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UNO [17]
3 years ago
6

When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0

g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Physics
1 answer:
adoni [48]3 years ago
3 0

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

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The motion of a piston in an auto engine is simple harmonic. If the piston travels back and forth over a distance of 10 cm and t
chubhunter [2.5K]

Answer:

Maximum force will be 29040 N

Explanation:

We have given mass of the piston m = 1.5 kg

Amplitude A = 10 cm = 0.1 m

Angular speed \omega =4200rpm=\frac{2\times 3.14\times 4200}{60}=440rad/sec

We know that angular speed is given by \omega =\sqrt{\frac{k}{m}}

440=\sqrt{\frac{k}{1.5}}

Squaring both side

193600=\frac{k}{1.5}

k = 290400 N/m

Now force is given by F=kA=290400\times 0.1=29040N

3 0
3 years ago
Wall-E the robot is resting when he randomly explodes into two pieces that fly off in opposite directions. His head has a mass o
Brut [27]

Answer:

<em>The body flies off to the left at 9.1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2\qquad\qquad[1]

Wall-E robot is initially at rest, its two parts together. His head has a mass of m1=0.75 kg and his body has a mass of m2=6.2 kg. Both parts have initial speeds of zero v1=v2=0.

After the explosion, his head flies off to the right at v1'=75 m/s. We are required to find the speed of his body v2'. Solving [1] for v2':

\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}

Substituting values:

\displaystyle v'_2=\frac{0.75*0+6.2*0-0.75*75}{6.2}

\displaystyle v'_2=-9.1 \ m/s

The body flies off to the left at 9.1 m/s

3 0
3 years ago
Help on springs,I klai
ra1l [238]
K = ∆F / ∆s = 4N / 0.08m = 50N/m
3 0
3 years ago
A child bounces a 51 g superball on the sidewalk. The velocity change of the super bowl is from 22 m/s downward to 14 m/s upward
noname [10]

Answer:

F=1.02x10^{-3} N

Explanation:

From the exercise we know:

m=51g*\frac{1kg}{1000g}=0.051kg

v_{1}=-22m/s

v_{2}=14m/s

t_{2}-t_{1}=1800s

So, the average acceleration is:

a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{(14-(-22))m/s}{1800s}=0.02m/s^2

The average force is:

F=m*a=(0.051kg)(0.02m/s^2)=1.02x10^{-3} N

8 0
3 years ago
An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su
ra1l [238]

Answer:

Acceleration due to gravity will be g=17.3m/sec^2

Explanation:

We have given gauge pressure P = 3.8 atm = 3.8×101325 = 385035 Pa

Depth h = 24.3 m

Density \rho =1000kg/m^3

We have find the acceleration due to gravity at the surface of planet

We know that pressure is given by

P=\rho gh

So 385035=1000\times g\times 24.3

g=17.3m/sec^2

Acceleration due to gravity will be g=17.3m/sec^2

4 0
4 years ago
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