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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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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of
NikAS [45]

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

v=at

Where, v = speed

a = acceleration

t = time

Put the value into the formula

v=4.2\times6.5

v=27.3\ m/s

Hence, The gazelles top speed is 27.3 m/s.

6 0
3 years ago
A ____________ galaxy has arms of gas and dust; where as _____________ galaxy is a category of galaxy that does not have a disti
Jet001 [13]

Answer:

spiral, an irregular

Explanation:

i took the test :)

3 0
3 years ago
The speed of sound is 346 m/s. If a sound wave travels at a frequency of 55 Hz, what would its wavelength be?​
Korvikt [17]

Answer:

6.29 meters.

Explanation:

, where v is the speed of wave and f is the frequency of wave.

We are given that ,

The speed of sound is 346 m/s.

i..e v=346 m/s

A sound wave travels at a frequency of 55 H.

i..e f=55

the wavelength would be 6.29 meters.

This is based on another brainly answer

Link: brainly.com/question/12538018

3 0
3 years ago
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