A trip is taken that passes through the following points in order

Why the greater the depth, the greater the pressure?

Tatiana [17]

Well, the reason behind this is because water depth (I assume you're talking about water/liquid) increases the pressure because of the weight and volume of the water above. I hope this helps! ~Mia

7 0

4 years ago

Briefly explain the difference between being self-centered and

tensa zangetsu [6.8K]

Answer:

Explanation:

Self centered means you think about yourself and can sometimes mean that you're arrogant. Self aware means you are aware of certain things you do. Self centered has a negative connotation while self aware has a positive one.

6 0

4 years ago

Read 2 more answers

A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.

kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be 372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V = 372.77 m/s

Thus, the speed of transverse waves on the wire comes out to be 372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

#SPJ1

6 0

2 years ago

You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta

Arte-miy333 [17]

Answer:

1) 3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes: $\frac{Distance}{Speed}= \frac{22}{6} = 3.66$

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes: $Speed * Time = 34 * 3.66 = 124.44$

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : $s = u*t + \frac{1}{2}(a * t^2)$

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

$22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)$

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1 = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

4 0

3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago