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4 years ago

10

Durante un intento juego de croquet, una bola de 0.52 kg en reposo sobre el cesped es golpeada por un mazo con una fuerza promed

io de 190 n, si el mazo esta en contacto con la bola durante 7.2 x 10^-3 s, cual es la rapidez de la bola justo despues de ser golpeada?

1 answer:

Romashka [77]4 years ago

6 0

Answer: 2.63 m/s

Explanation:

The question in english is written below:

During an attemp of croquet game, a 0.52 kg ball at rest on the grass is hit by a mallet with an average force of 190 N, if the mallet is in contact with the ball for $7.2(10)^{-3} s$ , Whay is the speed of the ball just after being hit?

According to Newton's second law of motion, the force $F$ exerted on an object is directly proportional to its mass $m$ and acceleration $a$ :

$F=m.a$ (1)

On the other hand, acceleration is defined as the variation of velocity $V$ in time $t$ :

$a=\frac{V}{t}$ (2)

Substituting (2) in (1):

$F=m\frac{V}{t}$ (3)

Where:

$F=190 N$

$m=0.52 kg$

$t=7.2(10)^{-3} s$

Isolating the velocity $V$ from (3):

$V=\frac{Ft}{m}$ (4)

$V=\frac{(190 N)(7.2(10)^{-3} s)}{0.52 kg}$

Finally:

$V=2.63 m/s$

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Eva8 [605]

Answer:

The frequency does not depend on the amplitude for any (ideal) mechanical or electromagnetic waves.

In electromagnetism we have that the relation is:

Velocity = wavelenght*frequency.

So the amplitude of the wave does not have any effect here.

For a mechanical system like an harmonic oscillator (that can be used to describe almost any oscillating system), we have that the frequency is:

f = (1/2*pi)*√(k/m)

Where m is the mass and k is the constant of the spring, again, you can see that the frequency only depends on the physical properties of the system, and no in how much you displace it from the equilibrium position.

This happens because as more you displace the mass from the equilibrium position, more will be the force acting on the mass, so while the "path" that the mass has to travel is bigger, the mas moves faster, so the frequency remains unaffected.

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3 years ago

Read 2 more answers

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

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4 years ago

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If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the fi

denis-greek [22]

Answer:

d. 500 m/s

Explanation:

Momentum: This is the product of the mass of a body to its velocity, The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv....................... equation 1

Where M = momentum, m = mass, v = velocity

deduced from the question,

Momentum of the car = momentum of the barrier.

MV = mv ............................. Equation 1

Where M = mass of the car, V = velocity of the car, m = mass of the barrier, v = velocity of the barrier.

making v the subject of the equation,

v = MV/m........................ Equation 2

Given: M = 1000 kg, V = 10 m/s, m = 20 kg.

Substitute into equation 2

v = 1000(10)/20

v = 500 m/s.

Hence the speed of the barrier = 500 m/s

The right option is d. 500 m/s

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4 years ago

what is the light that appears when you touch something metal while you have alot of static electricity

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3 years ago

A ball is thrown directly upward from a height of 3 ft with an initial velocity of 28 ft/sec. the function s(t)equalsminus16ts

borishaifa [10]

Answer:

$t = 0.875 s$

$s_{max} = 15.25 ft$

Explanation:

Position of the ball as a function of time is given as

$s(t) = s_o + v_y t + \frac{1}{2}at^2$

$s(t) = 3 + 28 t - 16 t^2$

now we know that when ball will attain maximum height then the differentiation of the position with respect to time will become zero

so we have

$\frac{ds}{dt} = 0$

$0 = 0 + 28 - 32 t$

$t = 0.875 s$

now the maximum height is given as

$s_{max} = 3 + 28(0.875) - 16(0.875)^2$

$s_{max} = 15.25 ft$

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3 years ago