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3 years ago

The wave frequency is where 1 hertz equals 1 wave passed a fixed point in 1 second.

Physics

1 answer:

ahrayia [7]3 years ago

3 0

   (10 waves) / (5 sec)

   = (10/5) (wave/sec)

   = 2 per sec

   = 2 Hz .

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50 g of liquid Y at 10 Celcius and 200 g of liquid Y at 40Celcius are mixed. Final temperature of the mixture is measured as 15

Sholpan [36]

50 g of liquid X at 10 Celcius and 200 g of liquid Y

mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx) * (ty - t) / (t-tx)
cx/cy = 200/50*(40-15)/(15-10) = 20
cx/cy = 20

5 0

3 years ago

A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same di

Vilka [71]

Answer:

(a)2.865 s

(b)2.865 s

Explanation:

We are given that

Acceleration,a= $3.49 m/s^2$

a.Initial speed,u=29 m/s

Final speed,v=39 m/s

We know that

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

Using the formula

$t=\frac{39-29}{3.49}=2.865 s$

b.Initial speed,u=59 m/s

Final speed,v=69 m/s

Again using the formula

$t=\frac{69-59}{3.49}=2.865 s$

7 0

3 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

How much time does it take for an eagle flying at a speed of 50 kilometers per hour to travel a distance of 2000 kilometers?

gtnhenbr [62]

Answer:40 hour

Explanation:

8 0

3 years ago

A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

n200080 [17]

Answer:

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

Work done = $m*g*h$

where m is mass

g is acceleration due to gravity

and h is the height

Work done in terms of force and distance is = mg

where m is mass of bucket and

g is acceleration due to gravity

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = $\Delta P.E$

W = m*g*h

where g = 9.9 m/ $s^2$

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0

3 years ago