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2 years ago

Which units are used to measure both velocity and speed?

2 answers:

5 0

(Any unit of distance)
divided by
(any unit of time)

becomes a unit of speed. Some favorites are mile/hour,
meter/second, and foot/second.

Speed is a part of velocity, so the same units are involved
in velocity. But velocity needs more than that.

When you have the distance and the time, you know the
speed completely ... there's no more to know about it.

But speed is only part of velocity. In addition to speed,
there's more to know in order to have a velocity. That's
the direction of the speed.

irina1246 [14]2 years ago

4 0

The SI unit for velocity and speed is meters per second

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The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the co

-BARSIC- [3]

Sunday, July 20, marked 45 years since the United States put the first two astronauts safely on the moon. The cost for the Mercury, Gemini and Apollo programs was more than $25 billion at the time more like $110 billion in today’s world. The ensuing U.S. space efforts have cost an additional $196 billion for the shuttle and $50 billion for the space station. NASA’s total inflation-adjusted costs have been more than $900 billion since its creation in 1958 through 2014 (more than $16 billion per year). Looking back, have we gotten our money’s worth from the investment?

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2 years ago

In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m

ahrayia [7]

Answer:

The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²

Explanation:

Given;

density of the liquid, ρ = 1500 kg/m³

frequency of the wave, F = 410 Hz

wavelength of the sound, λ = 7.80 m

The speed of the wave is calculated as;

v = Fλ

v = 410 x 7.8

v = 3,198 m/s

The bulk modulus of the liquid is calculated as;

$V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2$

Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²

5 0

3 years ago

Calculate the density of a substance whose mass is 160,000kg and volume 20m

Semmy [17]

I’m assuming that’s m^3? If so then simply divide 160,000 by 20 and you get the answer.

8,000 kg/m^3

3 0

2 years ago

A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats

MAVERICK [17]

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

7 0

2 years ago

A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec

stiv31 [10]

Answer:

Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$ .)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$ .

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

$\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$ .
$\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$ .

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$ .

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$ .

$F_1$ , $A_1$ , and $A_2$ have all been found. The question is asking for $F_2$ . Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$ .

Evaluate this expression to obtain the value of $F_2$ , which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$ .

6 0

3 years ago