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adelina 88 [10]

3 years ago

12

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height

from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?

1 answer:

nikdorinn [45]3 years ago

4 0

Answer:

Explanation:

According to Equations of Projectile motion :

$Time\ of\ Flight = \frac{2vsin(x)}{g}$

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

$Maximum Height = \frac{(vsinx)^{2} }{2g}$

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

$Horizontal Range = vcosx * t$

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

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Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more

umka2103 [35]

Answer:

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Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

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As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.

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You are planning to make an open rectangular box from an 8-inch by 15-inch piece of cardboard by cutting congruent squares from

Phoenix [80]

Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

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It is 2x since we cut on two sides.

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V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

Taking the 1st derivative:

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Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

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3 years ago