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3 years ago

If an astronaut has a mass of 16 Kg on Earth, what would be his mass on the moon and on the space station

Physics

1 answer:

nlexa [21]3 years ago

3 0

Answer:

The astronaut's mass is 16 kg.

Explanation:

Mass can be defined as a measure of the amount of matter an object or a body comprises of. The standard unit of measurement of the mass of an object or a body is kilograms.

Irrespective of the location of an object or a body at a given moment in time, the mass (amount of matter that they're made up of) is constant. This ultimately implies that, whether you're in the moon, space, earth or any other place, your mass remains the same (constant).

Therefore, if an astronaut has a mass of 16 Kg on Earth, his mass on the moon and on the space station would remain the same, as his original mass of 16 Kg because mass is indestructible.

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When vibrational motion in an object increases, which is a true statement?

zubka84 [21]

A -..................................is the correct option

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3 years ago

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A karate master strikes a board with an initial velocity of 10.0 m/s, decreasing to 1.0 m/s as his hand passes through the board

kenny6666 [7]

The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>

Initial velocity (u) = 10 m/s
Final velocity (v) = 1 m/s
Time (t) = 0.002 s
Mass (m) = 1 Kg
Force (F) = ?

<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

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2 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$

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3 years ago

A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900

sweet-ann [11.9K]

Answer:

0.16 m

Explanation:

A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).

50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³

The volume of the rectangular tank is:

volume = width × length × depth

depth = volume / width × length

depth = 0.074 m³ / 0.500 m × 0.900 m

depth = 0.16 m

3 0

3 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$

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3 years ago

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