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4 years ago

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Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i

nitial velocity of 6.05 m/s, while the second player is 119 kg and has an initial velocity of −3.50 m/s. What is their velocity (in m/s) just after impact if they cling together? (Indicate the direction with the sign of your answer.)

1 answer:

romanna [79]4 years ago

7 0

Answer:

$v_f=0.825m/s$

Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

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Wavelength*frequency=velocity
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A reaction that releases energy as it occurs is classified as a(n ________.

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3 years ago

A copper wire (density equal to 8900 kg / m3) 0.3 mm in diameter "floats" due to the force of the earth's magnetic field, which

Andrews [41]

Answer: $I=112.094\ A$

Explanation:

Given

density $\rho =8900\ kg/m^3$

diameter $d=0.3\ mm$

Magnetic field $B=5.5\times 10^{-5}\ T$

Force on the current carrying conductor placed in a magnetic field

$F=BIL\sin \theta$

where L=length of conductor

$\theta$ =angle between magnetic field and current

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$Weight=mg=\rho ALg$

Therefore

$\rho ALg=BIL\times \sin 90$

$8900\times \frac{\pi}{4}(0.3\times 10^{-3})^2L\times 9.8=5.5\times 10^{-5}\times I\times L$

$I=\frac{8900\times \pi\times 9\times 10^{-8}\times 9.8}{4\times 5.5\times 10^{-5}}$

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3 years ago

A ball at rest rolls across a frictionless floor at 12.0 m/s/s. How far will it travel in

professor190 [17]

Answer:

The distance, d travelled by the ball is 768 metres.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.

Hence, if we subtract the final speed from the initial speed and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{initial \; speed - final \; speed}{time}$

$a = \frac{v - u}{t}$

Where,

a is acceleration measured in $ms^{-2}$

v and u is initial and final speed respectively, measured in $ms^{-1}$

t is time measured in seconds.

Given the following data;

Acceleration = 12.0m/s²

Time, t = 8secs

Velocity =?

First, we would calculate its velocity;

$a = \frac{v - u}{t}$

Since the ball rolls at rest, initial velocity is zero (0).

$V = a * t$

$V = 12 * 8$

$Velocity = 96ms^{-1}$

We can now solve for the distance;

$Velocity = \frac{distance}{time}$

Therefore,

$Distance = velocity * time$

$Distance = 96 * 8$

Distance = 768m.

3 0

3 years ago

Professor Stefanovic is spinning a bucket of water by extending his arm and rotating his shoulder in class to show the effects o

anyanavicka [17]

Answer:

a ) 2.368 rad/s

b) 3.617 rad/s

Explanation:

the minimum angular velocity that Prof. Stefanovic needs to spin the bucket for the water not to fall out can be determined by applying force equation in a circular path

i.e

$F_{inward } = F_G + F_T$ ------ equation (1)

where;

$F_{inward} = m *a_c$

$F_{inward} = m*r* \omega^2$

Also

$F_G = m*g$

$F_T = 0$ since; that is the initial minimum angular velocity to keep the water in the bucket

Now; we can rewrite our equation as :

$mr \omega^2= mg + 0\\\omega^2 = \frac{m*g}{m*r}\\\omega^2 = \frac{g}{r}\\\omega = \sqrt{\frac{g}{r} \ \ } ------ equation \ \ \ {2}$

So; Given that:

The rope that is attached to the bucket is lm long and his arm is 75 cm long.

we have our radius r = 1 m + 75 cm

= ( 1 + 0.75 ) m

= 1.75 m

g = acceleration due to gravity = 9.81 m/s²

Replacing our values into equation (2) ; we have:

$\omega = \sqrt{ \frac{9.81}{1.75}}\\\omega = 2.368 \ rad/s$

b) if he detaches the rope and spins the bucket by holding it with his hand ; then the radius = 0.75 m

∴

$\omega = \sqrt{ \frac{9.81}{0.75}}\\\omega = 3.617 \ rad/s$

4 0

4 years ago