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34kurt
3 years ago
13

You own a high speed digital camera that can take a picture every 0.5 seconds. You decide to take a picture every 0.5 seconds of

your physics teacher as she drops a watermelon off the top of the school. After you develop and analyze the pictures, you realize that the distance the watermelon falls from the teacher in each new picture is
A) the same in each picture.
B) gradually less in each picture.
C) gradually more in each picture.
D) proportional to the mass of the watermelon.
Physics
1 answer:
Vesnalui [34]3 years ago
8 0

-- There is no need to develop the pictures.  They are available immediately in a digital camera.

-- There is no change in the teacher from one picture to the next.

-- The distance the watermelon falls from the teacher in each new picture is more in each picture than in the picture before it. (C)


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An element carries a_____charge<br>​
ratelena [41]

Answer:

The charge of an element is equal to the number of protons minus the number of electrons. The number of protons is equal to the atomic number of the element given in the periodic table. The number of electrons is equal to the atomic number minus the charge of the atom.

Explanation:

3 0
3 years ago
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A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
3 years ago
A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
An airplane is pushed 22 degrees eastward off its northward course by a jet stream traveling at a speed of 136.73 km/hr. The new
uranmaximum [27]

As the speed of airplane is change due to jet stream

So the net speed is given as

v_{net} = v_{plane} + v_{stream}

now we can rearrange it as

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now by the formula of vector difference we have

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3 0
3 years ago
It is easy..plz help!
Aleks04 [339]
No because point A has to be higher for more KE to push the object and make the look
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4 years ago
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