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UNO [17]
2 years ago
15

A ball is launched from ground level at 20 m/s at an angle of 40° above the

Physics
1 answer:
DedPeter [7]2 years ago
8 0

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

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elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, B=2.7\ mT=2.7\times 10^{-3}\ T

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

B=\dfrac{\mu_o I}{2r}

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}

r = 0.00238 m

r=2.38\times 10^{-3}\ m

Hence, this is the required solution.

8 0
3 years ago
Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number o
lys-0071 [83]

Answer:

0.99 m

Explanation:

Parameters given:

Amplitude, A = 7.00cm

Wave number, k = 3.00m^-1

Angular Frequency, ω = 2.50Hz

Period = 6.00 s

Phase, ϕ = π/12 rad

Note: All parameters are the same for both waves except the phase.

Wave 1 has a wave function:

y1(x, t) = Asin(kx - ωt)

y1(x, t) = 7sin(3x - 2.5t)

Wave 2 has a wave function:

y2(x, t) = Asin(kx - ωt + ϕ)

y2(x, t) = 7sin(3x - 2.5t + π/12)

π is in radians.

When Superposition occurs, the new wave is represented by:

y(x, t) = 7sin(3x - 2.5t) + 7sin(3x - 2.5t + π/12)

y(x, t) = 7[sin(3x - 2.5t) + sin(3x - 2.5t + π/12)]

Using trigonometric function:

sin(a) + sin(b) = 2cos[(a - b)/2]sin[(a + b)/2]

Where a = 3x - 2.5t, b = 3x - 2.5t + π/12

We have that:

y(x, t) = (2*7)[cos(π/24)sin(3x - 2.5t + π/24)]

Therefore, when x = 0.53cm and t = 2s,

y(x, t) = (2*7)[cos(π/24)sin{(3*0.53) - (2.5*2)+ π/24}]

y(x, t) = 14 * 0.9914 * 0.0713

y(x, t) = 0.99 m

The height of the resultant wave is 0.99cm

5 0
2 years ago
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Semenov [28]
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3 years ago
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geniusboy [140]

Answer:

Explanation:

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Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the
Kaylis [27]

Answer:

The value is  F_{net} =  4444 lb

The force will be  outward

Explanation:

From the question we are told that

   The diameter of the disk is  d = 50.0 \ cm  = \frac{50}{100} = 0.5 \ m

   The external pressure on Mars  is P = 650 \ N/m^2

From the question we are told that  

   Internal pressure  =  External pressure

Generally the external Force on earth is

      F_E = P_{atm} * A

Here P_{atm} is the atmospheric pressure with value  P_{atm} = 1.013*10^{5}\ Pa

So

      F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}

=>   F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}

=>   F_E = 19893 \  N

Generally the external Force on Mars is  

       F= P * A

      F = 650 * \pi * \frac{d^2}{4}

=>   F = 650 *3.142 * \frac{0.5^2}{4}

=>   F = 127.6 \  N    

Net force is mathematically represented as

      F_{net} = F_E -F

=>    F_{net} =  19893  -127.6

=>    F_{net} =  19765.6 \ Nconverting to  pounds

    F_{net} = \frac{19765.6}{4.448}

=> F_{net} =  4444 lb

Given that that the value is positive then the force will be  outward

7 0
2 years ago
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