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4 years ago

Centripetal is a word used in science to mean

1 answer:

7 0

: the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation a string on the end of which a stone is whirled about exerts centripetal<span> force on the stone — compare centrifugal force.</span>

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You drop a stone down off a bridge. You are able to count to 4.0 seconds when it finally hits the water. How high is the bridge?

mart [117]

Answer:

The height of the bridge is 78.4 m.

Explanation:

Given;

time of the stone motion off the bridge, t = 4.0 s

acceleration due to gravity, g = 9.8 m/s²

The height of the bridge is given by;

h = ut + ¹/₂gt²

where;

u is the initial velocity of the stone, u = 0

h = ¹/₂gt²

h = ¹/₂(9.8)(4)²

h = 78.4 m

Therefore, the height of the bridge is 78.4 m.

7 0

3 years ago

Scientists might use a diagram to model the water cycle. What are two<br> benefits of this model?

Kitty [74]

Answer:

B, and D

Explanation:

3 0

3 years ago

Read 2 more answers

An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square cl

Talja [164]

Answer:

vavg = 0.37 m/s

Explanation:

The average speed is just the relationship between the total distance traveled, and the total time required for that travel , as follows:

$v_{avg} = \frac{\Delta x}{\Delta t} (1)$

We know that for the first leg of the journey, the ant crawls at a constant speed of 0.24 m/s, moving 3.0 m.
We can find the time required for this part, just applying the definition of average velocity, and solving for the time t (which we will call t₁), as follows:

$t_{1} =\frac{x_{1}}{v_{1} } = \frac{3.0m}{0.24m/s} = 12.5 s (2)$

From the givens, we know that the time for the second part is exactly the half of the value found in (2), so we can write the total time Δt as follows:

$\Delta t = t_{1} + \frac{t_{1} }{2} = 12.5 s + 6.25 s = 18.75 s (3)$

We also know that in the second leg of the journey, the ant traveled 4.0 m, which adds to the 3.0 m of the first part, making a total distance of 7.0 m.
Per definition of average speed, we can write the following expression as in (1) replacing Δx and Δt by their values, as follows:

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{7.0m}{18.75m} = 0.37 m/s (4)$

3 0

3 years ago

A particle has ~r(0) = (4 m) ˆ and ~v(0) = (2 m/s)ˆı. If its acceleration is constant and given by ~a = −(2 m/s 2 ) (ˆı +ˆ), a

Troyanec [42]

The particle has $\vec r(0)=(4\,\mathrm m)\,\vec\imath$ and $\vec v(0)=\left(2\frac{\rm m}{\rm s}\right)\,\vec\imath$ , and is undergoing a constant acceleration of $\vec a=-\left(2\frac{\rm m}{\mathrm s^2}\right)(\vec\imath+\vec\jmath)$ .

This means its position at time $t$ is given by the vector function,

$\vec r(t)=\vec r(0)+\vec v(0)t+\dfrac12\vec a t^2$

$\implies\vec r(t)=\left[4\,\mathrm m+\left(2\dfrac{\rm m}{\rm s}\right)t-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

The particle crosses the x-axis when the $\vec\imath$ component is 0 for some time $t>0$ , so we solve:

$4+2t-t^2=0\implies t^2-2t+1=5$

$\implies(t-1)^2=5$

$\implies t-1=\pm\sqrt5$

$\implies t=1\pm\sqrt5$

The negative square root introduces a negative solution that we throw out, leaving us with $\boxed{t=1+\sqrt5}$ or about 3.24 seconds after it starts moving.

3 0

4 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$