Ask question

Ask question

All categories

3 years ago

14

Hearing the siren of an approaching fire truck, you pull over to the side of the road and stop. As the truck approaches, you hea

r a tone of 470 Hz ; as the truck recedes, you hear a tone of 400 Hz .Howmuch time will it take for the truck to get from your position to the fire 5.0 km away, assuming it maintainsa constant speed?

1 answer:

11111nata11111 [884]3 years ago

6 0

Answer:

time = 182.68 s

Explanation:

given data

truck approaches = 470 Hz

truck recedes = 400 Hz

position away = 5.0 km

solution

if V is the speed of sound in air and v is the speed to truck

f is the frequency of sound

when truck approach

f1 = $\frac{V+v}{V} f$

and when truck recede

f2 = $\frac{V+v}{V} f$

so here we can say

$\frac{f1}{f2} = \frac{V+v}{V-v}$ ...........1

$\frac{470}{400} = \frac{340.278+v}{340.278-v}$

v = 27.37 m/s

so here 5 km

S = u×t + 0.5 ×a×t² ...........2

5 × 10³ = 27.37 t + 0

t = 182.68 s

You might be interested in

A car moving in the positive direction with an initial velocity of 26 m/s slows down at a constant rate of -3 m/s2. What is its

Greeley [361]

The answer is 5 m/s
$- 3 = \frac{x - 26}{ 7} \\ x = \frac{7 \times( - 3)}{1} + 26 = - 21 + 26 \\ = 5$

good luck

8 0

3 years ago

A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

Andreas93 [3]

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

$p=m_pv_p+m_bv_b=0$

Therefore,

$v_p=-\frac{m_b}{m_p} v_p$

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

$v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s$

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

8 0

3 years ago

The chart below shows the average surface temperature or temperature range for each of the eight planets.

Vaselesa [24]

Answer:

A) Earth and the other inner planets have higher average surface temperatures than the outer planets.

Explanation:

the earth and the other inner planets have higher average surface temperatures than the outer planets.

The reason for this response is due to the distance between the sun and the respective planet, the source of energy generation is the sun and the only way in which the temperature increase of each planet is guaranteed is by radiation, the further away a planet is from its star, its temperature will decrease. Although it is also important to highlight the atmospheric composition of the planet if this planet in its stratosphere has high density clouds that do not allow the entry of solar radiation, the temperature of the planet's surface will not increase, independent of the distance from the sun, but these are more complex cases where specialists in that area enter to perform a study in detail.

4 0

3 years ago

Read 2 more answers

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago