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3 years ago

14) What is the velocity of a wave 5.35 x 10 cm long with a frequency of 16kHz

Physics

1 answer:

JulijaS [17]3 years ago

6 0

Answer:

wavelength= 5.35×10cm=0.535m

f= 16000hz

v= wavelength × f= 0.535×16000=8560m/sec

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A rockets thruster provides a 300 N force to the right . Air resistance causes a 130 N to the left what is the net force?

mel-nik [20]

Answer:

+300N-130N=170N

Explanation:

because two forces in opposite direction one take +and the other takes -

6 0

3 years ago

The complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? Expres

k0ka [10]

We have to covert 512 cal of heat in kilo joules.

As, 1 cal = 0.004184 kJ = 4.184 joules.

Therefore,

$512 \ cal = 512 \times 4.184 \ J = 2142.208 \ J \\\\\ = 2.142 \times 10 ^3 J = 2.142 \ kJ$

Thus, combustion of a small wooden match produces approximately ( in kilo joules ) is 2.142 kJ .

5 0

3 years ago

Calculate the drop voltage for a battery of internal resistance 2 ohm and emf of 24 volt

Ivenika [448]

Answer:

<em>The drop voltage is 0.3 V</em>

Explanation:

Electromotive Force EMF

When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:

$\displaystyle i=\frac{\varepsilon }{R_e+R_i}$

The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:

$\displaystyle i=\frac{24 }{158+2}$

$\displaystyle i=\frac{24 }{160}$

i = 0.15 A

The drop voltage is the voltage of the internal resistance:

$V_i = i.R_i$

$V_i = 0.15*2$

$\boxed{V_i = 0.3\ V}$

The drop voltage is 0.3 V

3 0

3 years ago

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0