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3 years ago

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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.

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 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

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3 years ago

An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh

lara [203]

Answer:

$a_y = 4.9\ m/s^2$

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

$\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}$

$\dfrac{dz}{dy}=-0.5$

The differential form with the acceleration

$\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}$

$-0.5=-\dfrac{a_y}{a_z + g}$

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

$0.5=\dfrac{a_y}{0 + 9.8}$

$a_y = 9.8\times 0.5$

$a_y = 4.9\ m/s^2$

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

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3 years ago

Its ____ makes Sirius the brightest star in the night sky.

grigory [225]

With its apparent magnitude

7 0

3 years ago

Read 2 more answers

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago

A 2200-kg railway freight car coasts at 4.1 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

777dan777 [17]

Answer:

The answer is " $2.41 \times 10^3$ "

Explanation:

Given:

$m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\$

Using formula:

$\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}$

$p_i, p_f$ = system initial and final linear momentum.

$V_i, v_f$ = system original and final linear pace.

$m_i$ = original weight of the car freight.

$m_f$ = car's maximum weight

$= \frac{ 2000 \times 4.1}{3.4}\\\\= \frac{ 8.2\times 10^3}{3.4}\\\\= 2.41 \times 10^3$

$\boxed{m_f = 2.41 \times 10^3}$

8 0

2 years ago