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3 years ago

URGENT: write a simple equation relating the frequency (f) of a wave to its period (T)

Physics

1 answer:

EastWind [94]3 years ago

5 0

Answer:

f(x) = mx + b

Explanation:

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Two water balloons of mass 0.35 kg collide and bounce off of each other

sladkih [1.3K]

Answer:

Option D

2 m/s East

Explanation:

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

Since the masses are the same then

$m(u_1+u_2)=m(v_1+v_2)$

Therefore

$u_1+u_2=v_1+v_2$

where $v_1$ and $v_2$ are final velocities of objects while $u_1$ and $u_2$ are the initial velocities respectively

Taking East direction as positive then West as negative and by substitution

$2.5+-2.25=v_1-1.75$

Therefore

$v_1=2 m/s$ hence East since it's positive

6 0

3 years ago

What is the potential energy of a spring that is stretched 0.15 m from equilibrium and has a spring constant of 0.55 N/m?

yan [13]

Explanation:

EE = ½ kx²

EE = ½ (0.55 N/m) (0.15 m)²

EE = 0.62 J

8 0

3 years ago

1-D Kinematics, Constant Acceleration After falling a distance of 45.0 m from the top of a building, a box is landing on the top

Mrrafil [7]

Answer:

Explanation:

Given

Object fall from a height of $s=45\ m$

Considering initial velocity to be zero i.e. $u=0$

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-0=2(9.8)\cdot 45$

$v=29.69\ m/s\approx 29.7\ m/s$

(b)Average acceleration

After falling 45 m, object strike the car and comes to rest after covering a distance of 0.5 m

again using

$v'^2-u'^2=2as'$

here final velocity will be zero i.e. $v'=0$

initial velocity $u'=v$

$0-(29.7)^2=2\cdot a\cdot 0.5$

$a=-882.09\ m/s^2$

(c)time taken by it to stop

$v'=u'+a't$

$0=29.7-882.09\cdot t$

$t=0.034\ s$

5 0

3 years ago

Help me pleasee ill give brainliest

Alex

Answer:

1.Coronary heart disease. Coronary heart disease occurs when the flow of oxygen-rich blood to the heart muscle is blocked or reduced. ...

2.Strokes and TIAs. ...

3.Peripheral arterial

4. Aortic disease.

CREDIT TO THE GIRL ON TOP

Explanation:

6 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago