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3 years ago

7

A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 3 m and it

does so in 0.49 s. The ball accelerates all the way down; let vA be its speed as it passes the window’s top A and vB its speed as it passes the window’s bottom B. O A B 3 m bb b b b b b b b b b b x y How much did the ball speed up as it passed the window; i.e., calculate ∆vdown = vB −vA ? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s. 017 (part 2 of 2) 10.0 points Calculate the speed vA at which the ball passes the window’s top. Answer in units of m/s.

1 answer:

Usimov [2.4K]3 years ago

3 0

Answer:

Part a)

$\Delta v_{down} = 4.8 m/s$

Part b)

$v_a = 3.71 m/s$

Explanation:

Since ball is dropped under uniform gravity

so here we can say that the distance of 3 m moved by the ball under uniform acceleration is given as

$d = (\frac{v_f + v_i}{2})t$

so we have

$3 = (\frac{v_f + v_i}{2})(0.49)$

$v_f + v_i = 12.24$

also we know that

$v_f - v_i = at$

$v_f - v_i = (9.81)(0.49)$

$v_f - v_i = 4.81$

now we will have

$v_f = 8.52 m/s$

$v_i = 3.71 m/s$

Part a)

$\Delta v_{down} = v_b - v_a$

$\Delta v_{down} = 8.52 - 3.71$

$\Delta v_{down} = 4.8 m/s$

Part b)

speed at the top of the window is

$v_a = 3.71 m/s$

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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3 years ago

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m

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Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get:

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component.

(b)
To much the same end do we derive the vertical component:

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Which we then divide by acceleration, a_y, to derive the time till maximal displacement,

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Answer:

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(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

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2 years ago

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Answer:
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