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3 years ago

What is the speed of sound for a noise that travels 2km in 5.8s?

Physics

1 answer:

Gnesinka [82]3 years ago

5 0

Answer:

Explanation:

Speed is defined as the rate at which an object covers a particular distance. So the formula for determining speed is given as the ratio of distance to time taken for covering that distance.

Speed = Distance/Time

As here the distance is given in km units and time in s units, so the units of any one parameter should be changed. Since we know that speed of sound is always about 300 m/s. So it is better to convert the unit of distance from km to m.

Hence, now the distance traveled by the noise is 2000 m and time taken is 5.8 s.

So the speed of noise = Distance/Time = 2000/5.8=345 m/s.

Thus, the speed of noise is slightly greater than the speed of sound and it is found to be 345 m/s.

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A ball is launched from ground level at 20 m/s at an angle of 40° above the

DedPeter [7]

(a) The ball's height y at time t is given by

y = (20 m/s) sin(40º) t - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :

0 = (20 m/s) sin(40º) t - 1/2 g t ²

0 = t ((20 m/s) sin(40º) - 1/2 g t )

t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 g t

t = (40 m/s) sin(40º) / g

t ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So

0² - ((20 m/s) sin(40º))² = 2 (-g) y

where y in this equation refers to the maximum height of the ball. Solve for y :

y = ((20 m/s) sin(40º))² / (2g)

y ≈ 8.4 m

8 0

2 years ago

At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai

zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

Motion of 0.50 kg ball:

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

Motion of 0.25 kg ball:

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

$\bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

m₁ = 0.50 kg

x₁ = 5.38 m

m₂ = 0.25 kg

x₂ = 10.38 m

Substituting

$\bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m$

The center of mass of the two-ball system is 7.05 m above ground.

8 0

3 years ago

What component of an atom determines what element it is?

serg [7]

The number of protons in the element.

8 0

3 years ago

Please solve this question

lesantik [10]

Answer:

88200 Pa

it is because

height =9m

density=1000kg/m(cube)

gravity = 9.8m/s(square)

now,

P=d×g×h

= 1000×9.8×9

=88200pa

8 0

2 years ago

A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m/s. After 1.1 minutes, the finch ti

Romashka [77]

Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance = 990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

Average Speed = 6.37 m/s

4 0

2 years ago

What is the speed of sound for a noise that travels 2km in 5.8s?​

What is the speed of sound for a noise that travels 2km in 5.8s?