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vesna_86 [32]
4 years ago
15

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

ound. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur?
Physics
1 answer:
DiKsa [7]4 years ago
3 0

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

x_A = v_0 t + \frac{1}{2} at^2

v_0* t= velocity ( time ) = distance

since ball is at height, the above equation changes as

x_A = H - \frac{1}{2} gt^2

for ball B

xB = v0 t - \frac{1}{2} gt^2

the condition of collision is

xA = xB

vA = - 2vB (given)

from the kinematic equation

the speed of the ball A is

v_A = u- gt

since initial speed of the ball A is zero

, so

v_A = -gt

the speed of the ball B is

v_B = v_0 - gt

sincev_A = - 2v_B

   -gt = -2 ( v_0 - gt)

-gt = -2 v_0 +2gt

3gt =2 v_0

t = \frac{2v_0}{3g}

since x_A = x_B

H -  \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2

H = v_0 t

= v_0 (2v_0/3g)

= \frac{2 v_0^2}{ 3g}

x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2

  = 4 \frac{v_0^2}{9g} = (2/3) H

so, (2/3) times height collision occur

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two electronics students are discussing electrical units. student A says that electrical power is measured in units coulombs, st
Savatey [412]
The statements of both students are incorrect.

-- Electrical power, just like mechanical power, is expressed in units of watts.
-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms'  means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).

There are no 'following statements'.

All in all, a very disappointing question.

8 0
3 years ago
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
4 years ago
I dnt know how to do it
NNADVOKAT [17]

Here's what you need to know about waves:

Wavelength = (speed) / (frequency)

Now ... The question gives you the speed and the frequency,
but they're stated in unusual ways, with complicated numbers.

Frequency:  How many each second ?
The thing that's making the waves is vibrating 47 times in 26.9 seconds.
Frequency = (47) / (46.9 s) =  1.747... per second.  (1.747... Hz)

Speed:  How far a point on a wave travels in 1 second.
The crest of one wave travels 4.16 meters in 13.7 seconds.
Speed = (4.16 m / 13.7 sec) = 0.304... m/s

Wavelength = (speed) / (frequency)

Wavelength = (0.304 m/s) / (1.747 Hz)  =  0.174 meter per second


4 0
3 years ago
Hartman value profile
Grace [21]
What is your question exactly? I'm confused?
6 0
4 years ago
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