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solong [7]
2 years ago
7

A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the

way down to the ground below the balcony. What is the rock's displacement?
Physics
1 answer:
Dvinal [7]2 years ago
7 0
Displacement = distance and direction from the start-point
to the end-point, regardless of the route followed on the way.

From the throw to the 'plop', the displacement is 5 meters down.
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7 0
2 years ago
A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is
Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

\frac{\Delta l}{l_{o}}=\frac{F}{AY} (1)

T=2 \pi \sqrt{\frac{l_{o}}{g}} (2)

Where:

\Delta l=0.05 mm=5(10)^{-5} m is the length the steel wire streches (taking into account 1mm=0.001 m)

l_{o} is the length of the steel wire before being streched

F=mg=(2 kg)(9.8 m/s^{2})=19.6 N is the force due gravity (the weight) acting on the pendulum with mass m=2 kg

A is the transversal area of the wire

Y=2(10)^{11} Pa is the Young modulus for steel

T is the period of the pendulum

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

Knowing this, let's begin by finding A:

A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4} (3)

Where d=1.1 mm=0.0011 m is the diameter of the wire

A=\pi \frac{(0.0011 m)^{2}}{4} (4)

A=9.5(10)^{-7}m^{2} (5)

Knowing this area we can isolate l_{o} from (1):

l_{o}=\frac{\Delta l AY}{F} (6)

And substitute l_{o} in (2):

T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}} (7)

T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}} (8)

Finally:

T=1.39 s

3 0
3 years ago
How many grams are in 6.53 moles of Mn?
Westkost [7]

Answer:

359 g Mn

General Formulas and Concepts:

  • Dimensional Analysis
  • Reading the Periodic Table of Elements

Explanation:

<u>Step 1: Define</u>

6.53 mol Mn

<u>Step 2: Find conversion</u>

1 mol Mn = 54.94 g Mn

<u>Step 3: Dimensional Analysis</u>

<u />6.53 \hspace{3} mol \hspace{3} Mn(\frac{54.94 \hspace{3} g \hspace{3} Mn}{1 \hspace{3} mol \hspace{3} Mn} ) = 358.758 g Mn

<u>Step 4: Simplify</u>

<em>We are given 3 sig figs.</em>

358.758 g Mn ≈ 359 g Mn

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2 years ago
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3 years ago
An 1800 kg helicopter rises with an upward acceleration of 2.0 m/s?. What lifting force is supplied by its rotating blades?
Viktor [21]

Answer:

Lifting force, F = 21240 N

Explanation:

It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force  supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, F=1800\ kg(9.8\ m/s^2+2\ m/s^2)

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

4 0
3 years ago
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