Resistance: When you push on something, it pushes back. 

From yahoo answers
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An object that is accelerating may be
pickupchik [31]
The answer of this question is D. All of the above

Acceleration happen when an obeject change its velocity. It has nothing to do with speed.

The huge misconception about acceleration is when we thought it only aply if we increase our speed ( in a sport match, sportcaster often describe acceleration as an increase in players speed)

slower, faster, right , left, it does not matter, as long as that object change its velocity, it accelerates
8 0
2 years ago
What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ
borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

\omega = \sqrt{\frac{k}{m}}

now we have

\omega = \sqrt{\frac{k}{1200}}

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

a = \omega^2 A

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

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

here we know that

v = 85 km/h

v = 85 \times\frac{1000}{3600} = 23.61 m/s

now we have

(1200)(23.61^2) = kA^2

A^2 = \frac{6.68 \times 10^5}{k}

now from above equation of acceleration we have

5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}

5.0(9.81) = \sqrt{k}(0.68)

k = 5178.8 N/m

6 0
3 years ago
A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

3 0
3 years ago
Read 2 more answers
A hockey puck wont slide very far on concrete or asphalt, but it will slide for a very long time on ice. Why is this?
Ilya [14]
When hockey players push the puck along the ice it slides causing heat which melts the ice causing the friction against the ice to be less.
7 0
3 years ago
Read 2 more answers
You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m
slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock m=500 gm

diameter of circle d=2.2 m

radius r=\frac{2.2}{2}=1.1 m

At highest Point

mg+N=\frac{mv^2}{r}

At highest Point N=0 because mass is just balanced by centripetal Force

thus mg=\frac{mv^2}{r}

v=\sqrt{gr}

v=\sqrt{9.8\times 1.1}

v=\sqrt{10.78}

v=3.28 m/s

6 0
3 years ago
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