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2 years ago

If you stood on Mars and lifted a 15kg pack,you would be pulling with a force greater than...

Physics

1 answer:

Oxana [17]2 years ago

6 0

Answer: See answers below.

Explanation: In this problem, we must be clear about the concept of weight. Weight is defined as the product of mass by gravitational acceleration.

We must be clear that the mass is always preserved, that is, the mass of 15 [kg] will always be the same regardless of the planet where they are.

where:

W = weight [N] (units of Newtons)

m = mass = 15 [kg]

g = gravity acceleration [m/s²]

Since we have 9 places with different gravitational acceleration, then we calculate the weight in each of these nine places.

Mercury

Venus

Moon

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

In this problem, we must be clear about the concept of weight. Weight is defined as the product of mass by gravitational acceleration.

We must be clear that the mass is always preserved, that is, the mass of 15 [kg] will always be the same regardless of the planet where they are.

where:

W = weight [N] (units of Newtons)

m = mass = 15 [kg]

g = gravity acceleration [m/s²]

Since we have 9 places with different gravitational acceleration, then we calculate the weight in each of these nine places.

Mercury

Venus

Moon

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

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What is the relationship between thickness of lens and focal length?

Ymorist [56]

Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.

7 0

2 years ago

How much time does it take a dropped object to fall 180 m on Earth?

rusak2 [61]

Answer:

Explanation:

Assume it is dropped from rest and the gravitational acceleration is 10

By the equation of motion under constant acceleration:

$s=ut+\frac{1}{2} at^2$

180 = (0)t+10(t^2)/2

t = 6 or -6 (rejected)

t = 6 s

7 0

3 years ago

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

muminat

Answer:

The speed of the plank relative to the ice is:

$v_{p}=-0.33\: m/s$

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

$m_{g}v_{g}+m_{p}v_{p}=0$ (1)

Where:

m(g) is the mass of the girl
m(p) is the mass of the plank
v(g) is the speed of the girl
v(p) is the speed of the plank

Now, as we have relative velocities, we have:

$v_{g/b}=v_{g}-v_{p}=1.55 \: m/s$ (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

$45v_{g}+168v_{p}=0$

$v_{g}-v_{p}=1.55$

$v_{p}=-0.33\: m/s$

I hope it helps you!

8 0

2 years ago

The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in

Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

Angular acceleration = -3,080 rev per minute / sec

That's a perfectly good and true answer to the question, but the units are ugly. We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

AngAccel = -322.5 radian/s²

7 0

3 years ago

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

ω = 0.05 rad/s

7 0

3 years ago