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lutik1710 [3]
3 years ago
6

An object is thrown downward from the top of a 175 meter building with an initial speed of 10m/s

Physics
1 answer:
Snezhnost [94]3 years ago
6 0

Answer:

Explanation:

we can look for the final velocity of the object using the eqaution of motion as shown:

v² = u²+2gH

v is the  final velocity

u is the initial velocity = 10m/s

g is the acceleration due to gravity = 9.81m/s²

H is the height of the object = 175m

Subxtitute the given parameters inti the formula and get v:

v² = 10²+2(9.81)(175)

v² = 100+3433.5

v² = 3533.5

v = √3533.5

v = 59.44m/s

Hence the final velocity of the object is 59.44m/s

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The period T of a pendulum of length L is measured to determine g at the surface of Earth. The equation used is T=2π√L/g. The ma
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Answer:

C: Variation in the value of g as the pendulum bob moves along its arc.

Explanation:

The formula for period of a simple pendulum is given by;

T = 2π√(L/g)

Where;

L is length

g is acceleration due to gravity

Now, from this period equation, it is clear that the only thing that can affect the period of a simple pendulum are changes to its length and acceleration due to gravity.

Looking at the options, the only one that talks about either the length or gravity as being potential causes of the error is option C

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3 years ago
an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m
Ksenya-84 [330]

Answer:

7.87x10^5m/s^2

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How would the seasons change if the earth were tilted at 90 degrees instead of 23.5?
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6 0
3 years ago
Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro
LUCKY_DIMON [66]

Let \mathbf r_A denote the position vector of the ball hit by player A. Then this vector has components

\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}

where g=9.8\,\dfrac{\mathrm m}{\mathrm s^2} is the magnitude of the acceleration due to gravity. Use the vertical component r_{Ay} to find the time at which ball A reaches the ground:

1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s

The horizontal position of the ball after 0.49 seconds is

\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector \mathbf r_B of the ball hit by player B has

\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}

Again, we solve for the time it takes the ball to reach the ground:

1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s

After this time, we expect a horizontal displacement of 12 meters, so that v_0 satisfies

v_0(0.57\,\mathrm s)=12\,\mathrm m

\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}

5 0
4 years ago
An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5
dexar [7]

Answer:

The alligator will take t = 10 s to reach the final speed of 35 m/s

Explanation:

As we know that the initial speed of the alligator is 5 m/s

then it accelerate by given acceleration to reach the final speed of 35 m/s

so we will have

v_i = 5 m/s

v_f = 35 m/s

a = 3m/s^2

now we have

v_f = v_i + at

35 = 5 + 3 t

t = 10 s

5 0
3 years ago
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