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2 years ago

An astronaut is weighed before he sets off to the Moon. He has a mass of 80 kg.

Physics

1 answer:

denpristay [2]2 years ago

8 0

formula

weight= mass×gravity

weight=80×10=800

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Test questions!!!!!!!!

Pani-rosa [81]

Answer:

B. surface wave

8 0

3 years ago

The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.

Eva8 [605]

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

8 0

3 years ago

With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as

Nonamiya [84]

Answer:

|Δf| = 37.3 kHz

Explanation:

given,

peak velocity = 4 m/s

speed of the sound = 1500 m/s

frequency = 7 MHz

$v = C\dfrac{\pm \dlta f}{2 f_0}$

$\delta f = \pm 2 f_0 (\dfrac{V}{C})$

$\delta f = \pm 2\times 7 (\dfrac{4}{1500})$

$=\pm 0.0373 MHz$

= 37.3 kHz

|Δf| = 37.3 kHz

hence, frequency shift between the opening and closing valve is 37.3 kHz

4 0

3 years ago

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

$P = \rho h$

Here,

h = Depth average

$\rho$ = Density

Moreover,

$\text{Density of water}= \rho = 62.4lb/ft^3$

Replacing,

$P = (62.4lb/ft^3)(\frac{35}{2}ft)$

$P = 1092 lb/ft^2$

Finally the force

$\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}$

$F = 1092lb/ft^2(101ft*52ft)$

$F = 5.735*10^6lbf$

6 0

3 years ago

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

8 0

3 years ago