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3 years ago

11

A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the

magnitude of the average force acting on the ball during the collision

2 answers:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

-371.73 N

Explanation:

From Newton's Second Law of motion,

F = m(v-u)/t..................... Equation 1

Where F = Force acting on the ball, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t= time.

Note: Let the direction of the initial velocity be positive

Given: m = 544 g = (544/1000) kg = 0.544 kg, u = 14.3 m/s, v = -14.4 m/s (rebounds), t = 0.042 s.

Substitute into equation 1

F = 0.544(-14.4-14.3)/0.042

F = 0.544(-28.7)/0.042

F = -371.73 N

Note: The force is negative because it act against the direction of the initial motion of the ball

Nezavi [6.7K]3 years ago

3 0

Answer:

$F = 371.738\,N$

Explanation:

Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:

$(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )$

$F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}$

The average force acting on the ball during the collision is:

$F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}$

$F = 371.738\,N$

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3 years ago

A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr

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Answer:

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Explanation:

The given values are:

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or,

$=0.009 \ kg$

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speed,

$s=0.390 \ m$

Coefficient of kinetic friction,

$\mu=0.20$

As we know,

The Kinematic equation is:

⇒ $v^2=u^2+2as$

then,

Initial velocity will be:

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On substituting the given values, we get

⇒ $u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}$

$=\sqrt{-1.5288}$

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As we know,

The conservation of momentum is:

⇒ $mu=m'u'$

or,

⇒ Initial speed, $u'=\frac{mu}{m'}$

On substituting the values, we get

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What is the momentum of a .005kg bumble bee that is traveling at a velocity of 3.0m/s?

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3 years ago

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

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Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

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Calculate the magnitude of the total impulse applied to the car to bring it to rest

n200080 [17]

Answer:

Total impulse = $mv$ = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

$|J|=|p_f-p_i|$

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

$p_i=mv$

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$p_f=m(0)=0$

Therefore, the impulse is given as:

$|J|=|p_f-p_i|=|0-mv|=|-mv|=mv$

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

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