calculate the average speed of talias car during the trip

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

Answer:

112.63km/hr

Explanation:

The given dimension is :

         70mph

We are to convert this to km/hr

         1 mile  = 1.609km

so;

     70mph x 1.609    = 112.63km/hr

So,

  The solution is 112.63km/hr

Answer:

0.25 kg

Explanation:

p = mv

2 = m(8)

2/8 = m(8)/8 *cancels

m = 1/4 OR 0.25 kg

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

[tex]v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}[/tex]

[tex]v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}[/tex]

[tex]v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}[/tex]

[tex]v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}[/tex]

[tex]\mathbf{v_{avg} =12.31 \ m/s}[/tex]

Answer:

the tack's tangential speed is  5.59 m/s

Explanation:

Given that;

R = 0.331 m

wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so

ω = 2.69 rev/s × 2π/1s = 16.9 rad/s

using the relation of angular speed with tangential speed

tangential speed v of the tack is expressed as;

v = R × ω

so we substitute

v = 0.331 m × 16.9 rad/s

v = 5.59 m/s

Therefore, the tack's tangential speed is  5.59 m/s

Answer:

2 m/s²

Explanation:

From the given information:

The first mass m_1 = 0.6 kg

The second mass m_2 = 0.3 kg

The magnitude for the acceleration of 0.3 kg is:

a = net force/ effective mass

Mathematically, it can be computed as follows:

[tex]a = \dfrac{F}{m}[/tex]

[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]

[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]

a ≅ 2 m/s²

Answer:

Hello your question is incomplete attached below is the missing part of the question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I

answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L

              part A = attached below

Explanation:

Part A :

Assuming that mass of swing is negligible

α = T/I

where ; T = torque, I = inertia,

hence T =  L/2*9*(M1 - M2)

also;  I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]=  ( M1 + M2) * (L/2)^2

Finally the magnitude of the angular acceleration α

α = 2*[(M1 - M2)/(M1 + M2)]*g/L

Part B attached below

Answer:

C. A heat engine must deposit some energy in a cold reservoir.

Explanation:

The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."

This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.

Then we have the equation:

Q = W + q

From this we can conclude that the correct option is:

C. A heat engine must deposit some energy in a cold reservoir.

There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.

C. A heat engine must deposit some energy in a cold reservoir.

Therefore, option C is correct.

Learn more:

brainly.com/question/17172535

Answer:

15 seconds

Explanation:

If car was moving at 20m/s for 3 sec.

if car traveled 100m = 15 sec total

Answer:

reducing temperature of the surrounding

Explanation:

combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer

Answer:

8.34 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1  = 100kg

Mass 2 = 200kg

Distance of separation  = 40cm = 0.4m

Unknown:

Gravitational force of attraction between them  = ?

Solution:

To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:

           Fg  = [tex]\frac{G x mass 1 x mass 2}{d^{2} }[/tex]

G is the universal gravitation constant = 6.67 x 10⁻¹¹

d is the separation

 Now;

  Fg = [tex]\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }[/tex]   = 8.34 x 10⁻⁶N

Answer:

Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

Explanation:

Answer:False

Explanation:The Earth rotates faster at the equator than at the poles.