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Arlecino [84]
2 years ago
15

You want to determine whether the mass of an object attached to a parachute affects the time it takes to fall to the ground. In

this investigation, would "the material the parachute is made out of" be a suitable control variable? no or yes​
Physics
1 answer:
hammer [34]2 years ago
5 0

Answer:

The Answer is below!!

Explanation:

The larger the area of the parachute, the more air needs to be pushed out of the way, and so the slower it descends. the independent variable is the shape of the parachutes' canopies. The dependent variable is the drop speeds of the parachutes. How long does it take for the parachutes to reach the ground? Measure this using a stopwatch.

Hope I Helped!!

:)

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A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc
ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

8 0
3 years ago
a watermelon is balanced by a dog, a pumpkin, a flowerpot, and a baseball as shown below. Is the weight of the watermelon equal
alukav5142 [94]

Answer:

pumpkin

Explanation:

watermelon and pumpkins are close to shape and size

6 0
2 years ago
Read 2 more answers
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
The record for the highest speed achieved in alaboratory for a
Ira Lisetskai [31]

Answer:

0.000234 seconds

Explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

\omega = \frac{v}{r} = \frac{2010}{0.075} = 26800 rad/s

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds

3 0
3 years ago
What number on this diagram refers to the tick marks for the dependent variable?
givi [52]

Answer:

I'm pretty sure it's 3.

Explanation:

Because if you look at your options the only that would be relevant to tick marks would be either 4 or 3. And it said in the question that we're looking for the one for the dependent variable. And the dependent variable is on the Y- Axis and the 3 is the tick marks for the y-axis. So your answer is 3.

4 0
3 years ago
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