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1 month ago

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n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration

on Zuton is 3.0 m/s 2 (down). At the moment of liftoff the acceleration of the space ship is 0.50 m/s 2 (up). What is the magnitude of the force of the space ship on the astronaut

1 answer:

serg [3.5K]1 month ago

5 0

The answer is 0.29 kN. For the astronaut's weight on Earth, we have free fall acceleration due to gravity on Earth and on Zuton, which assists in determining the force the spaceship exerts on the astronaut during liftoff.

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A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Sav [3153]

Response:

A=0.199

Clarification:

We know that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation=

$\nu=1.2Hz$

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Therefore, the amplitude of oscillation=A=0.199

4 0

1 month ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

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2 months ago

An airplane flies with a velocity of 55.0 m/s [35o N of W] with respect to the air (this is known as air speed). If the velocity

ValentinkaMS [3465]

V - wind speed;
53° - 35° = 18°
v² = 55² + 40² - 2 · 55 · 40 · cos 18°
v² = 3025 + 1600 - 2 · 55 · 40 · 0.951
v² = 440.6
v = √440.6
v = 20.99 ≈ 21 m/s
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The charge on the plastic cube is determined as follows.

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Softa [3030]

Answer:

I do not communicate in Spanish

Explanation:

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