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3 years ago

A person's _____________ will change if they move from the Earth to the moon.

Physics

1 answer:

Diano4ka-milaya [45]3 years ago

5 0

1. A person's weight will change if they move from the Earth to the moon.

In fact, the weight of a person is given by:

$W=mg$

where m is the mass of the person and g is the gravitational acceleration. The mass of the person, m, is the same on the Earth and on the moon, but the value of g is different on the Moon (about 1/6 of the Earth's value), so the weight also changes.

2. An astronaut is launched into space. The mass of the astronaut did not change. This is measured in Kg.

The mass of an object (or of a person, as in this case) is an intrinsec property of the object, that depends on the amount of matter inside the object: therefore, this quantity does not depend on the location of the object, so it is the same on the Earth, on the Moon and in space.

3. What is the weight of a ring tailed lemur that has a mass of 10 kg? -98 N

The weight of the lemur is given by:

$W=mg$

where m=10 kg is the lemur's mass and g=-9.8 m/s^2 is the gravitational acceleration. Using these numbers, we find

$W=(10 kg)(-9.8 m/s^2)=-98 N$

and the negative sign simply means that the direction of the weight is downward.

4. What is the mass of the lemur from the previous question if it was on the International Space Station? 10 kg

As we said in question 1), the mass of an object does not depend on the location, so the mass of the lemur is still 10 kg, as in the previous exercise.

5. A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton's laws? Third's Newton Law

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the fuel and the rocket. The fuel is pushed backward by the rocket, so the fuel exerts an equal and opposite force on the rocket, which then moves forward.

6. When a cannon is fired, the projectile moves forward. According to Newton's 3rd law, the cannon will want to travel backward.

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the cannon and the projectile.The projectile is pushed forward by the cannon, so the projectile exerts an equal and opposite force on the cannon, which moves backward.

7. An object has a weight of 21,532 N on Earth. What is the mass of the object? 2,197 kg

The weight of the object is given by: $W=mg$

If we re-arrange the formula and we use W=21,532 N, we can find the mass of the object:

$m=\frac{W}{g}=\frac{21,532 N}{9.8 m/s^2}=2,197 kg$

8. What is the mass of the object from the previous question if we put it on the moon? The force of gravity on the moon is 1.62 m/s2. 2,197 kg

As we said in question 4), the mass of an object does not change if we move it to another location, so its mass is still 2,197 kg.

9. How much force is exerted if a 250 kg object has an acceleration of 750 m/s2 ? 187,500 N

The force exerted on the object is given by Newton's second law:

$F=ma$

where F is the force, m=250 kg is the mass and a=750 m/s^2 is the acceleration. By using these numbers, we find

$F=(250 kg)(750 m/s^2)=187,500 N$

10. A resting soccer ball moving after it is kicked is an example of which of Newton's laws? Newton's second law

Newton's second law states that when an object is acted upon unbalanced force, the object has an acceleration, given by the law

$F=ma$

So, in this case, the ball is kicked and so an unbalanced force is applied to it, and for this reason the ball has an acceleration (in fact, it starts from rest, but then its velocity increases since it starts moving).

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Goryan [66]

Yeah that’s is correct

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3 years ago

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.

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3 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight W, with magnitude W = 100 N, pointing directly downward

• two tension forces T₁ and T₂ with equal magnitude T₁ = T₂ = T = 1000 N, both making an angle of θ with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ F = T₁ sin(θ) + T₂ sin(180° - θ) - W = 0

We have sin(180° - θ) = sin(θ) for all θ, so the above reduces to

2T sin(θ) = W

2 (1000 N) sin(θ) = 100 N

sin(θ) = 0.05

θ ≈ 2.87°

If y is the vertical distance between the stoplight and the ground, then

tan(θ) = (15 m - y) / (100 m)

Solve for y :

tan(2.87°) = (15 m - y) / (100 m)

y = 15 m - (100 m) tan(2.87°)

y ≈ 9.99 m

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2 years ago

The shifting of the observed wavelength of light due to the motion of the source toward or away from the observer is called the

alexdok [17]

Answer:

doppler effect

Explanation:

When the relative motion of two bodies results in the wavelength becoming shorter this means that the bodies are getting closer. This is known as blue shift.

When the relative motion of two bodies results in the wavelength becoming longer this means that the bodies are getting farther. This is known as red shift.

Collectively this phenomenon is known as the Doppler effect.

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3 years ago

What is the motion of the particles in this kind of wave?

Musya8 [376]

The answer to this question is B I think

7 0

2 years ago

Read 2 more answers