The bulk modulus of water is 2.2 x 109 Pa, and its density is 1.0 x 103 kg/m3. What is the speed of sound in water?

Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long woul

Tcecarenko [31]

Answer:

The time taken in years is $x = 125 \ years$

Explanation:

From the question we are told that

The speed is $v = 540.00 \ km /hr = \frac{540 *1000}{3600} = 150\ m/s$

The distance from the sun to Pluto is $d = 5.9*10^{9} \ k m = 5.9*10^9 * 1000 = 5.9*10^{12} \ m$

Generally the time taken is mathematically represented as

$t = \frac{d}{v}$

=> $t = \frac{5.9*10^{11}}{150}$

=> $t = 3.933*10^{9}$

Converting to years

$1 year \to 3.154*10^7 \ s$

$x \ years \to 3.933*10^{9}$

=> $x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}$

=> $x = 125 \ years$

7 0

3 years ago

How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Artyom0805 [142]

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

$\text { 1 miles }=\frac{1}{0.621} \mathrm{km}$

So, 2.50 miles will be equal to

$2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}$

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km. So,

$1 \mathrm{km}=1000 \mathrm{m}$

Thereby,

$4.026 \mathrm{km}=4026 \mathrm{m}$

7 0

3 years ago

A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul

Marina CMI [18]

The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.

5 0

3 years ago

Which of the following is an obstacle to creating computer-based models for tracking a hurricane?

iren [92.7K]

Answer:

4. All of the above I think, not to sure about 1. but the rest are right so im like 90.99999 percent sure good luck

5 0

3 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago