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Kaylis [27]
3 years ago
8

Let's say that we have a pilot that is dropping a package from a plane that is flying horizontally at a constant speed. If we ne

glect air resistance when the package hits the ground what would be the horizontal location of the package with respect to the plane?
Physics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

The package will be directly below the location of the plane.

Explanation:

Look up projectile motion for more information. The horizontal speed of the package is separate from the vertical speed of the package. The vertical speed of the falling package will be based on the rate of acceleration and the height of the package when dropped. The horizontal speed of the package will be the same as the plane so the package will remain directly below the plane the entire time until the package hits the ground.

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Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equa
alukav5142 [94]

Answer:

d. equal to one-fourth the acceleration at the surface of the asteroid.

Explanation:

The explanation is attached as a picture with this answer

Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.

as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.

As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.

It is  taken 2r for the top is the path. hence we obtain (1/4)g as our answer.

4 0
3 years ago
Read 2 more answers
Which is correct? I need to know!!
Mnenie [13.5K]
What website are you using?
6 0
3 years ago
You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh
ASHA 777 [7]

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt

The time of motion of the object from the given height is calculated as;

h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

3 0
2 years ago
You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil
Step2247 [10]

Answer:

C. it will not change.

Explanation:

While combing, the rubbing of the comb with the hair, transfer of electron takes place from the hair to the comb and the comb becomes negatively charged. But, this transfer of electron does not make any considerable change in the mass of the hair. This is because the mass of an electron is highly negligible. Now, neglecting the mass of an electron, the transfer of the electrons from the hair to the comb makes charging of the comb, but no loss of mass in the hair. So, the mass of hair will no change.

7 0
3 years ago
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