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Jobisdone [24]
3 years ago
15

To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1

522 m/s. Find the period of the wave.
Physics
1 answer:
dedylja [7]3 years ago
4 0

Answer:

2.2\times 10^{-5} s

Explanation:

We are given that  

The wavelength of sound wave=\lambda=3.3 cm=3.3\times 10^{-2}m/s

1 cm/s=10^{-2}m/s

Speed of sound wave,v=1522 m/s

We have to find the period of the wave.

We know that

Frequency=\nu=\frac{v}{\lambda}

Using the formula

Frequency =\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4} Hz

Time period=\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s

Using identity:\frac{a^x}{a^y}=a^{x-y}

Hence, the time period of the wave=2.2\times 10^{-5} s

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A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The
Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is W=2430\ pound

mass of roller coaster m=\frac{W}{g}=\frac{2430}{32.2}=75.45

Distance traveled by roller coaster d=396\ ft

drag force f_d=85\ pounds

velocity at top v=59 mph\approx 86.53\ ft/s

Suppose E is the initial energy

Conserving Energy at bottom and top

E=\frac{1}{2}mv^2+mgh+f_d\cdot d

E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396

E=2.9\times 10^5\ foot-pound

5 0
3 years ago
Describe asexual reproduction
Brrunno [24]
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4 0
3 years ago
A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.
Alex Ar [27]

Good.  You can do some very interesting experiments with that equipment.

3 0
3 years ago
a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired
blondinia [14]
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
6 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
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