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2 years ago

Suppose a ball is dropped from a building. What is the change of velocity between the first and fifth seconds of its flight

Physics

1 answer:

Advocard [28]2 years ago

4 0

Answer: 39.2m/s

Explanation:

Due to the free falling body, the acceleration of the ball is 9.8 m/s^2, and its initial velocity is 0 m/s.

V(t) = g*t

The change of velocity from t=1 to t=5:

Delta V = 9.8*(5-1) = 39.2 m/s

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Streak is a reliable identifier of a mineral. true or false

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Answer:

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Explanation:

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3 years ago

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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

A pendulum is transported from sea-level, where the acceleration due to gravity g is 9.80 m/s2, to the bottom of Death Valley. A

tatyana61 [14]

Answer: the value of g in Death Valley is 10.417 m/s²

Explanation:

Given that;

acceleration due to gravity at the point is g = 9.8 m/s²

Lets say the acceleration due to gravity at the bottom of Death valley is g'

as the period of the pendulum is decreased by 3.00%

T' = 0.97 T

T is the period of the pendulum at sea level and T' is the period of the pendulum at bottom of Death valley

therefore from the relation

T = 2π√(l/g)

g'/g = T²/T'²

g' = (T²/ (0.97T)²)g

g' = 1.063g

g' = 10.417 m/s²

therefore the value of g in Death Valley is 10.417 m/s²

7 0

2 years ago

A coffee filter of mass 1.2 grams dropped from a height of 1 m reaches the ground with a speed of 0.8 m/s. How much kinetic ener

iren [92.7K]

<u>Answer:</u> The energy gained by the air molecules is 0.011 J.

<u>Explanation:</u>

Law of conservation of energy states that energy can neither be created nor be destroyed but it can only be transformed from one form to another form.

Here, the potential energy of the coffee filter is getting converted into kinetic energy of the coffee filter and some energy is lost by it which is gained by the air molecules in the form of kinetic energy.

So, calculating the potential energy of coffee filter, we use the equation:

P = mgh

where,

m = mass of coffee filter = 1.2 g = $1.2\times 10^{-3}kg$ (Conversion used: 1 kg = 1000 g)