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photoshop1234 [79]
3 years ago
14

According to Newton's Second Law, the force of the club hitting the golf ball will cause it to accelerate. At the moment of impa

ct, according to Newton's Third Law,
A) the club hits the ball with twice as much force as the golf ball pushes back
B) the club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club
C) the ball puts more force on the club than the club on the ball
D) the force of the club on the ball is a balanced force, making the golf ball accelerate
Physics
2 answers:
vazorg [7]3 years ago
8 0

Answer:

Option B

Explanation:

<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>

Here in this case the force of the club hitting the golf ball will be in one direction and the force acting on club due to golf ball will be in opposite direction and magnitude of this force will be same as the magnitude of the force of the club hitting the golf ball

In this case the action will be the force of the club hitting the golf ball and reaction will be the force acting on club due to golf ball

∴ The club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club  

ZanzabumX [31]3 years ago
5 0

Answer:

the club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club

Explanation:

At the moment of impact of the club and the golf ball according to Newton's Third Law, the club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club.

According to newton's third law of motion, actions and reaction are equal and opposite e.g recoil of a gun. The force (bullet) acts in the forward direction while the reaction(gun cocked backwards) acts in the opposite direction. Similarly for the club and the golf ball, the force exerted on the golf ball by the club will create an opposing force due to the effect of the golf ball on the club itself.

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In the case of liquids , Owing to contact forces between the edge of the surface and the vessel, the surface acquires a curvature, and if the liquid rises up at the edges where it meets the vessel, the pressure will be less in the liquid than in the air, for points just below and just above the surface. The vessel exerts an upward force on the liquid. This is simply a matter of looking at the directions of forces acting, knowing that the surface is under tension.

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evalating the soltion

Explanation:

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The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's
Ivan

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity v>0.

The acceleration a is the rate of change of the velocity v. This means that if v is <em>increasing during</em> time, then a must be positive. But if v is <em>decreasing over</em> time, then a will be negative (even though the velocity is positive).

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4 years ago
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If a steel cylindrical specimen is stressed nominally to 53 MPa, what stress level exists at the tip of an elliptical surface fl
ElenaW [278]

Answer:

227.9MPa

Explanation:

Length of the flaws is given by

2b = 5.8microns

b = 2.9 × 10⁻⁶m

The relation between the radius of curvature and length and width of the elliptical flaw

r = \frac{a^2}{b}

a = \sqrt{rb}

Equation for stress at the tip of an elliptical surface flaw

\sigma _t = \sigma(1 + 2\frac{b}{a}  )\\\\\sigma _t = \sigma(1 + 2\frac{b}{\sqrt{rb} })\\\\\sigma _t = \sigma(1 + 2\frac{\sqrt{b} }{\sqrt{r} })\\\\\sigma _t = \sigma(1 + 2\sqrt{\frac{b}{r} })

\sigma _t = 53 \times 10^6 (1 + 2\sqrt{\frac{2.9\times10^-^6 }{1065 \times 10^-^9} } \\\\\sigma _t = 227.9 \times 10^6\\\\\sigma _t  = 227.9MPa

4 0
3 years ago
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